PHY301 — Midterm Summary (Lectures 1–22)

📘 Lecture 1 — Introduction-Some Basic Concepts

📖 Overview: This lecture introduces the fundamental building blocks of circuit theory, beginning with the International System of Units (SI) and its seven base units. It then explores the atomic structure of matter, explaining how electrons, protons, and neutrons form atoms, and how the behavior of electrons determines whether a material is a conductor, insulator, or semiconductor. Finally, it establishes the core electrical concepts of current, voltage, resistance, and power, along with their units and the fundamental relationship defined by Ohm's Law.

🗂️ Topics Covered

The lecture begins with a detailed look at the seven base SI units (meter, kilogram, second, ampere, Kelvin, mole, candela), their definitions, and the decimal prefixes used to scale them. It then delves into subatomic particles (electrons, protons, neutrons) and the structure of the atom, including electron shells, valence, and atomic number. Following this, the concepts of conductors, insulators, and semiconductors are introduced. The lecture then formally defines the coulomb as the unit of electric charge and explains static electricity and charge polarity. Finally, it covers the fundamental principles of electricity in a circuit: the nature of electric current, the role of a battery, and the key measurements of current (in amperes), voltage (in volts), and resistance (in ohms), culminating in Ohm's Law and the Power Formula.

📝 Lecture Summary

International System of Units

The International System of Units (SI) is built upon seven basic units: meter (length), kilogram (mass), second (time), ampere (electric current), Kelvin (thermodynamic temperature), mole (amount of substance), and candela (luminous intensity). The meter is defined as the distance traveled by light in a vacuum in 1/299,792,458 of a second. The second is defined as the duration of 9,192,631,770 cycles of radiation from a specific transition of the Cesium-133 atom. The kilogram is the only base unit still defined by a physical object, a platinum-iridium cylinder. The Kelvin is defined as 1/273.16 of the thermodynamic temperature of the triple point of water, which is 0.01°C or 32.02°F; 0°K is called absolute zero. The ampere is defined as the constant current that, if maintained in two long parallel wires one meter apart, would produce a force of 2 x 10⁻⁷ newtons per meter between them.

🔑 Definition — Joule: The energy consumed in moving an object of one kilogram through a distance of one meter. One joule is equivalent to 0.7376 foot pound-force and 0.2388 calories.

🔑 Definition — Watt: The rate of doing work. One watt equals one joule per second (1 W = 1 J/s) and is equivalent to 0.7376 ft-lbf/s or 1/745.7 horsepower.

The SI system uses a decimal system with prefixes to signify powers of 10. Examples include milli (m, 10⁻³), micro (µ, 10⁻⁶), nano (n, 10⁻⁹), pico (p, 10⁻¹²), kilo (k, 10³), mega (M, 10⁶), and giga (G, 10⁹).

📌 Example 1: A laser emits light at a wavelength of 248 nm. This is the same as: (a) 0.0248 millimeter (b) 2.48 micrometer (c) 0.248 micrometer () (d) 24800 angstrom

📌 Example 2: A logic gate switches from on to off in 10 nanoseconds. This corresponds to: (a) 0.1 microsecond (b) 10 microseconds (c) 0.001 microsecond () (d) 0.01 microsecond

Sub-atomic Elements and the Atom

The three stable particles in an atom are the electron (negatively charged), the proton (positively charged), and the neutron (neutral, no charge). An atom is the smallest particle of an element. Protons and neutrons are in the nucleus, while electrons orbit in shells. The electrical force attracting the electron to the proton is balanced by the mechanical (centrifugal) force directing it outwards. In a neutral atom, the number of electrons equals the number of protons. The distribution of electrons in orbital rings determines the atom's electrical stability.

🔑 Definition — Free Electrons: The outermost electrons in a material like copper, which are not tightly bound to any one atom and can migrate easily from one atom to another at random.

The modern planetary model of the atom was proposed by Niels Bohr in 1913, incorporating the nuclear atom by Lord Rutherford and the quantum theory by Max Planck and Albert Einstein. The orbits for planetary electrons are also called shells or energy levels.

🔑 Definition — Electron Valence: The number of electrons in an incomplete outermost shell. A completed outer shell has a valence of zero. Copper has a valence of 1.

All shells except the K shell are divided into sub-shells, which account for different types of orbits (e.g., circular vs. elliptical) and the magnetic properties of the atom.

🔑 Definition — Atomic Number: The number of protons (or electrons in a neutral atom) in an atom of an element. For example, a hydrogen atom has an atomic number of 1.

The planetary electron shells are called K, L, M, N, O, P, and Q at increasing distances from the nucleus. Each shell has a maximum number of electrons for stability, corresponding to the inert gases (e.g., K=2 for Helium, L=8 for Neon). The electronic configuration for copper (29 protons) is K=2, L=8, M=18, N=1.

📌 Example 1: An element with 16 protons and 16 electrons has atomic number: (c) 16

📌 Example 2: The electron valence of an element with atomic number 5 (electronic config: K=2, L=3) is: (b) 3

Conductors, Insulators, and Semiconductors

A conductor is a material where electrons can move easily from one atom to another. All metals are good conductors. An insulator (or dielectric) is a material where electrons tend to stay in their own orbits and cannot conduct electricity easily. A semiconductor (e.g., carbon, germanium, silicon) conducts less than a conductor but more than an insulator.

Coulomb: The Unit of Electric Charge

The fundamental unit of electric charge is the coulomb (C). One coulomb equals the charge of 6.25 × 10¹⁸ electrons or protons. The analysis of static charges and their forces is called electrostatics. The symbol for electric charge is Q or q. Negative polarity refers to the static charge on rubber and resin, while positive polarity refers to the charge on glass. The law of electrical charges states: Like charges repel, opposite charges attract.

The Atom (Advanced)

An ion is an atom that has gained or lost electrons, giving it a net positive or negative charge. Isotopes are atoms of the same element (same atomic number) but with different numbers of neutrons. The designations of positive and negative were made by Benjamin Franklin.

What is Electricity?

Electricity is the flow of free electrons in a wire, a flow called current. Electrons flow because they are attracted to a positive charge and repelled by a negative charge at opposite ends of a wire.

Batteries

A battery is a device that creates a positive charge at one terminal and a negative charge at the other, causing a current to flow. When a closed circuit is formed by connecting a wire across the terminals, electrons flow from the negative terminal to the positive terminal. However, by convention, current is said to flow from positive to negative.

Measuring Current

Current (I) is measured in amperes (A). One ampere is defined as 1 coulomb per second (1 A = 1 C/s).

Measuring Voltage

Voltage (V) is a measure of the "electrical pressure" pushing electrons through a wire. It is measured in volts (V). A voltage of 1 volt means the battery delivers 1 joule of energy for each coulomb of charge that flows.

Measuring Resistance

Resistance (R) is a measure of how much a wire impedes the flow of current. It is measured in ohms (Ω). A resistor is a circuit component that provides a specific amount of resistance.

🔑 Definition — Ohm's Law: The relationship between voltage (V), current (I), and resistance (R). The formula is: 📐 Formula: I = V / R → Current equals voltage divided by resistance. 📌 Example: If a 2-volt battery is connected to a 3-ohm resistor, the current is I = 2 V / 3 Ω = 2/3 amperes.

Measuring Power in Watts

Power (P) is the rate at which energy is delivered or consumed. It is measured in watts (W), where 1 watt equals 1 joule per second (1 W = 1 J/s).

📐 Formula: P = I × V → Power equals current times voltage. 📌 Example: With a battery voltage of 2 volts and a current of 3 amps, the power delivered to the resistor is P = 3 A × 2 V = 6 watts.

⭐ Key Takeaways

This lecture establishes the absolute foundation for circuit theory. You must be fluent in the definitions and symbols for the three fundamental circuit quantities: voltage (V, in volts), current (I, in amperes), and resistance (R, in ohms). The core relationship linking them is Ohm's Law (I = V/R), which you will use constantly. The concept of power (P = I×V, in watts) is equally critical, as it describes energy transfer. Finally, understanding the atomic basis for conductors (like copper with free electrons), insulators, and semiconductors is essential for grasping why materials behave as they do in circuits. Remember the conventional current direction (positive to negative) even though electron flow is opposite.

🧠 Quick Revision Questions

1. What is the difference between the flow of electrons and the direction of conventional current in a circuit?
2. State Ohm's Law in words and as a formula. If you know the voltage and current in a circuit, how would you calculate the resistance?
3. Define a "coulomb." How many electrons does it contain?
4. What is the electron valence of an atom, and why is it important for determining if a material is a conductor?
5. If a 9-volt battery delivers a current of 2 amperes to a light bulb, what is the power consumed by the bulb and the resistance of the bulb?

📘 Lecture 2 — Introduction-some basic concepts

📖 Overview: This lecture introduces fundamental electrical concepts including battery polarities, potential difference, voltage, current, resistance, and power. It explains the relationship between these basic quantities and their practical significance in circuit analysis, making it essential foundational knowledge for all subsequent circuit theory topics.

🗂️ Topics Covered

The lecture covers negative and positive polarities of battery, potential difference and its calculation between different charges, voltage conventions and the volt unit, current definition and mathematical expression (I = dq/dt), the ampere unit, resistance and conductance concepts, ohm unit, closed/open/short circuits, power (P = V×I), passive sign conventions, and the importance of earth ground versus chassis ground with safety implications.

📝 Lecture Summary

NEGATIVE AND POSITIVE POLARITIES OF BATTERY

All materials contain two basic particles of electric charge: electrons (negative polarity) and protons (positive polarity). To use electrical forces associated with these charges, work must be done to separate electrons and protons. A battery uses chemical energy to separate electric charges, producing an excess of negative charge at its negative terminal and an excess of positive charge at its positive terminal. With separate and opposite charges at the two terminals, electric energy can be supplied to a circuit connected to the battery.

POTENTIAL DIFFERENCE

Potential difference is the difference between electric potentials at two terminals of a circuit element. Potential refers to the possibility of doing work. Any charge has the potential to do work of moving another charge by attraction or repulsion. When we consider two unlike charges, they have a difference of potential. A charge results from work done in separating electrons and protons. The work of producing charge causes a condition of stress in protons, which try to attract electrons and return to the neutral condition.

POTENTIAL BETWEEN DIFFERENT CHARGES

For a positive charge of 3 C, work must be done to move electrons. Assuming a charge of 1C can move three electrons: a +3C charge can attract 9 electrons toward right, while a +1C charge can attract 3 electrons toward left. The net result is that 6 electrons move toward the more positive charge. When one charge is 2C and the other is neutral (0C), for the difference of 2C, 6 electrons can be attracted to the positive side.

🔑 Definition — Potential Difference: The difference between electric potentials at two terminals of a circuit element.

POTENTIAL DIFFERENCE BETWEEN THE TERMINALS

A voltage can exist between the terminals of a battery whether current is flowing or not. An automobile battery, for example, has 12 volts across its terminals even if nothing is connected to the terminals.

VOLTAGE CONVENTIONS

The volt is named after Alessandro Volta. Fundamentally, the volt measures the work needed to move an electric charge. When 0.7376 foot-pound of work is required to move 6.25×10¹⁸ electrons between two points, the potential difference is one volt. 6.25×10¹⁸ electrons make up one coulomb. 0.7376 ft-lb of work equals 1 joule.

📐 Formula: 1V = 1J/1C → One volt equals one joule per coulomb (the work per unit charge).

The symbol of potential difference is V for voltage. The volt unit is used so often that potential difference is called voltage.

CURRENT

When the potential difference between two charges forces a third charge to move, the charge in motion is called current. To produce current, charge must be moved by a potential difference. In solid materials like copper wire, free electrons can be forced to move by a potential difference. This current is a drift of electrons, from the point of negative charge at one end to the positive charge at the other end. All electrons are identical; the drift of free electrons results in charge moving through the wire. Current is the constant flow of electrons. Only the electrons move, not the potential difference. The current must be the same at all points of the wire at all times.

Let q(t) be the total charge that has passed a reference point since time t=0, moving in the defined direction. The current at a specific point flowing in a specified direction is the instantaneous rate at which net positive charge is moving past that point.

📐 Formula: I = dq/dt → Current equals the time rate of change of charge (coulombs per second or amperes).

Current is symbolized as I or i.

POTENTIAL DIFFERENCE IS NECESSARY FOR CURRENT

The number of free electrons forced to drift through a wire depends upon the amount of potential difference across the wire. With more applied voltage, the forces of attraction and repulsion make more free electrons drift, producing more current. With zero potential difference across the wire, there will be no current.

THE AMPERE OF CURRENT

Since current is the movement of charge, the unit for stating the amount of current is defined as the rate of flow of charge.

🔑 Definition — Ampere: When charge moves at the rate of 6.25×10¹⁸ electrons flowing past a given point per second, the value of current is one Ampere.

RESISTANCE

The fact that a wire conducting current can become hot shows that work is done against some form of opposition. This opposition, which limits current, is called resistance. Copper wire has many free electrons that move easily, giving low resistance. Carbon has fewer free electrons than copper, so when the same voltage is applied, fewer electrons flow. Carbon opposes current more than copper, therefore having higher resistance.

🔑 Definition — Ohm: A resistance that develops 0.24 calorie of heat when one ampere of current flows through it for one second has opposition of one ohm.

The symbol for resistance is R. The abbreviation for ohm is the Greek letter OMEGA (Ω). In diagrams, resistance is indicated by a zigzag line.

CONDUCTANCE

The reciprocal of resistance is called conductance. The lower the resistance, the higher the conductance. Its symbol is G and the unit is siemens (formerly mho).

📐 Formula: G = 1/R → Conductance equals the reciprocal of resistance.

THE CLOSED CIRCUIT

For a closed circuit, three conditions must be met: (1) There must be a source of voltage; without applied voltage, current cannot flow. (2) There must be a complete path for current flow from one side of the voltage source, through the external circuit, returning to the other side. (3) The current path normally has resistance, either for generating heat or limiting current.

OPEN CIRCUIT

When any part of the path is broken, the circuit is open because there is no continuity in the conducting path. The resistance of an open circuit is infinitely high, resulting in no current.

SHORT CIRCUIT

In a short circuit, the voltage source has a closed path across its terminals, but the resistance is practically zero. The result is too much current. Usually, the short circuit is a bypass across the load resistance.

POWER

Rate of doing work is called power. The unit of electric power is watt. One watt equals the work done in one second by one volt moving one coulomb of charge. One coulomb per second is an ampere.

📐 Formula: P = V × I → Power in watts equals volts times amperes (joules per coulomb × coulombs per second = joules per second or watts).

If one terminal of an element is v volts positive with respect to the other, and current i enters the element through that terminal, then power p = vi is absorbed by the element.

PASSIVE SIGN CONVENTIONS: If the current arrow is directed into the + marked terminal of an element, then p = vi yields the absorbed power. A negative value indicates the power is actually being generated by the element.

📌 Example: A 12V lead-acid battery delivers total energy of 460.8 watt-hours over 8 hours.

(a) Power delivered = (460.8 W-hr) / (8 hr) = 57.6 W

(b) Current through headlight = Power / Voltage = (57.6 W) / (12V) = 4.8 A

Importance of Earth Ground

Voltage cannot be defined at a single point; it is by definition the difference in potential between two points. Many schematics use the convention of taking earth as defining zero volts, so all other voltages are implicitly referenced to this potential. This concept is called earth ground and is fundamentally tied to safety regulations to prevent fires, fatal electrical shocks, and related hazards.

The earth ground symbol represents a common node. Circuits using earth ground as a common terminal are equivalent in terms of voltage values.

Chassis Ground: The common terminal of every circuit in equipment may be tied together and electrically connected to the conducting equipment chassis. This terminal is often denoted using the chassis ground symbol. The electrical connection to earth may have nonzero resistance. The fact that "ground" is not always "earth ground" can cause serious safety and electrical noise problems. If a person's equivalent resistance is significantly less than the resistance of all other possible paths to ground, dangerous electric shock can occur.

⭐ Key Takeaways

Current (I = dq/dt) flows only when a potential difference exists across a conductor, with the ampere defined as 6.25×10¹⁸ electrons per second passing a point. Voltage (1V = 1J/1C) represents work per unit charge and exists even without current flow. Resistance opposes current flow, with conductance (G = 1/R) as its reciprocal in siemens. Power (P = V×I) follows passive sign convention where current entering the positive terminal indicates power absorption. Earth ground (zero volts reference) differs from chassis ground, and this distinction is critical for safety—improper grounding can create lethal shock hazards when low-resistance paths to ground exist through a person.

🧠 Quick Revision Questions

1. What is the mathematical relationship between current and charge, and what is the unit of current?
2. How is one volt defined in terms of work and charge?
3. What is the difference between an open circuit and a short circuit?
4. Using passive sign convention, when does p = vi represent power absorbed versus power generated?
5. Why is it dangerous to assume that chassis ground is the same as earth ground?

📘 Lecture 3 — (Resistance in Series)

📖 Overview: This lecture introduces the fundamental concepts of series and parallel resistor connections and how to simplify complex resistor networks into a single equivalent resistance. Mastering these simplification techniques is crucial for analyzing any circuit, as they form the basis for calculating current, voltage, and power in more complex systems.

🗂️ Topics Covered

The lecture begins by defining resistors in series, where the end of one resistor connects to the start of another, and derives the formula for equivalent resistance as the sum of individual resistances. It then introduces resistors in parallel, where corresponding terminals are joined, and presents the reciprocal formula for calculating the equivalent resistance. The bulk of the lecture is dedicated to numerous worked examples that demonstrate the step-by-step process of simplifying circuits that combine both series and parallel resistors, progressively reducing them to a single equivalent resistor.

📝 Lecture Summary

Resistance in Series

If we connect resistors across a source such that the ending point of one resistor is joined with the starting point of the other resistor, then they are said to be connected in series. The combined effect of all the resistors will be equal to the sum of individual resistances.

Consider two resistances R1 and R2 with terminals A, B and C, D. They will be in series if we connect B with C. The combined effect of these two resistances will be: Req = R1 + R2

Therefore, if we connect N resistances in series then: R_eq = R1 + R2 + R3 + ... + RN

🔑 Definition — Series Connection: A configuration where components are connected end-to-end so that the same current flows through each component.

📐 Formula: R_eq = R1 + R2 + ... + RN → The total resistance of a series circuit is the arithmetic sum of all individual resistances.

📌 Example 1: Simplify the given circuit with two 1kΩ resistors in series. Solution: R_AB = R1 + R2 = 1k + 1k = 2kΩ

📌 Example 2: Simplify the given circuit. Solution: 1kΩ and 4kΩ are in series so they will be combined as 5kΩ. Now 5kΩ and 3kΩ are also in series, resulting in 8kΩ. 2kΩ and 8kΩ are also in series. So R_AB = 2k + 8k = 10kΩ

Resistance in Parallel

Consider two resistances with terminals A, B and C, D. If we connect A with C and B with D, they are said to be connected in parallel.

The equivalent of these two resistances will be: 1/R_eq = 1/R1 + 1/R2 = (R1 × R2) / (R1 + R2)

If we connect N number of resistances in parallel, their equivalent will be: 1/R_eq = 1/R1 + 1/R2 + 1/R3 + ... + 1/RN

🔑 Definition — Parallel Connection: A configuration where all components are connected across the same two points, providing multiple paths for current to flow.

📐 Formula: 1/R_eq = 1/R1 + 1/R2 + ... + 1/RN or R_eq = (R1 × R2)/(R1 + R2) for two resistors → The total resistance of a parallel circuit is less than the smallest individual resistance.

📌 Example 3: Simplify the given circuit with two 1kΩ resistors in parallel. Solution: R_eq = (R1 × R2)/(R1 + R2) = (1 × 1)/(1 + 1) = 1/2 = 0.5kΩ

📌 Example 4: Simplify the given circuit. Solution: 4kΩ is parallel with 4kΩ so 4k||4k = (4 × 4)/(4 + 4) = 16/8 = 2kΩ. 2kΩ is parallel with 2kΩ. So R_eq = (2 × 2)/(2 + 2) = 4/4 = 1kΩ

📌 Example 5: Simplify the given circuit. Solution: 12kΩ is parallel with 4kΩ so 12k||4k = (12 × 4)/(4 + 12) = 48/16 = 3kΩ. Now 2kΩ is in series with 3kΩ so R_AB = 2k + 3k = 5kΩ

📌 Example 6: Simplify the given circuit. Solution: 4kΩ is in series with 8kΩ so the combined effect = 12kΩ. 12kΩ is in parallel with 12kΩ so 12k||12k = (12 × 12)/(12+12) = 144/24 = 6kΩ. 4kΩ is in series with 6kΩ so their combined effect = 4k + 6k = 10kΩ. 6kΩ is in parallel with 10kΩ so R_AB = (6 × 10)/(6+10) = 3.75kΩ 💡 Why this matters: This example shows how circuits with multiple series and parallel combinations can be simplified step-by-step by repeatedly applying the two fundamental formulas.

📌 Example 7: Simplify the given circuit. Solution: 3kΩ is in series with 6kΩ, therefore, their combined effect = 3k + 6k = 9kΩ. 9kΩ is in parallel with 18kΩ so 9k||18k = (9 × 18)/(9+18) = 162/27 = 6kΩ. 6kΩ is in series with 10kΩ. So their combined effect = 6k + 10k = 16kΩ. 6kΩ is in series with 16kΩ so R_AB = 6 + 16 = 22kΩ

📌 Example 8: Simplify the given circuit. Solution: 1kΩ is in series with 2kΩ so their combined effect = 1+2 = 3kΩ. 3kΩ is in parallel with 6kΩ. 3k||6k = (3 × 6)/(6+3) = 18/9 = 2kΩ. 10kΩ is in series with 2kΩ, therefore, their combined effect = 10k+2k = 12kΩ. 12kΩ is in parallel with 6kΩ, hence 12k||6k = (12 × 6)/(12+6) = 4kΩ. 2kΩ is in series with 4kΩ, combined effect = 2+4 = 6kΩ. 6kΩ is in parallel with 6kΩ, therefore, 6k||6k = (6 × 6)/(6+6) = 36/12 = 3kΩ. 3kΩ is in series with 9kΩ, therefore, combined effect = 3k+9k = 12kΩ. 12kΩ is in parallel with 4kΩ so 12k||4k = (12 × 4)/(12+4) = 48/16 = 3kΩ. 2kΩ is in series with 3kΩ so R_AB = 2k+3k = 5kΩ

⭐ Key Takeaways

The most critical skills from this lecture are: recognizing whether resistors are connected in series (same current path, end-to-end) or in parallel (same voltage across them, connected to the same two nodes); correctly applying the simple additive formula (R_eq = R1 + R2 + ...) for series circuits and the reciprocal formula (1/R_eq = 1/R1 + 1/R2 + ... or the product-over-sum rule for two resistors) for parallel circuits; and systematically simplifying complex resistor networks by combining pairs of series or parallel resistors step-by-step, working from the innermost combinations outward until only a single equivalent resistor remains between the two terminals of interest.

🧠 Quick Revision Questions

1. A 5Ω resistor and a 7Ω resistor are connected in series. What is their equivalent resistance?
2. A 12Ω resistor and a 4Ω resistor are connected in parallel. What is their equivalent resistance?
3. What is the general formula for the equivalent resistance of N resistors connected in parallel?
4. In a series circuit, how does the equivalent resistance compare to the largest individual resistance?
5. A circuit has a 6kΩ resistor in series with a parallel combination of a 6kΩ and a 3kΩ resistor. What is the total equivalent resistance seen by the source?

📘 Lecture 4 — Series Parallel Combination

📖 Overview: This lecture introduces the concept of inductance and capacitance in series and parallel combinations. It also covers Ohm's Law, its three forms, and power dissipation in resistance. Understanding these fundamentals is essential for analyzing electrical circuits.

🗂️ Topics Covered

The lecture covers inductance and its units, series and parallel combinations of inductors with examples, capacitance and its units, series and parallel combinations of capacitors with examples, Ohm's Law in three forms (I=V/R, V=IR, R=V/I), the concepts of high voltage with low current and low voltage with high current, volt-ampere characteristics, and power dissipation in resistance with formulas P=I²R and P=V²/R.

📝 Lecture Summary

Inductance

Inductance is the resistance offered by an inductor in a circuit. The unit of inductance is Henry, and it is denoted by L.

🔑 Definition — Inductance (L): The property of an inductor to oppose changes in current, measured in Henrys.

Inductance in Series

If we connect n inductances in series, the combined effect of all these inductances is equal to the sum of individual inductances.

📐 Formula: L_eq = L₁ + L₂ + L₃ + ... + Lₙ → Total series inductance is the sum of all individual inductances.

Inductance in Parallel

If we connect n inductances in parallel, the reciprocal of the combined effect of all these inductances is equal to the sum of reciprocals of individual inductances.

📐 Formula: 1/L_eq = 1/L₁ + 1/L₂ + 1/L₃ + ... + 1/Lₙ → The reciprocal of total parallel inductance equals the sum of reciprocals of each inductance.

📌 Example: Simplify the given inductance circuit. 1mH and 1mH are in series, so their combined effect = 1 + 1 = 2mH.

📌 Example: Simplify the given inductance circuit. 1mH is in series with 4mH and with 3mH, therefore, their effect = 1 + 4 + 3 = 8mH.

📌 Example: Simplify the given inductance circuit. 8mH is in parallel with 8mH, so L_AB = (8 × 8) / (8 + 8) = 64/16 = 4mH.

📌 Example: Simplify the given inductance circuit. 6mH is in parallel with 3mH so 6mH || 3mH = (6 × 3) / (6 + 3) = 18/9 = 2mH. Then, 2mH is in parallel with 4mH, therefore, 2mH || 4mH = (2 × 4) / (2 + 4) = 8/6 = 1.33mH.

📌 Example: Simplify the given inductance circuit. 3mH is in parallel with 6mH so 3mH || 6mH = (3 × 6) / (3 + 6) = 18/9 = 2mH. 2mH is in series with 2mH, therefore, the combined effect of these two = 2 + 2 = 4mH. 4mH is in parallel with 4mH so 4mH || 4mH = (4 × 4) / (4 + 4) = 16/8 = 2mH. 1mH is in series with 2mH so L_AB = 1 + 2 = 3mH.

Capacitance

Capacitance is the resistance offered by a capacitor in a circuit. The unit of capacitance is Farad, and it is denoted by C.

🔑 Definition — Capacitance (C): The ability of a capacitor to store electric charge, measured in Farads.

Capacitance in Parallel

If we connect n capacitances in parallel, the combined effect of all these capacitances is equal to the sum of individual capacitances.

📐 Formula: C_eq = C₁ + C₂ + C₃ + ... + Cₙ → Total parallel capacitance is the sum of all individual capacitances.

Capacitance in Series

If we connect n capacitances in series, the reciprocal of the combined effect of all these capacitances is equal to the sum of reciprocals of individual capacitances.

📐 Formula: 1/C_eq = 1/C₁ + 1/C₂ + 1/C₃ + ... + 1/Cₙ → The reciprocal of total series capacitance equals the sum of reciprocals of each capacitance.

📌 Example: Simplify the given circuit. 2μF capacitor is in series with another 2μF capacitor, their combined effect will be = (2 × 2) / (2 + 2) = 4/4 = 1μF.

📌 Example: Simplify the given circuit. 12μF is in series with 4μF, so their combined effect will be = 48/16 = 3μF. 2μF is in series with 3μF so C_AB = (2 × 3) / (2 + 3) = 1.2μF.

📌 Example: Simplify the given circuit. 4μF is in parallel with 12μF so 12 || 4 = 12 + 4 = 16μF. And 16μF is in series with 3μF so = (3 × 16) / (3 + 16) = 48/19 = 2.5μF. 12μF is in parallel with 2.5μF so their combined effect will be = 12 + 2.5 = 14.5μF. The capacitors are in series so C_AB = (4 × 14.5) / 18.5 = 3.13μF.

OHM’S LAW

If a voltage across a conductor is applied, the current passing through the conductor is directly proportional to the voltage provided temperature remains constant.

🔑 Definition — Ohm's Law: V ∝ I, or V = IR, where 'R' is the resistance of the conductor. Resistance ('R') depends upon the material of the conductor.

The Current I = V/R

If we keep the same resistance in a circuit but vary the voltage, the current will vary. For the general case, for any values of 'V' and 'R', Ohm's Law is I = V/R. Where 'I' is the amount of current through resistance 'R', which is connected across a potential difference 'V'. Volt (V) is the practical unit of potential difference and Ohm (Ω) for resistance, therefore: Ampere = Volts / Ohms. This formula tells us to calculate the Amperes of Current through 'R', simply divide the voltage across 'R' by the Ohms of Resistance.

High Voltage but Low Current

It is important to realize that with high voltage the current can have a low value when there is a very high resistance in the circuit. For example, 1000 (1k) volts applied across 1,000,000 (1M) Ω results in a current of only 0.001 (1m) A. The practical fact is that high voltage circuits usually do have a small value of current in electronic equipment; otherwise, a tremendous amount of power would be necessary for operation.

Low Voltage but High Current

At the opposite extreme, a low value of voltage in a very low resistance circuit can cause a very high current to flow. For example, a 6-volt battery connected across a resistance of 0.001Ω causes 600 A of current to flow: I = V/R = 6v / 0.01Ω = 600 A. Similarly, more 'R' will result in less 'I'.

📌 Example: A heater with a resistance of 8Ω is connected across the 220-volt power line. How much is the current 'I' flowing through the heater coil? Solution: I = V/R = 220/8 = 27.5 A.

📌 Example: A small light bulb with a resistance of 2400Ω is connected across the same 220-volt power line. How much is the current 'I' through the bulb filament? Solution: I = V/R = 220/2400 = 0.09 A = 90 mA.

The Voltage V = IR

It is the other form of the same formula. Besides the numerical calculations possible with the 'IR' formula, it is useful to consider that the 'IR' product means voltage. Whenever there is current through a resistance, it must have a potential difference across its two terminals equal to the product 'IR'. As studied in the last lecture, if there was no potential difference, no electrons could flow to produce the current.

The Resistance R = V/I

As a third and final version of Ohm's Law, three factors V, I, and R are related by the formula R = V/I. As studied in the last lecture, physically, a resistance can be considered as some material with elements having an atomic structure that allows free electrons to drift through it. Electrically, a more practical and general way of considering resistance is simply as a V/I ratio.

📌 Example: A conductor allows 1 A of current with 10 volts applied at its ends. How much will be the resistance of the conductor? Solution: R = V/I = 10/1 = 10 ohm.

The Linear Proportion Between V & I

The Ohm's law formula V = IR states that V and I are directly proportional for any one value of R. This relation is true for constant values of R.

Volt-Ampere Characteristics

The graph in the figure is called the volt-ampere characteristic of R. It shows how much current the resistor allows for different voltages.

💡 Why this matters: The volt-ampere characteristic visually demonstrates the linear relationship between voltage and current for a fixed resistance, which is the foundation of Ohm's Law.

Power Dissipation in Resistance

When current flows through a resistance, heat is produced due to friction between the moving free electrons and the atoms which obstruct the path of electron flow. The heat is evidence that power is used in producing current. The electric energy converted to heat is considered to be dissipated or used up because the calories of work cannot be returned to the circuit as electric energy. Since power is dissipated in the resistance of a circuit, it is convenient to express the power in terms of resistance R. The formula P = V × I can be arranged as follows:

• Substituting IR for V: P = V × I = IR × I = I²R. This is the common form of the formula because of heat produced in a resistance due to current R.
• For another form, substitute V/R for I: P = V × I = V × V/R = V²/R.

In all the formulae, V is the voltage across R in ohms, producing the current in amperes, for power in Watts.

⭐ Key Takeaways

The key takeaways from this lecture are that inductors in series add directly while inductors in parallel follow the reciprocal sum formula, exactly opposite to resistors. Similarly, capacitors in parallel add directly while capacitors in series follow the reciprocal sum formula, also opposite to resistors. Ohm's Law provides three fundamental relationships—V=IR, I=V/R, and R=V/I—which are essential for circuit analysis, and the volt-ampere characteristic for a fixed resistance is a straight line showing linear proportionality. Finally, power dissipated in a resistance can be calculated using either P=I²R or P=V²/R, both derived from the basic power formula P=VI.

🧠 Quick Revision Questions

1. What is the formula for total inductance when two inductors of 4mH and 6mH are connected in series?
2. What is the formula for total capacitance when two capacitors of 3μF and 6μF are connected in parallel?
3. State Ohm's Law in words and write its three mathematical forms.
4. Using Ohm's Law, what is the current through a 10Ω resistor connected to a 5V battery?
5. What are the two formulas for calculating power dissipated in a resistor, and from which basic formula are they derived?

📘 Lecture 5 — AC, DC Current, voltages

📖 Overview: This lecture introduces the fundamental concepts of ideal and dependent sources in circuit theory, distinguishing between direct current (DC) and alternating current (AC) quantities. It also covers the practical applications of Ohm's law and introduces voltage divider principles for series circuits, which are essential for circuit analysis.

🗂️ Topics Covered

The lecture begins by defining ideal voltage and current sources with their graphical representations, then explains direct and alternating quantities. It covers dependent or controlled sources including voltage-controlled and current-controlled voltage and current sources. Several examples illustrate Ohm's law calculations, and the lecture concludes with the concept of voltage dividers and series voltage division formulas with practical examples.

📝 Lecture Summary

TYPES OF SOURCES:

An ideal voltage source is a source in which terminal voltage remains the same independent of the amount of current drawn. The graph shows V as independent of I — whatever the value of x (current), the value of y (voltage) will remain the same.

🔑 Definition — Ideal Voltage Source: a source where terminal voltage remains constant regardless of current drawn 📐 Graph: V is independent of I → voltage is constant across all current values

The dc voltage source is symbolically represented with a battery symbol showing positive and negative terminals.

An ideal current source is a source which gives constant current independent of the terminal voltage. The graph shows I is independent of V.

🔑 Definition — Ideal Current Source: a source that gives constant current regardless of terminal voltage 📐 Graph: I is independent of V → current is constant across all voltage values

DIRECT VOLTAGE AND QUANTITIES

Direct voltage is the voltage which is independent of time, and its magnitude and direction do not change with time.

Direct quantities are quantities whose magnitude and direction do not change with time (for example V or I).

ALTERNATING QUANTITIES

Alternating quantities are quantities whose magnitude and direction change with respect to time (for example V or I).

DEPENDENT OR CONTROLLED SOURCES

There are two types of dependent voltage sources:

1. Voltage controlled voltage source (VCVS) — the source whose magnitude is controlled by voltage
2. Current controlled voltage source (CCVS) — a source whose voltage is controlled by current

For current sources: 3. Voltage controlled current source (VCCS) — if magnitude of current is controlled by input voltage 4. Current controlled current source (CCCS) — if magnitude of the current is controlled by input current

💡 Why this matters: Dependent sources model real-world devices like transistors and operational amplifiers where output is controlled by an input signal.

EXAMPLES USING OHM'S LAW

Example: A battery of 10 V is applied across a resistor of 3kΩ.

• By Ohm's law: I = V/R = 10/3k = 3.33 mA

Example: Calculate the current I through a circuit with 2kΩ and 3kΩ in series, with 10V applied.

• Combined resistance = 2k + 3k = 5kΩ
• I = V/R = 10/5k = 2 mA

Example: Calculate the current through resistors of 25Ω, 10Ω, and 5Ω in series with 40V.

• Combined resistance = 25 + 10 + 5 = 40Ω
• I = V/R = 40/40 = 1 A

Example: Find R₁ when current is 5mA and voltage is 10V.

• R₁ = V/I = 10V/5mA = 2kΩ

Example: Find V when current through 10kΩ resistor is 3mA.

• V = IR = 10k × 3mA = 30V

Example: Find V when 5kΩ and 10kΩ are in series with 2mA current.

• Combined resistance = 5k + 10k = 15kΩ
• V = IR = 2mA × 15k = 30V

Example: Calculate current through all resistors in a complex circuit.

• 3kΩ || 6kΩ = (3×6)/(3+6) = 18/9 = 2kΩ
• 2kΩ in series with 2kΩ = 4kΩ
• 3kΩ in series with 1kΩ = 4kΩ
• 4kΩ || 4kΩ = 16/8 = 2kΩ
• Final: 6kΩ + 2kΩ + 4kΩ = 12kΩ
• I = 12V/12kΩ = 1mA

Example: Find current through a circuit with 20V.

• 3kΩ || 6kΩ = 2kΩ
• 4kΩ + 2kΩ = 6kΩ
• 6kΩ || 12kΩ = (6×12)/(12+6) = 72/18 = 4kΩ
• Final: 2kΩ + 4kΩ + 4kΩ = 10kΩ
• I = 20V/10kΩ = 2mA

VOLTAGE DIVIDERS AND CURRENT DIVIDERS

Any series circuit is a voltage divider. The IR drops are proportional parts of the applied voltage. Special formulae can be used for voltage and current division as shortcuts in calculations. The voltage division formula gives the series voltage even when the current is not known.

SERIES VOLTAGE DIVIDERS

• The current is the same in all resistances in a series circuit
• The voltage drop equals the product IR
• IR voltages are proportional to series resistances
• A higher resistance has a greater IR voltage than a lower resistance
• Equal resistances have the same IR drop across each resistance
• If R₁ is double R₂, then V₁ will be double V₂

📐 Formula: Voltage Division Rule: V = (R/Rₜ) × Vₜ

• Where V is voltage across resistor R, Rₜ is total resistance, Vₜ is total applied voltage
• Meaning: The voltage across any resistor in a series circuit equals its proportional share of the total resistance multiplied by the total voltage

Example: Calculate voltage drop across 4kΩ resistor (circuit with 10V across 4kΩ and 6kΩ in series).

• V₄ₖ = (4/10) × 10 = 4 volts

Now calculate voltage drop across 6kΩ resistor:

• V₆ₖ = (6/10) × 10 = 6 volts

Example: Calculate the power dissipated in a circuit with 12V source.

• Step 1: 12kΩ || 12kΩ = 6kΩ
• Step 2: 6kΩ + 2kΩ = 8kΩ
• Step 3: 8kΩ || 4kΩ = 2.66kΩ
• Step 4: 2.66kΩ || 8kΩ = 2kΩ
• Step 5: 2kΩ || 4kΩ = 1.33kΩ
• Total current: I = 12V/1.33kΩ = 9.022 mA
• Power dissipation: P = VI = 12 × 9.022mA = 108.26 mW

⭐ Key Takeaways

The key distinction between ideal voltage and current sources must be memorized: an ideal voltage source maintains constant voltage regardless of current drawn, while an ideal current source maintains constant current regardless of terminal voltage. Dependent sources are controlled by either voltage or current at another point in the circuit and exist in four types (VCVS, CCVS, VCCS, CCCS). Direct quantities have constant magnitude and direction, while alternating quantities change with time. Ohm's law (V=IR) is applied repeatedly to simplify complex resistor networks by combining series and parallel resistors. The voltage division rule V = (R/Rₜ) × Vₜ is essential for finding voltage drops across individual resistors in series without calculating current first, and power dissipation is calculated as P = VI.

🧠 Quick Revision Questions

1. What is the key difference between an ideal voltage source and an ideal current source?
2. List the four types of dependent or controlled sources and explain what controls each one.
3. What distinguishes direct quantities from alternating quantities?
4. A 10V source is connected across resistors of 2kΩ, 3kΩ, and 5kΩ in series. Using the voltage division rule, find the voltage drop across the 3kΩ resistor.
5. If two resistors of 4kΩ each are in parallel, what is their equivalent resistance? If this combination is in series with a 2kΩ resistor and 12V is applied, what is the total current?

📘 Lecture 6 — Voltage divider, Current divider

📖 Overview: This lecture covers the voltage divider rule for series circuits and the current divider rule for parallel circuits. It explains how to calculate voltage drops across individual resistors in series and current through individual branches in parallel using simple ratio formulas, with multiple worked examples demonstrating both principles.

🗂️ Topics Covered

The lecture introduces voltage division rule for calculating voltage drops across series resistors, including the method for two voltage drops in series where one can be found by subtraction. It then covers current division for two parallel resistances, explaining how currents divide inversely as branch resistance. Multiple examples demonstrate voltage division across various series resistor combinations and current division in parallel circuits, including cases with short circuits and combined series-parallel configurations. The lecture concludes with an example calculating power absorbed by each element in a circuit.

📝 Lecture Summary

Voltage divider rule for series circuits

The voltage drop across a resistor in a series circuit can be found by multiplying the total voltage by the ratio of that resistor's value to the total series resistance. The formula is V = (R/Rt) × Vt, where V is the voltage across the resistor, R is the resistor value, Rt is total series resistance, and Vt is total applied voltage.

🔑 Definition — Voltage divider rule: The voltage across a resistor in a series circuit equals the total voltage multiplied by the ratio of that resistor's resistance to the total series resistance.

📐 Formula: Vₓ = (Rₓ / Rₜₒₜₐₗ) × Vₜₒₜₐₗ → The voltage across a specific resistor is proportional to its share of the total resistance.

📌 Example: For a 9kΩ and 3kΩ resistor in series with a 12V source: V₁ (across 9kΩ) = (9kΩ / (9kΩ+3kΩ)) × 12V = (9/12) × 12 = 9 volts V₂ (across 3kΩ) = (3/12) × 12 = 3 volts.

📌 Example: For 50kΩ, 30kΩ, and 20kΩ resistors in series with 200V: V₅₀ₖ = (50k/100k) × 200 = 100V V₃₀ₖ = (30/100) × 200 = 60V V₂₀ₖ = (20/100) × 200 = 40V

📌 Example: In a circuit with 10V source in parallel with 10kΩ, the same 10V appears across the 10kΩ resistor. When 4kΩ is in series with another 4kΩ, voltage division gives V = (4/8) × 10 = 5 volts.

💡 Why this matters: The voltage divider is fundamental for designing reference voltages, sensor circuits, and biasing networks in electronics.

Two voltage drops in series

For two series resistances, it is not necessary to calculate both voltages separately. After finding one voltage drop, subtract it from the total voltage to find the other. For example, if Vt is 48V across two series resistors and V₁ is 18V, then V₂ must be 48V - 18V = 30V.

Current divider with two parallel resistances

It is often necessary to find individual branch currents in a circuit without knowing the branch voltage. This is solved using the fact that currents divide inversely as branch resistance. The formula is I₁ = Is × R₂/(R₁ + R₂), where Is is the total source current entering the parallel combination.

🔑 Definition — Current divider rule: In a parallel circuit, the current through one branch equals the total current multiplied by the ratio of the opposite branch resistance to the sum of both resistances.

📐 Formula: I₁ = Iₛ × R₂/(R₁ + R₂) → Current through R₁ depends on the value of R₂ (the other branch), not R₁.

📌 Example: For 4Ω and 2Ω resistors in parallel with 30A total current: I through 4Ω = (2Ω/(2Ω+4Ω)) × 30A = (2/6) × 30 = 10A I through 2Ω = (4Ω/(2Ω+4Ω)) × 30A = (4/6) × 30 = 20A

📌 Example: For a circuit where 30A enters a node with 1kΩ and 4kΩ parallel branches, the current through 1kΩ is I = (4k/(4k+1k)) × 30 = (4/5) × 30 = 24A.

📌 Example: To find current through 4kΩ resistor when current divides at Node A between 4kΩ and series combination of 2kΩ+2kΩ=4kΩ: I = (4k/(4k+4k)) × 12 = (4/8) × 12 = 6A.

📌 Example: For a circuit with -6A current source (direction downward) feeding 4kΩ and 8kΩ parallel combination: I through 8kΩ = (4k/(4k+8k)) × (-6) = (4/12) × (-6) = -2A.

💡 Why this matters: Current division is essential for designing current measurement circuits and analyzing parallel loads in power systems.

Finding voltage using current division

To find voltage across a resistor, first find the current through it using current division, then apply Ohm's Law (V = IR).

📌 Example: Find voltage across 3kΩ resistor. At point A, current divides between 12kΩ and series combination of 3kΩ+1kΩ=4kΩ. First find current through series combination: I = (12k/(4k+12k)) × 1A = (12/16) × 1 = 0.75A Same current flows through both series resistors, so voltage across 3kΩ: V = 0.75 × 3k = 2250 volts.

Circuit reduction and voltage calculation

Complex circuits can be simplified by combining series and parallel resistors step by step to find voltages or currents.

📌 Example: Find V in a circuit with 1A source feeding 2kΩ in series with a parallel combination. Working from right to left:

• 4kΩ + 14kΩ = 18kΩ (series)
• 18kΩ || 9kΩ = (18×9)/27 = 6kΩ (parallel)
• 6kΩ || 12kΩ = (6×12)/18 = 4kΩ (parallel)
• 2kΩ + 4kΩ = 6kΩ (series total) Voltage V = IR = 1A × 6kΩ = 6000 volts.

Effect of short circuits on current division

A short circuit (zero resistance path) will divert all current away from parallel resistors.

📌 Example: Find current through 6kΩ resistor. At point C, current should divide between paths, but due to a short circuit between C and D, all current flows through the short. By current division: I = (6k/12k) × 1 = 0.5A.

📌 Example: Calculate current through 5kΩ resistor. Since 5kΩ has zero resistance in parallel (short circuit), current arriving at node C takes the zero resistance path and no current passes through 5kΩ.

Voltage across resistor with current source direction change

When a current source direction is reversed, the current value becomes negative for calculations.

📌 Example: Calculate voltage across 4kΩ resistor with negative current source (-2A). Current divides between 3kΩ and series combination of 2kΩ+4kΩ=6kΩ: I through 6kΩ = (3k/(3k+6k)) × (-2) = (3/9) × (-2) = -0.66A Voltage across 4kΩ = 0.66 × 4k = 2640 volts.

Voltage division with parallel simplification

Parallel resistors can be reduced to an equivalent resistance before applying voltage division.

📌 Example: Calculate voltage across 12kΩ resistor. First, 4kΩ is in parallel with 6V source so same voltage appears (omit from circuit). Then 12kΩ || 6kΩ = (12×6)/(12+6) = 4kΩ. Apply voltage division: V₄ₖ = (4k/(4k+2k)) × 6V = (4/6) × 6 = 4V. Since 4kΩ is the parallel combination of 6kΩ and 12kΩ, same voltage appears across 12kΩ, so V₀ = 4V.

Power absorbed by each element

Power in each element can be calculated using P = VI, P = V²/R, or P = I²R after finding all voltages and currents.

📌 Example: For a circuit with 20V source and resistors (1.5Ω, 14Ω, 2Ω, 4Ω, 2.5Ω): P₂₀ᵥ = -(20)(4) = -80W (negative indicates power supplied by source) V₁.₅ = 4(1.5) = 6V, p₁.₅ = (6²)/1.5 = 24W V₁₄ = 20 - 6 = 14V, p₁₄ = 14²/14 = 14W i₂ = (6/1.5) - (14/14) = 3A, V₂ = 2(3) = 6V, p₂ = 6²/2 = 18W V₄ = 14 - 6 = 8V, p₄ = 8²/4 = 16W i₂.₅ = (6/2) - (8/4) = 1A, V₂.₅ = 2.5(1) = 2.5V, p₂.₅ = 2.5²/2.5 = 2.5W Is = -1A

⭐ Key Takeaways

The voltage divider rule (Vₓ = Rₓ/Rₜ × Vₜ) allows quick calculation of voltage drops across individual resistors in series circuits, while the current divider rule (I₁ = Iₛ × R₂/(R₁+R₂)) finds branch currents in parallel circuits where currents divide inversely to resistance. When two resistors are in series, finding one voltage drop and subtracting from the total gives the other drop. Complex circuits should be simplified by combining series and parallel resistors before applying divider rules. Short circuits (zero resistance) will divert all current away from parallel paths. Power calculations require finding all voltages and currents first, then applying P=VI, P=V²/R, or P=I²R.

🧠 Quick Revision Questions

1. If three resistors of 10kΩ, 20kΩ, and 30kΩ are in series with a 60V source, what is the voltage across the 30kΩ resistor?
2. In a parallel circuit with 12A total current and two branches of 3Ω and 6Ω, what current flows through the 3Ω resistor?
3. Why does a short circuit in parallel with a resistor cause no current to flow through that resistor?
4. If V₁ is 15V across a series resistor and the total voltage is 50V, what is V₂?
5. When a current source direction is reversed in a circuit, what happens to the sign of the current values used in calculations?

📘 Lecture 7 — Kirchhoff's Laws Kirchhoff's Current Law (KCL)

📖 Overview: This lecture introduces Kirchhoff's Current Law (KCL) and demonstrates its application in solving circuit problems. Understanding KCL is fundamental for analyzing complex circuits as it governs how current flows at junctions. The lecture also covers related circuit analysis techniques including voltage division, current division, and series-parallel reduction.

🗂️ Topics Covered

The lecture covers Kirchhoff's Current Law (KCL) with definitions and assumptions, node and branch analysis techniques, and voltage/current division rules. It includes multiple worked examples demonstrating KVL and KCL applications for calculating unknown voltages, currents, and source values in resistive circuits. Key concepts like ground reference, loop definition, and the formula for writing node equations (N-1 equations) are also explained.

📝 Lecture Summary

Example: Calculate source voltage Vs given voltage between nodes A and B is 4V

The voltage between node A and B is given as 4V. Using the voltage division rule, the source voltage can be calculated. The formula used is V = (R/Rt)Vs, where V is the voltage across a specific resistor, R is that resistor value, and Rt is the total series resistance.

🔑 Definition — Voltage Division Rule: In a series circuit, the voltage across any resistor is proportional to its resistance relative to the total series resistance. 📐 Formula: V = (R/Rt)Vs → The voltage across a resistor equals its resistance divided by total resistance, multiplied by the source voltage. 📌 Example: Given V = 4V across 4kΩ resistor with total series resistance of 12kΩ (4kΩ + 8kΩ): 4 = (4/12)Vs, so Vs = (12 × 4)/4 = 12 Volts.

Example: Calculate the source current Is

The voltage across a 3kΩ resistor is 12V. Using Ohm's Law, the current through it is calculated. This current flows through the series combination of 3kΩ and 9kΩ resistors. The current division rule is then applied to find the source current.

💡 Why this matters: Current division allows us to find how current splits between parallel branches, which is essential for analyzing complex networks.

📐 Formula: I = V/R → Current equals voltage divided by resistance. 📌 Example: I = 12/3k = 4mA through the 3kΩ resistor. After combining series resistors (3k+9k=12k and 2k+4k=6k), by current division: I = (6k/(12k+6k))Is, so 4mA = (6k/18k)Is, giving Is = 12mA.

Example: Calculate the source voltage Vs

Starting from a known 4V across a 2kΩ resistor, the current through it is found using Ohm's Law. This same current flows through the series combination of 2kΩ and 4kΩ resistors, allowing calculation of the voltage across 4kΩ. Series and parallel combinations are used to simplify the circuit and find Vs using voltage division.

📌 Example: I = 4/2k = 2mA. Voltage across 4kΩ = 2mA × 4kΩ = 8V. Total voltage across 2kΩ and 4kΩ = 4V + 8V = 12V. After combining series resistors (4k+2k=6k) and parallel (6k||6k = 3k), Vs = 12 × (12/3) = 48V.

Example: Calculate the voltage across 4kΩ resistor

This example demonstrates multiple series-parallel reductions to find an equivalent circuit. Using successive parallel combinations (12k||4k = 3k, 12k||6k = 4k, 12k||4k = 3k), the circuit is simplified. Voltage division is applied twice to find the voltage across the target resistor.

📌 Example: After reducing the circuit, by voltage division: V across equivalent 3kΩ = (3/6) × 12 = 6V. Then V across 3kΩ = (3/12) × 6 = 1.5 volts. This same voltage appears across the 4kΩ resistor.

Kirchhoff's Current Law (KCL)

Kirchhoff's Current Law states that the sum of all currents entering a node equals the sum of all currents leaving the node. It can also be defined as the sum of entering currents plus the sum of leaving currents equals zero.

🔑 Definition — Node: The junction of two or more than two elements; a point of connection between circuit elements. 🔑 Definition — Branch: The distance or link between two nodes. 🔑 Definition — Loop: The closed path for current flow in which no node is encountered more than once.

ASSUMPTIONS:

• All entering currents are taken as negative.
• All leaving currents are taken as positive.

Formula for Writing Equations: Number of equations in node analysis = N - 1, where N is the number of nodes.

🔑 Definition — Ground: A common or reference point among all nodes without insertion of any component between them.

Example: Find values of I₁, I₂, I₃, I₄

Assuming currents leaving the node are positive, KCL equations are written for each node.

📌 Example:

• Node 1: -I₁ + 0.06 + 0.02 = 0, so I₁ = 0.08A
• Node 2: I₁ - I₄ + I₆ = 0, so 0.08 - I₄ + I₆ = 0, therefore -I₄ + I₆ = -0.08A
• Node 3: -0.06 + I₄ - I₅ + 0.04 = 0, so I₄ - I₅ = 0.02A
• Node 4: -0.02 + I₅ - 0.03 = 0, so I₅ = 0.05A
• Substituting I₅ in node 3: I₄ - 0.05 = 0.02, so I₄ = 0.07A
• Substituting I₄ in node 2: -0.07 + I₆ = -0.08, so I₆ = -0.01A

Example: Find Vx, Iin, and Is

This example combines Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL) to find unknown values in a circuit containing dependent sources.

📌 Example: (a) By KVL: -2 + Vx + 8 = 0, so Vx = -6V (b) By KCL at the top right node: Is + 4Vx = 4 - Vx/4, so Is + 4(-6) = 4 - (-6/4), therefore Is = 29.5A (c) Iin = 1 + Is + Vx/4 - 6 = 1 + 29.5 + (-6/4) - 6 = 23A

⭐ Key Takeaways

Kirchhoff's Current Law is the fundamental principle that current entering a node equals current leaving it, with entering currents taken as negative and leaving as positive. The number of node equations needed equals the number of nodes minus one (N-1). Voltage division (V = R/Rt × Vs) and current division are essential tools for simplifying circuits with series and parallel combinations. Ground serves as a common reference point for voltage measurements without any component between nodes. Combining KVL and KCL enables solving for unknown voltages, currents, and source values even in circuits containing dependent sources.

🧠 Quick Revision Questions

1. State Kirchhoff's Current Law in two different ways.
2. What is the formula for the number of equations in node analysis?
3. How is ground defined in circuit analysis?
4. In KCL, what sign convention is used for entering and leaving currents?
5. When combining KVL and KCL, what additional circuit elements can be solved for that KCL alone cannot handle?

📘 Lecture 8 — (Application of Nodal Analysis)

📖 Overview: This lecture focuses on applying Kirchhoff's Current Law (KCL) to write nodal equations for various circuits. Through numerous examples, it demonstrates how to systematically formulate KCL equations at different nodes to solve for unknown currents and voltages, emphasizing the practical use of nodal analysis in circuit theory.

🗂️ Topics Covered

The lecture covers writing KCL equations for multiple nodes in various circuits, calculating unknown currents using nodal analysis, solving for node voltages to find specific currents like I₀, determining voltage values that satisfy given conditions, and computing power absorbed by resistors using nodal analysis results.

📝 Lecture Summary

Example: Write KCL equations for all nodes.

For node 1: I₁ + I₂ – I₅ = 0
For node 2: -I₂ + I₃ - 50I₂ = 0
For node 3: -I₁ + 50I₂ + I₄ = 0
For node 4: I₅ – I₃ – I₄ = 0

Example Write KCL equations for all nodes.

For node A: 10mA is entering the node and the source current Iₜ is leaving and 60mA is also entering.
So for node A: Iₜ - 60mA – 10mA = 0
For node B: 60mA – 40mA – 20mA = 0

Example Write KCL equation for node A.

By KCL the equation for node A: -12mA + 4mA + I = 0

Example Write KCL equation for node A and node B.

For node A: -I₁ + I₂ + 3mA = 0
For node B: -12mA + 4mA + I₁ = 0

Example Write KCL equation for node A.

For node A: -10Iₓ + Iₓ + 44mA – 12mA = 0

Example Write KCL equation for node A.

For node A equation will be by KCL: Iₓ + 10Iₓ – 44mA = 0

Example Calculate the values of I₁ and I₂.

We want to calculate the values of I₁ and I₂.
For node A: 4mA + 8mA - I₁ = 0 → I₁ = 12mA
Now for I₂, for node B: -8mA + 2 mA + I₂ = 0 → I₂ = 6mA

Example Calculate the values of I₁, I₂, and I₃.

We want to calculate the values of I₁, I₂, and I₃ using node analysis.
For node A: -I₁ – I₂ + 8mA = 0 → -I₁ – I₂ = -8mA
For node B: I₂ + I₃ + 4mA = 0 → I₂ + I₃ = -4mA
For node C: -I₃ + 2mA – 8mA = 0 → I₃ = -6mA
Putting the value of I₃ in equation of node B: I₂ - 6mA = -4mA → I₂ = 2mA
Putting the value of I₂ in equation of node A: -I₁ – 2mA = -8mA → I₁ = 6mA

Example: Find the KCL equations for node A, node B, node C and node D.

For node A: -5mA + 8mA + 4mA = 0
For node B: I₁ – I₂ + 5mA = 0
For node C: -I₁ – 2mA + 3mA – 8mA = 0
For node D: -4mA – 3mA + I₃ = 0

Example Calculate the current I₀.

At node 1: (V₁/12k) + ((V₁ – V₂)/10k) = 6mA
5V₁ + 6V₁ – 6V₂ = (60k)(6mA) → 11V₁ – 6V₂ = 360

At node 2: (V₂/3k) + (V₂/6k) + ((V₂ - V₁)/10k) = 0
10V₂ + 5V₂ + 3V₂ - 3V₁ = 0 → 18V₂ - 3V₁ = 0 → 6V₂ – V₁ = 0

Equating equations:
33V₁ - 18V₂ = 1080
-3V₁ + 18V₂ = 0
30V₁ = 1080 → V₁ = 36 volts
6V₂ - 36 = 0 → V₂ = 6 volts
I₀ = V₂/6k = 6/6k = 1mA

🔑 Definition — I₀: The current through the 6kΩ resistor, calculated as V₂/6k.

Example: Use nodal analysis to determine the value of V₂ that will result in V₁ = 0

If V₁ = 0, the dependent source is a short circuit and we may redraw the circuit.

At NODE 1: 4 - 6 = V₁/40 + (V₁ - 96)/20 + (V₁ - V₂)/10
Since V₁ = 0, this simplifies to: -2 = -96/20 - V₂/10
So that V₂ = -28V

Example Use nodal analysis to find V₁ and V₂, compute the power absorbed by 6Ω resistance.

Designate the node between the 3Ω and 6Ω resistors as node X, and the right hand node of the 6Ω resistor as node Y. The bottom node is chosen as the reference node.

Writing the two nodal equations:
NODE X: -10 = (Vₓ - 240)/3 + (Vₓ - Vᵧ)/6
NODE Y: 0 = (Vᵧ - Vₓ)/6 + Vᵧ/30 + (Vᵧ - 60)/12

Simplifying: -180 + 1440 = 9Vₓ – 3Vᵧ [1]
10800 = -360Vₓ + 612Vᵧ [2]

Solving: Vₓ = 181.5V and Vᵧ = 124.4V
Thus, V₁ = 240 - Vₓ = 58.50V and V₂ = Vᵧ - 60 = 64.40V

The power absorbed by the 6Ω resistor is (Vₓ - Vᵧ)² / 6 = 543.4 W

📐 Formula: Power absorbed by a resistor = V²/R = (Vₓ - Vᵧ)² / 6 → The power dissipated in the 6Ω resistor, calculated using the voltage difference across it.

💡 Why this matters: This example demonstrates how nodal analysis can be used not only to find node voltages but also to compute power dissipation in circuit elements, a critical calculation in circuit design and analysis.

⭐ Key Takeaways

KCL states that the sum of currents entering a node equals the sum of currents leaving that node, which forms the foundation of nodal analysis. When writing nodal equations, you must account for all sources (current and voltage) and all resistors connected to each node, carefully considering current directions. Nodal analysis is a systematic method: identify all nodes, designate one as reference, write KCL equations for each non-reference node, and solve the resulting system of equations. The voltage difference across any resistor can be found by subtracting node voltages, and this voltage difference is essential for calculating branch currents and power. Node voltages are always measured with respect to the reference node, and dependent sources must be handled carefully as they introduce additional constraints into the equations.

🧠 Quick Revision Questions

1. What is the KCL equation for node A in the circuit with -12mA, +4mA, and I?
2. For the circuit calculating I₁ and I₂, what value of I₁ is obtained from the node A equation?
3. In the example finding I₀, what are the final values of V₁ and V₂ after solving the nodal equations?
4. What value of V₂ makes V₁ = 0 in the example with the dependent source?
5. What two node voltages Vₓ and Vᵧ are found when analyzing the circuit with the 6Ω resistor, and what power is absorbed by that resistor?

📘 Lecture 9 — Application of Nodal Analysis

📖 Overview: This lecture demonstrates the application of nodal analysis to solve complex circuit problems involving multiple nodes, voltage sources, and current sources. It reinforces the practical technique of setting up and solving simultaneous equations to determine unknown voltages and currents, which is fundamental for circuit analysis.

🗂️ Topics Covered

This lecture presents a series of seven detailed examples that apply nodal analysis to find unknown currents and voltages in various circuits. The examples progress from simple two-node circuits to more complex four-node circuits, including cases with dependent sources and circuits where voltage divider rule is applied after nodal analysis. Problems cover finding current through specific resistors, voltage across resistors, and the value of a dependent source constant.

📝 Lecture Summary

Example 1: Find current flowing through 6k ohm resistor

The circuit has two unknown nodes, node 1 and node 2. Apply Kirchhoff’s Current Law (KCL) at each node to establish equations.

At node 1: ((V1 - V2)/6k) + (V1/3k) = 2mA. This simplifies to V1 – V2 + 2V1 = 6k x 2mA → -V2 + 3V1 = 12. Multiply both sides by 3: 9V1 - 3V2 = 36. This is equation (A).

At node 2: V2/12k + 4mA + (V2 - V1)/6k = 0. This simplifies to V2 + 48 + 2V2 - 2V1 = 0 → 3V2 – 2V1 = -48. This is equation (B).

Add equations (A) and (B): (9V1 - 3V2) + (3V2 – 2V1) = 36 + (-48) → 7V1 = -12 → V1 = -12/7 V.

Substitute V1 into equation (A): V2 = 3V1 - 12 = 3(-12/7) - 12 = (-36/7) - 12 = (-36-84)/7 = V2 = -120/7 V.

The voltage across the 6k resistor is Vo = V2 – V1 = (-120/7) + (12/7) = -108/7 V. The current Io = Vo/6k = (-108/7) x (1/6k) = -108/42 = Io = -2.57 mA.

📌 Example: For the circuit with a 6k resistor between two nodes, V1 = -12/7 V, V2 = -120/7 V, and the current through the 6k resistor is 2.57 mA (flowing opposite to assumed direction).

Example 2: Find voltage across 2k ohm resistor

Combine the two series 2k resistors on the right side into a single 4k resistor. Then apply KCL at node 1 and node 2.

At node 1: V1/3k + 4mA + 2mA + (V1 - V2)/6k = 0. Multiply by 6k: 2V1 + 24 + 12 + V1 – V2 = 0 → 3V1 – V2 = -36.

At node 2: V2/4k + V2/12k - 2mA + (V2 – V1)/6k = 0. Multiply by 12k: 3V2 + V2 – 24 + 2V2 – 2V1 = 0 → 6V2 – 2V1 = 24.

Solve the two equations: from node 1 equation, -V2 + 3V1 = -36. Multiply by 2: -2V2 + 6V1 = -72. Add to node 2 equation (6V2 – 2V1 = 24): 4V2 = -48 → V2 = 0 V.

The voltage across the 2k resistor (using voltage divider rule on the series combination): V0 = (2/(2+2)) x V2 = 2/4 x 0 = 0 V.

🔑 Definition — Voltage Divider Rule: For two resistors in series, the voltage across one resistor equals the ratio of that resistor to the total resistance times the total voltage across the series combination. Formula: V_Rx = (Rx / (R1 + R2)) × V_total.

📌 Example: Two 2k resistors in series with 0V across them produce 0V across each resistor.

Example 3: Find current flowing through 4k ohm resistor

The circuit has a single unknown node V1. Apply KCL at node 1.

At node 1: (V1 - 12)/12k + V1/6k + (V1 + 6)/4k = 0. Multiply by 12k: V1 – 12 + 3V1 + 3(6) + 2V1 = 0 → 6V1 = -6 → V1 = -1 V.

The current through the 4k resistor: Io = (V1 – (-6))/4k. Note the negative sign is due to the negative reference of the battery. Io = (-1 + 6)/4k = 5/4k = Io = 1.2 mA.

📌 Example: For the circuit with a 12V battery and a -6V battery, V1 = -1V and the current through the 4k resistor connected to the -6V source is 1.2 mA.

Example 4: Find voltage across 1k ohm resistor

The 2k and 1k resistors are in series, so combine them into a single 3k resistor. Apply KCL at the single unknown node V1.

At node 1: (V1 + 6)/6k + (V1 + 3)/2k + V1/3k = 0. Multiply by 6k: 6 + V1 + 9 + 2V1 + 3V1 = 0 → 6V1 + 15 = 0 → V1 = -15/6 = -2.5 V.

Using the voltage divider rule on the original 2k and 1k series combination: V0 = (1k/(2k+1k)) x V1 = 1/3 x (-15/6) = V0 = -5/6 V ≈ -0.833 V.

📌 Example: The voltage across the 1k resistor in a series combination with a 2k resistor is -5/6 V when the node voltage is -15/6 V.

Example 5: Find the voltage across a 12K ohm resistance

This circuit has three nodes: reference, node between 6V battery and 12K resistance (neglected), and node V1. The node equation is written for V1.

At node 1: (V1 – 12)/6k + (V1 - 6)/12k + V1/6k = 0. Multiply by 12k: 2(V1 - 12) + (V1 - 6) + 2V1 = 0 → 2V1 - 24 + V1 - 6 + 2V1 = 0 → 5V1 = 30 → V1 = 6 V.

The voltage across the 12K resistor is between the 6V battery (Vs) and V1: V0 = V1 - Vs = V1 - 6 = 6 - 6 = 0 V.

📌 Example: When V1 equals the battery voltage (6V), the voltage across the 12K resistor connected between them is 0V.

Example 6: Find the current Io through the 12K ohm resistor

This circuit has two unknown nodes, V1 and V2. Apply KCL at both nodes.

At node 1: (V1+6)/12 + V1/12 + (V1 – V2)/12 = 0. Multiply by 12: V1 + 6 + V1 + V1 – V2 = 0 → 3V1 – V2 = -6. Multiply by 2: 6V1 – 2V2 = -12 (Equation A1).

At node 2 (both resistors are 12k, so resistances are in kΩ): (V2 – V1)/12 + V2/12 = 2mA. Multiply by 12: V2 – V1 + V2 = 24 → 2V2 – V1 = 24 (Equation B).

Add equation A1 and B: (6V1 – 2V2) + (2V2 – V1) = -12 + 24 → 5V1 = 12 → V1 = 12/5 = 2.4 V.

Substitute into equation A: V2 = 3V1 + 6 = 3(12/5) + 6 = (36 + 30)/5 = V2 = 66/5 = 13.2 V.

The current through the 12K resistor: Io = (V1 – V2)/12 = (12/5 – 66/5)/12 = (-54/5) x (1/12) = -54/60 = Io = -0.9 mA.

📌 Example: For the circuit with nodes at 2.4V and 13.2V, the current through the 12K resistor connecting them is 0.9 mA flowing from node 2 to node 1.

Example 7: Use nodal analysis to find Vp

The bottom node has the largest number of branch connections, so it is chosen as the reference node. Working from left to right, name nodes 1, P, 2, and 3.

At node 1: 10 = V1/20 + (V1 - Vp)/40. Multiply by 40: 400 = 2V1 + V1 - Vp → 3V1 – Vp = 400. Multiply by 20: 60V1 - 20Vp = 8000 [1].

At node P: 0 = (Vp - V1)/40 + Vp/100 + (Vp - V2)/50. Multiply by 200: 0 = 5(Vp - V1) + 2Vp + 4(Vp - V2) → 0 = 5Vp - 5V1 + 2Vp + 4Vp - 4V2 → -5V1 + 11Vp - 4V2 = 0. Multiply by 10: -50V1 + 110Vp - 40V2 = 0 [2].

At node 2: -2.5 + 2 = (V2 - Vp)/50 + (V2 - V3)/10. Simplify: -0.5 = (V2 - Vp)/50 + (V2 - V3)/10. Multiply by 50: -25 = (V2 - Vp) + 5(V2 - V3) → -25 = V2 - Vp + 5V2 - 5V3 → -Vp + 6V2 - 5V3 = -25 [3].

At node 3: 5 - 2 = V3/200 + (V3 - V2)/10. Simplify: 3 = V3/200 + (V3 - V2)/10. Multiply by 200: 600 = V3 + 20(V3 - V2) → 600 = V3 + 20V3 - 20V2 → -20V2 + 21V3 = 600. Multiply by 10: -200V2 + 210V3 = 6000 [4].

Solving the system of equations: Vp = 171.6 V.

💡 Why this matters: This example demonstrates that nodal analysis can handle circuits with four unknown nodes, requiring the solution of four simultaneous equations—a common real-world circuit analysis scenario.

📌 Example: In a complex circuit with four nodes, nodal analysis yields Vp = 171.6V after solving a system of linear equations.

Example 8: Use nodal analysis to find K that will cause Vy to be zero

This circuit has a dependent voltage source with gain K. Apply KCL at nodes x and y.

At node x: Vx/4 + (Vx - Vy)/2 + (Vx - 6)/1 = 0. Multiply by 4: Vx + 2(Vx - Vy) + 4(Vx - 6) = 0 → Vx + 2Vx - 2Vy + 4Vx - 24 = 0 → 7Vx - 2Vy = 24. With Vy = 0: 7Vx = 24 → Vx = 48/14 = 3.429 V (Equation [1] simplified).

At node y: (Vy - KVx)/3 + (Vy - Vx)/2 = 2. Multiply by 6: 2(Vy - KVx) + 3(Vy - Vx) = 12 → 2Vy - 2KVx + 3Vy - 3Vx = 12 → -2KVx - 3Vx + 5Vy = 12. With Vy = 0: -2KVx - 3Vx = 12.

From equation [2]: -2KVx - 3Vx = 12 → -2K(3.429) - 3(3.429) = 12 → -6.858K - 10.287 = 12 → -6.858K = 22.287 → K = -3.250.

📌 Example: For a dependent source with gain K, setting Vy = 0 requires K = -3.25, with Vx = 3.429V.

Example 9: Use nodal analysis to find I5

Choose the bottom node as ground. Name nodes: "1" (left-most), "2" (top), "3" (central), "4" (between 4Ω and 6Ω resistors).

At node 1: -3 = v1/2 + (v1 - v2)/1. Multiply by 2: -6 = v1 + 2(v1 - v2) → -6 = v1 + 2v1 - 2v2 → 3v1 - 2v2 = -6 [1].

At node 2: 2 = (v2 - v1)/1 + (v2 - v3)/3 + (v2 - v4)/4. Multiply by 12: 24 = 12(v2 - v1) + 4(v2 - v3) + 3(v2 - v4) → 24 = 12v2 - 12v1 + 4v2 - 4v3 + 3v2 - 3v4 → -12v1 + 19v2 - 4v3 - 3v4 = 24 [2].

At node 3: 3 = v3/5 + (v3 - v4)/7 + (v3 - v2)/3. Multiply by 105: 315 = 21v3 + 15(v3 - v4) + 35(v3 - v2) → 315 = 21v3 + 15v3 - 15v4 + 35v3 - 35v2 → -35v2 + 71v3 - 15v4 = 315 [3].

At node 4: 0 = v4/6 + (v4 - v3)/7 + (v4 - v2)/4. Multiply by 84: 0 = 14v4 + 12(v4 - v3) + 21(v4 - v2) → 0 = 14v4 + 12v4 - 12v3 + 21v4 - 21v2 → -21v2 - 12v3 + 47v4 = 0. Multiply by 2: -42v2 - 24v3 + 94v4 = 0 [4].

Solving the system of four equations: v3 = 6.76 V. Therefore, I5 = v3/5 = 1.352 A.

📌 Example: In a complex four-node circuit, solving the system yields v3 = 6.76V and the current through the 5Ω resistor (I5) is 1.352A.

⭐ Key Takeaways

Nodal analysis is a systematic method for determining unknown voltages by applying KCL at each non-reference node and solving the resulting system of linear equations. The number of equations equals the number of unknown node voltages. When resistors are in series, they can be combined to simplify the circuit before analysis. The voltage divider rule is a useful tool after nodal voltages are found to determine voltages across individual series resistors. Circuits with dependent sources require additional constraint equations but follow the same nodal analysis procedure.

🧠 Quick Revision Questions

1. What is the first step in applying nodal analysis to a circuit?
2. In Example 1, how many equations were needed to solve for two unknown node voltages?
3. How does the voltage divider rule work for two resistors in series?
4. In Example 8, what constraint was imposed to find the value of K?
5. What is the formula for calculating current through a resistor between two nodes using nodal voltages?

📘 Lecture 10 — (Super Node - Constraint or Coupling Equation)

📖 Overview: This lecture introduces the concept of the super node, a powerful technique for analyzing circuits containing voltage sources between two non-reference nodes. It explains how to formulate constraint (or coupling) equations to mathematically describe super nodes, enabling the calculation of node voltages and currents in complex circuits. This topic is crucial for simplifying circuit analysis where standard nodal analysis would be difficult.

🗂️ Topics Covered

The lecture begins with examples calculating current through resistors using standard node voltage analysis, then introduces the super node concept as a combination of two ordinary nodes around a voltage source. It explains the constraint or coupling equation that mathematically describes a super node, and works through multiple examples calculating currents, voltages, and power in circuits with dependent and independent sources using this technique.

📝 Lecture Summary

Example: Calculate the current I₀ through the 3k ohm resistor.

Solution: At node A: ((V₁ – V₂)/3k) + (V₁/6k) = 2mA where V₂ = 6V

Substituting V₂: ((V₁ – 6)/3k) + (V₁/6k) = 2mA

Multiplying by 6k: 2V₁ - 12 + V₁ = 12 3V₁ = 24 V₁ = 8 volts

By Ohm's law: I₀ = (V₁ - V₂)/3k = (8 - 6)/3k = 0.667 mA

I₁ = 6/4k = 1.5 mA

Example: Calculate the current I₀ through the 10k ohm resistor also find V₁ and V₂.

Solution: I₀ = V₁/10k ----------------- (A)

For node 1: (V₁/10k) + ((V₁ - V₂)/10k) = 4mA (V₁/10k) + (V₁/10k) – (V₂/10k) = 4mA

Substituting I₀: I₀ + I₀ – (V₂/10k) = 4mA 2I₀ - (V₂/10k) = 4mA 20I₀ – V₂ = 40mA ------------- (B)

For node 2: (V₂/10k) + ((V₂ – V₁)/10k) = -2I₀ (V₂/10k) + (V₂/10k) – (V₁/10k) = -2I₀ (V₂/10k) + (V₂/10k) – I₀ = -2I₀

Equating equations of node 1 and 2: 2V₂ + 10I₀ = 0 ---------------- (C)

Solving (B) and (C): 50I₀ = 80mA Hence I₀ = 8/5 mA

From equation (A): I₀ = V₁/10k 8/5 mA = V₁/10k V₁ = 16 volts

Substituting I₀ from (A) into (C): 2V₂ + V₁ = 0 2V₂ = -V₁ V₂ = -16/2 = -8 volts

SUPER NODE:

A super node is a node which emerges as a result of combination of two ordinary nodes around a voltage source.

CONSTRAINT OR COUPLING EQUATION:

This is an equation which describes a super node mathematically, instead of writing equations for individual ordinary nodes of the super node.

Example: Calculate the power supplied by the current source.

Solution: Power = 1A(V₁)

Applying KCL to node 1: (V₁/10) - Iₓ = 1

Applying KCL to node 2: 0 = Iₓ + (V₂/5) + ((V₂-2)/7)

Now, 3 unknown quantities cannot be calculated from 2 equations, so try super node technique.

Equation for super node: 1 = (V₁/10) + (V₂/5) + ((V₂-2)/7)

Simplifying: 35V₁ + 120V₂ = 450 --------- (A)

Constraint or coupling equation: V₁ - V₂ = 6

Solving the two equations simultaneously: V₂ = V₁ – 6

Substituting into (A): 35V₁ + 120(V₁ – 6) = 450 35V₁ + 120V₁ – 720 = 450 155V₁ = 1170 V₁ = 7.55V

Therefore: V₁ = 7.55 V and V₂ = 1.55 V

Hence, power: P = 1(A)(V₁) = 7.55 W

Example: Calculate I₁ from the given circuit.

Solution: Redrawing the circuit:

Constraint equation for super node: V₁ – V₂ = 3

KCL equation for super node: (V₁/3k) + (V₂/6k) = 2 × 10⁻³ 2V₁ + V₂ = 12

From constraint equation: V₂ = V₁ - 3

Substituting into KCL equation: 2V₁ + V₁ – 3 = 12 3V₁ = 15 V₁ = 5 volts

Now: I₁ = V₁/3k = 5/3k = 1.6 mA

Example: Find I₁ from the circuit.

Solution: Redrawing the circuit:

Constraint equation for super node: V₂ – V₁ = 6V

KCL equation for super node: ((V₁ – 3)/6k) + (V₁/12k) + ((V₂ + 3)/12k) + (V₂/12k) = 0

2V₁ - 6 + V₁ + V₂ + 3 + 2V₂ = 0 3V₁ + 3V₂ = 3 V₂ + V₁ = 1

Adding constraint equation and KCL equation: V₂ – V₁ = 6 V₂ + V₁ = 1

2V₂ = 7 V₂ = 3.5V

I₁ = 3.5/6k = 35/60k = 7/12 mA

Example: Calculate the value I₁.

Solution: For node 1: (V₁/12k) + (V₁/4k) + (V₁/6k) – (6/6k) = 2 mA

((1/12k) + (1/4k) + (1/6k))V₁ – (6/6k) = 2 mA

(V₁/2k) – (1/1k) = 2 × 10⁻³ V₁ – 2 = 4 V₁ = 6V

I₁ = V₁/4k = 6/4k = 1.5 mA

⭐ Key Takeaways

The super node technique is essential when a voltage source connects two non-reference nodes in a circuit. Instead of writing separate KCL equations for each node, you combine them into one super node equation and add a constraint (coupling) equation that expresses the voltage difference between the two nodes (V₁ - V₂ = source voltage). This approach reduces the number of unknowns and simplifies the analysis. Always identify the super node around voltage sources between non-reference nodes, write the KCL equation for the entire super node, and then add the constraint equation based on the voltage source value. The sign convention in the constraint equation must match the polarity of the voltage source.

🧠 Quick Revision Questions

1. What is a super node, and when do you use it in circuit analysis?
2. Write the general form of a constraint or coupling equation for a super node containing a voltage source of value Vₛ.
3. In the example with the 10k ohm resistor, why was the super node technique not needed, and how were V₁ and V₂ found?
4. For the circuit with power supplied by the current source, how many equations were needed and what were they?
5. If a super node has two ordinary nodes with a 6V source between them and the KCL equation gives V₂ + V₁ = 1, what are the values of V₁ and V₂?

📘 Lecture 11 — Examples of Nodal Analysis - Super Node technique

📖 Overview: This lecture demonstrates the application of nodal analysis and the super node technique through multiple worked examples. It shows how to handle circuits containing voltage sources between nodes by forming a super node and solving for unknown voltages and currents.

🗂️ Topics Covered

The lecture covers five complete examples of nodal analysis: finding output voltage using constraint equations, calculating voltage and current using super node analysis, applying voltage division with super node results, solving circuits with dependent current sources, and determining current direction through negative sign interpretation.

📝 Lecture Summary

Example: Find the output voltage V₀

The circuit contains a voltage source between nodes. By redrawing the circuit, V₁ is identified as 6V. The constraint equation is V₁ – V₃ = 3V.

🔑 Definition — Constraint Equation: An equation that expresses the voltage relationship between two nodes due to a voltage source connected between them.

📐 Formula: V₁ – V₃ = 3V → The voltage across the source equals the difference between the two node voltages.

📌 Example: Given V₁ = 6V, the constraint equation 6 – V₃ = 3V gives V₃ = 3V. Since V₃ = V₀, therefore V₀ = 3 Volts.

Example: Calculate V₀ and I₀

Node 2 and Node 3 form a super node because a 6V voltage source is connected between them.

🔑 Definition — Super Node: A technique used when a voltage source is connected between two non-reference nodes. The two nodes are combined into one "super node" for KCL analysis.

📐 Constraint Equation: V₂ – V₃ = 6 --------------------- (A)

📐 KCL at Super Node: Sum of all currents leaving the super node = 0

📌 Example: KCL: (V₂-12)/6k + (V₂/3k) + (V₃/6k) + (V₃-12)/12k = 0 Simplifies to: 2V₂ + V₃ = 12 ......(B)

From constraint (A): V₃ = V₂ – 6 Substitute into (B): 2V₂ + (V₂ – 6) = 12 3V₂ = 18 → V₂ = 6 Volts Therefore V₃ = 0 Volt

V₀ = 0 Volt I₀ = V₂/3k = 6/3k = 2mA

Example: Calculate the value of V₀

The circuit contains a 6V voltage source between Node 2 and Node 3, forming a super node.

📐 Constraint Equation: V₃ – V₂ = 6 volts

📌 Example: KCL at super node: (V₂ – 6)/6k + V₂/12k + (V₃-6)/4k + V₃/12k + V₃/6k = 0 Simplifies to: V₂ + 2V₃ = 10 ----- (A)

From constraint: V₂ = V₃ – 6 Substitute into (A): (V₃ – 6) + 2V₃ = 10 → 3V₃ = 16 V₃ = 16/3 volts V₂ = (16/3) – 6 = –2/3 V

💡 Why this matters: These node voltages allow us to find V₀ using voltage division.

Voltage Division Rule: V₀ = (4k × V₃)/(2k + 4k) = 64/18 = 3.55 Volts

Example: Find V₀ (Circuit with Dependent Source)

This circuit has a dependent current source 2Iₓ controlled by current Iₓ.

📌 Example: At Node 1: V₁/10k = 3mA – 2Iₓ Given Iₓ = V₁/10k, substitute: V₁/10k = Iₓ = 3mA – 2Iₓ → 3Iₓ = 3mA → Iₓ = 1mA

At Node 2: V₂/10k = 2Iₓ = 2(1mA) → V₂ = 20 Volts Therefore V₀ = 20 Volts

Example: Find I₀

A straightforward nodal analysis problem without super nodes.

📌 Example: KCL at Node 1: (V₁ – 6)/4k + V₁/10k + (V₁ – 12)/2k = 0 Multiply through: 5V₁ – 30 + 2V₁ + 10V₁ – 120 = 0 17V₁ = 150 → V₁ = 8.82 Volts

I₀ = (V₁ – 12)/2k = (8.82 – 12)/2k = –1.58 mA

💡 Why this matters: The negative sign indicates the actual current flows in the opposite direction to what was assumed in the circuit diagram.

⭐ Key Takeaways

The super node technique is essential when a voltage source connects two non-reference nodes — these nodes are combined and their constraint equation is written as V₂ – V₃ = source voltage. When writing the KCL equation at the super node, sum all currents leaving the combined super node region, treating internal connections as part of the super node. Always substitute the constraint equation into the KCL equation to solve for node voltages. For circuits with dependent sources, express the controlling variable in terms of node voltages and solve the system of equations. A negative current result indicates the actual current direction is opposite to the assumed direction in the circuit diagram.

🧠 Quick Revision Questions

1. What is a super node and when is it used in nodal analysis?
2. In the super node technique, what form does the constraint equation take when a 6V source connects Node 2 and Node 3?
3. If V₂ = 6V and the constraint equation is V₂ – V₃ = 6V, what is the value of V₃?
4. How is voltage division rule applied to find V₀ from node voltages and resistor values?
5. What does a negative current value like –1.58mA indicate about current direction?

📘 Lecture 12 — (Kirchhoff's Voltage Law (KVL) - Loop - Examples)

📖 Overview: This lecture introduces Kirchhoff's Voltage Law (KVL), a fundamental principle in circuit theory that governs voltage relationships in closed loops. It explains how to write KVL equations, the assumptions involved, and demonstrates the application through worked examples.

🗂️ Topics Covered

The lecture covers the statement and meaning of Kirchhoff's Voltage Law, the definition of a loop, the assumptions for sign conventions (energy level increases and decreases, current direction relative to source), and the rule for the number of equations needed. Detailed step-by-step examples show how to write KVL equations for circuits with multiple loops and voltage sources, and demonstrates the concept of linear dependence among loop equations.

📝 Lecture Summary

KIRCHHOFF’S VOLTAGE LAW:

This law states that the algebraic sum of the voltages around any loop is zero. OR Sum of voltages rises and voltage drops around any closed path or loop is equal to zero.

LOOP:

It is the closed path for the flow of current in which no node is encountered more than once. Loop can be considered as closed path in which work done in moving a unit charge is equal to zero.

ASSUMPTIONS:

Any increase in energy level is taken as positive and any decrease in energy level is taken as negative. Current leaving the source is taken as positive and current entering the voltage source is taken as negative.

NO. OF EQUATIONS:

No. of equations to be written are equal to the no. of loops or closed paths.

Example:

We want to calculate the value of VR₃ where values of VR₁ and VR₂ are known by using KVL.

Solution: Now we will take increase in energy level positive and decrease in energy level negative. By using this assumption and KVL the equation of this loop will be:

+VR₁ - 5 + VR₂ – 15 + VR₃ – 30 = 0

which can be written as: VR₁ + VR₂ + VR₃ = 5 + 15 + 30 = 50

Now suppose that VR₁ and VR₂ are known to be 18V and 12V respectively:

VR₃ = 50 - 18 – 12 = 20V

🔑 Definition — Loop: The closed path for the flow of current in which no node is encountered more than once. It is a closed path in which work done in moving a unit charge is equal to zero.

📐 Formula: KVL: Sum of voltage rises = Sum of voltage drops, or ΣV = 0 around any closed loop.

📌 Example: In a loop with voltage sources 5V, 15V, 30V and resistors with voltage drops VR₁=18V, VR₂=12V, VR₃ unknown: The KVL equation gives VR₁ + VR₂ + VR₃ = 5 + 15 + 30 = 50, so VR₃ = 50 – 18 – 12 = 20V.

💡 Why this matters: This example shows how KVL provides a simple algebraic method to find an unknown voltage when other voltages in the loop are known.

Example:

Note that this network has three closed paths: the left loop, the right loop and outer loop.

Applying KVL to left loop: VR₁ + VR₄ – 16 – 24 = 0

KVL equation for right loop starting at point B: VR₂ + VR₃ + 8 + 16 – VR₄ = 0

The equation for outer loop starting at point A: VR₁ + VR₂ + VR₃ + 8 – 24 = 0

Note that if we add first two equations, we obtain the third equation. So these three equations are not linearly independent. We will discuss this issue in next lectures that we will use only linearly independent equations to solve for all voltages.

📌 Example: A circuit has three closed paths (left loop, right loop, and outer loop). Writing KVL for each:

• Left loop: VR₁ + VR₄ – 16 – 24 = 0
• Right loop (starting at B): VR₂ + VR₃ + 8 + 16 – VR₄ = 0
• Outer loop (starting at A): VR₁ + VR₂ + VR₃ + 8 – 24 = 0

Adding the left and right loop equations gives the outer loop equation, showing they are not all independent.

💡 Why this matters: This example demonstrates that not all loop equations are independent; some are linear combinations of others. For solving circuits, we need to select only the linearly independent equations.

⭐ Key Takeaways

Kirchhoff's Voltage Law states that the algebraic sum of voltages around any closed loop equals zero, meaning the sum of voltage rises equals the sum of voltage drops. In writing KVL equations, increases in energy level are taken as positive and decreases as negative; current leaving the source is positive and entering is negative. The number of KVL equations to write equals the number of loops in the circuit. However, not all loop equations are linearly independent — some may be derived from adding others, so only independent equations should be used for solving circuit unknowns.

🧠 Quick Revision Questions

1. State Kirchhoff's Voltage Law in your own words.
2. What is the sign convention for energy level changes when applying KVL?
3. How is current direction treated relative to a voltage source in KVL sign conventions?
4. In a circuit with three possible loops, how many independent KVL equations would you expect?
5. If the left loop equation is VR₁ + VR₄ = 40 and the right loop equation is VR₂ + VR₃ – VR₄ + 24 = 0, what would be the outer loop equation?

📘 Lecture 13 — (Applications of Loop Analysis)

📖 Overview: This lecture demonstrates how to apply Kirchhoff's Voltage Law (KVL) to solve for unknown voltages between two points that are not physically close in a circuit. It teaches students how to use imaginary arrow notation and multiple closed paths to determine voltages like (V_{AE}) and (V_{EC}), and how to systematically write KVL equations for circuits with multiple loops.

🗂️ Topics Covered

This lecture covers three worked examples: calculating voltage (V_{AE}) and (V_{EC}) using multiple closed paths (AEFA, ABCDEA, CDEC, CEABC, CEFABC); writing KVL equations for a two-loop circuit (loops ABCA and BDCB) with resistors and a voltage source; and calculating voltage (V_{bd}) across an open branch using closed paths abda and bcdb.

📝 Lecture Summary

Example: Calculate the voltages (V_{AE}) and (V_{EC})

To find voltages between points that are not physically close, we draw imaginary arrows across the points A, E, and C. The circuit is redrawn with these arrows.

Our approach is to apply KVL in a closed path that includes the unknown voltage. For (V_{AE}), we can use either path AEFA or path ABCDEA.

For path AEFA: [ V_{AE} + 10 - 24 = 0 ] [ V_{AE} = 14 \text{ volts} ]

For path ABCDEA: [ 16 - 12 + 4 + 6 - V_{AE} = 0 ] [ V_{AE} = 14 \text{ volts} ]

Solving both loops gives (V_{AE} = 14) volts.

To calculate (V_{EC}), we can use paths CDEC, CEFABC, or CEABC (since (V_{AE}) is now known).

For loop CDEC: [ 4 + 6 + V_{EC} = 0 ] [ V_{EC} = -10 \text{ volts} ]

For loop CEABC: [ -V_{EC} + V_{EA} + 16 - 12 = 0 ] where (V_{EA} = -V_{AE} = -14) volts [ -V_{EC} - 14 + 16 - 12 = 0 ] [ V_{EC} = -10 \text{ volts} ]

For loop CEFABC: [ -V_{EC} + 10 - 24 + 16 - 12 = 0 ] [ V_{EC} = -10 \text{ volts} ]

Each of the three paths gives the same result: (V_{EC} = -10) volts.

💡 Why this matters: When a voltage is not directly measurable between two distant points in a circuit, you can still find it by applying KVL around a closed loop that contains the unknown voltage.

🔑 Definition — Imaginary Arrow Notation: A method used to represent a voltage between two points that are not physically adjacent in a circuit, allowing KVL to be applied in a closed loop. 📐 Formula: (V_{AE} + V_{EA} = 0) → The voltage from A to E is the negative of the voltage from E to A. 📌 Example: In the path AEFA, the loop includes (V_{AE}), a 10V source, and a 24V source. Applying KVL: (V_{AE} + 10 - 24 = 0) gives (V_{AE} = 14V).

Example: Write KVL equations for the given circuit

The circuit has two loops: left loop ABCA and right loop BDCB.

KVL equation for left hand loop ABCA: [ V_{R1} + V_{R2} - V_s = 0 ]

KVL equation for right hand loop BDCB: [ 20V_{R1} + V_{R3} - V_{R2} = 0 ]

Note: The notation (20V_{R1}) suggests a dependent source or a scaling factor applied to the voltage across R1.

🔑 Definition — KVL (Kirchhoff's Voltage Law): The sum of all voltages around any closed loop in a circuit must equal zero. 📐 Formula: (\sum V = 0) around a closed loop → The algebraic sum of voltage rises and drops must be zero. 📌 Example: In loop ABCA, the voltage source (V_s) is a rise, while (V_{R1}) and (V_{R2}) are drops, giving (V_{R1} + V_{R2} - V_s = 0).

Example: Calculate the voltage (V_{bd})

We want (V_{bd}) between points b and d. Since there is no direct closed path between b and d, we redraw the circuit to show the open branch.

We consider two closed paths: abda and bcdb.

KVL equation for path abda: [ 2 + V_{bd} - 9 = 0 ] [ V_{bd} = 7 \text{ volts} ]

KVL equation for path bcdb: [ 4 + 3 - V_{bd} = 0 ] [ V_{bd} = 7 \text{ volts} ]

Both paths give the same result: (V_{bd} = 7) volts.

💡 Why this matters: Even when two points are in an open branch with no component connecting them directly, you can still find the voltage between them by applying KVL to a closed loop that goes around through other components.

🔑 Definition — Open Branch Voltage: The voltage across two points that are not connected by a direct component path can be found using KVL in a closed loop that includes those points. 📐 Formula: (V_{bd} = V_{ba} + V_{ad}) → Voltage across an open branch equals the sum of voltages along any closed path connecting the points. 📌 Example: In path abda, starting at point a, we have a 2V rise to b, then unknown (V_{bd}) to d (in the same direction as the loop), then a 9V drop back to a. KVL gives (2 + V_{bd} - 9 = 0), so (V_{bd} = 7V).

⭐ Key Takeaways

The most critical skill from this lecture is the ability to find a voltage between any two points in a circuit by applying KVL to a closed loop that contains those points as nodes, even if they are not physically adjacent or directly connected. When a voltage appears with a negative sign in the solution, it simply means the actual polarity is opposite to the assumed arrow direction. Students must be comfortable using multiple different closed paths to verify the same unknown voltage, as all paths must yield the same result if KVL is correctly applied. The concept of imaginary arrow notation is essential for handling voltages between distant nodes in complex circuits.

🧠 Quick Revision Questions

1. For the first example, what value does each of the three paths give for (V_{EC})?
2. How do you define an "imaginary arrow" when measuring voltage between two distant points?
3. In the second example, what is the KVL equation for the left hand loop ABCA?
4. In the third example, what are the two closed paths used to find (V_{bd})?
5. If (V_{AE} = 14) volts, what is (V_{EA})?

📘 Lecture 14 — Applications of Loop Analysis

📖 Overview: This lecture demonstrates practical applications of loop (mesh) analysis and nodal analysis for solving complex circuits. It shows how to calculate power dissipation in resistors and find unknown voltages across specific network points using multiple analysis techniques.

🗂️ Topics Covered

The lecture covers three worked examples: finding power dissipated by a 3Ω resistor using both mesh and nodal analysis, calculating voltage Vcf across a network using KVL along different paths, and determining voltages Vad and Vce in a network by redrawing the circuit and applying KVL to multiple paths.

📝 Lecture Summary

Example: Find the power dissipated by 3Ω resistance.

The circuit has two meshes with currents I₁ and I₂ both flowing clockwise. The current through the 3Ω resistor is I₃ = I₁ - I₂.

For mesh 1:

• The voltage rises across the -2V source: -(-2) = +2V
• Voltage across 5Ω: V₅ = 5I₁
• Voltage across 3Ω: Vₓ = 3(I₁ - I₂)
• KVL: 2 + 5I₁ + 3(I₁ - I₂) = 0
• Equation (A): 8I₁ - 3I₂ = -2

For mesh 2:

• Voltage across 3Ω: 3(I₂ - I₁)
• Voltage across 2Ω: 2I₂
• Voltage across 3V source (rising): +3
• Equation (B): -3I₁ + 5I₂ = -3

Solving equations (A) and (B):

• I₁ = -612.9 mA
• I₂ = -967.7 mA
• I₃ = I₁ - I₂ = 354.8 mA

📌 Example: Power dissipated in 3Ω resistor = 3 × (0.3548)² = 377.7 mW
Note: The direction of subtraction (I₁ - I₂ or I₂ - I₁) doesn't matter due to the square term.

Verification using Nodal Analysis:

• Node voltage V at the top of 3Ω resistor
• Three currents leaving the node: through 5Ω toward -2V, through 3Ω to ground, through 2Ω toward +3V
• Equation: 0 = (V - (-2))/5 + V/3 + (V - 3)/2
• Solving: V = 1.065V
• 💡 Why this matters: Confirms mesh analysis is correct.
• Power₃Ω = V²/3 = (1.065)²/3 = 378.1 mW (slight difference due to rounding)

🔑 Definition — Mesh Analysis: A circuit analysis method that assigns loop currents to each independent mesh and applies KVL to solve for unknown currents.

📐 Formula: Power dissipated = I²R = V²/R → where I is current through the resistor or V is voltage across it.

Example: Calculate the voltage Vcf in the network.

The network is redrawn to show clear paths between points c and f.

Path abcfa (upper path):

• Voltage drops: -4V (across left element) + 9V (source rise) + Vcf - 6V = 0
• Vcf = 1 volt — Equation (A)

Path cdefc (lower path):

• Voltage drops: -5V - 6V - 12V + Vfc = 0
• Vfc = -1 volt — Equation (B)

🔑 Definition — Vcf and Vfc: Vcf is the voltage at point c with respect to point f; Vfc is the voltage at f with respect to c. They are related by Vcf = -Vfc.

From equation (B): Since Vcf = -Vfc, we get Vcf = -(-1) = 1 volt

💡 Why this matters: The same voltage can be found using two different paths, validating Kirchhoff's Voltage Law and providing redundancy in circuit analysis.

Example: Calculate the voltage Vad and Vce in the network.

The circuit is redrawn because there is no physical closed path between these points.

For Vad: Two paths exist where Vad is the only unknown:

Path adea:

• Vad - 1V - 4Vx = 0
• Given Vx = 2V
• Vad - 1 - 8 = 0
• Vad = 9 volts

Path adcba (for verification):

• Vad - 12 + 2 + 1 = 0
• Vad = 9 volts (confirms the calculation)

💡 Why this matters: When circuits lack direct connections, redrawing them into equivalent topological layouts helps identify valid KVL paths.

⭐ Key Takeaways

The lecture demonstrates three critical skills: solving simultaneous mesh equations with opposite-direction currents to find individual branch currents; verifying mesh analysis results using nodal analysis for cross-validation; and finding voltages between arbitrary points in a network by choosing appropriate KVL paths. The sign convention for currents (I₁ - I₂ vs I₂ - I₁) determines the direction but not the magnitude of power calculation. Multiple paths between two points should yield the same voltage, providing a built-in verification method. Circuit redrawing is essential when no direct closed path exists between measurement points.

🧠 Quick Revision Questions

1. In the first example, why does the direction of subtraction (I₁-I₂ or I₂-I₁) not matter when calculating power?
2. What are the two mesh equations derived for the 3Ω resistor power example, and what do they represent?
3. How does nodal analysis confirm the results from mesh analysis in the first example?
4. Why must Vcf equal -Vfc, and how does this relationship help verify calculations?
5. In the third example, what are the two paths used to calculate Vad, and why do both give the same answer?

📘 Lecture 15 — Applications of Loop Analysis

📖 Overview: This lecture demonstrates practical applications of loop analysis (KVL) to calculate unknown voltages across various points in circuits. The examples show how to systematically apply Kirchhoff's Voltage Law to different paths in a circuit, even when no physical closed path exists between two points, and how to use mesh current analysis to find specific voltages.

🗂️ Topics Covered

The lecture covers three detailed examples of applying loop analysis techniques. The first example calculates voltages Vad, Vac, and Vbd in a simple circuit with multiple voltage sources. The second example extends the technique to find Vac and Vdb across different branches. The third example demonstrates mesh current analysis with a current source to calculate V0 across a resistor combination.

📝 Lecture Summary

Example: Calculate the voltages V_ad, V_ac and V_bd.

This example demonstrates calculating voltages between points that may not have a direct physical path between them. The circuit contains voltage sources of 12V, 4V, 6V, and 2V arranged in a rectangular configuration.

To find V_ad, we observe there is only a voltage source between points a and d, so V_ad = 12 volts directly.

For V_ac, there is no physical closed path between points a and c. We consider two possible closed paths:

• Path acba: KVL equation → V_ac - 4 - 6 = 0 → V_ac = 10 Volts
• Path acda: KVL equation → V_ac + 2 - 12 = 0 → V_ac = 10 Volts

Both paths give the same result, verifying the calculation.

🔑 Definition — V_db: The voltage from point d to point b, which is the negative of V_bd (V_db = -V_bd).

💡 Why this matters: When calculating voltages between arbitrary points in a circuit, we can use any closed path that includes both points, as long as we maintain consistent sign conventions.

Example: Calculate the voltages V_ac and V_db.

This example extends the technique to a more complex circuit with voltage sources of 10V, 12V, 6V, and 8V arranged in a rectangular configuration with a diagonal branch.

For V_ac, the two paths used are:

• Path acda: V_ac + 8 - 12 = 0 → V_ac = 4 volts
• Path acba: V_ac + 6 - 10 = 0 → V_ac = 4 volts

For V_bd, the two paths are:

• Path bdcb: V_bd - 8 + 6 = 0 → V_bd = 2 volts, therefore V_db = -2 volts
• Path bdab: V_bd - 12 + 10 = 0 → V_bd = 2 volts, therefore V_db = -2 volts

📐 Formula — Voltage sign convention: When traversing a voltage source from negative to positive terminal, add its voltage; when going from positive to negative, subtract its voltage.

📌 Example: In path bdcb: Starting at b, go to d (unknown V_bd), then to c (-8V because we encounter negative terminal first), then to b (+6V because we encounter positive terminal first). Thus: V_bd - 8 + 6 = 0 → V_bd = 2V.

Example: Calculate the voltage V_0.

This example introduces mesh current analysis combined with a practical application to find the voltage V0 across specific components.

The circuit contains:

• A 12V voltage source in series with a 3kΩ resistor
• A 6kΩ resistor between mesh 1 and mesh 2
• A 2mA current source in mesh 2
• A 2kΩ resistor in mesh 2

Given I₂ = 2 mA (from the current source), we write the KVL equation for mesh 1:

3k(I₁) + 6k(I₁ - I₂) - 12 = 0

9k(I₁) - 6k(I₂) = 12

9k(I₁) - 6k(2mA) = 12

9k(I₁) - 12 = 12

9k(I₁) = 24

I₁ = 8/3 mA

Voltage across 6kΩ resistor: V = 6k(I₁ - I₂) = 6k(8/3 - 2) = 6k(8/3 - 6/3) = 6k(2/3) = 4 volts

Voltage across 2kΩ resistor: V₂ₖ = 2k × I₂ = 2k × 2mA = 4 volts

V₀ = V₆ₖ - V₂ₖ = 4 - 4 = 0 volts

📐 Formula — Mesh current: For a resistor shared between two meshes, the voltage drop = R × (I_mesh1 - I_mesh2), where the direction is determined by the mesh under analysis.

🔑 Definition — Supermesh: When a current source is present between two meshes, we can combine those meshes into a supermesh to avoid having to write KVL through the current source. However, in this example, the current source is on the outer edge of mesh 2, so it directly determines I₂.

💡 Why this matters: The result V₀ = 0 volts shows that the voltage drops across the 6kΩ and 2kΩ resistors exactly cancel each other, demonstrating that different circuit paths can produce equal and opposite voltages.

⭐ Key Takeaways

The most critical concepts to remember are: (1) When calculating voltage between two points without a direct connection, you can use any closed path (KVL loop) that contains both points — the result will be the same regardless of the path chosen, which serves as a verification method. (2) Voltage polarity matters: V_db = -V_bd, so always pay attention to the order of subscripts and the direction of traversal. (3) In mesh analysis with current sources, the current source directly determines the mesh current in that mesh (if it's on the outer edge) or creates a constraint between meshes. (4) The voltage across a shared resistor between two meshes is R times the difference of the mesh currents. (5) Always verify your results by using at least two different paths or methods — if they agree, your answer is likely correct.

🧠 Quick Revision Questions

1. In the first example, what two different paths were used to verify V_ac = 10 volts?
2. Why does V_db equal -V_bd, and what does this tell us about voltage polarity conventions?
3. In the third example, how did the 2mA current source determine I₂ directly, and what effect did this have on setting up the mesh equations?
4. What formula is used to find the voltage across the 6kΩ resistor that is shared between meshes 1 and 2?
5. Why did V₀ calculate to 0 volts in the third example, and what does this result tell us about the circuit's behavior?

📘 Lecture 16 — Applications of Loop Analysis

📖 Overview: This lecture demonstrates practical applications of Kirchhoff's Voltage Law (KVL) and loop analysis to solve for unknown voltages and currents in various circuit configurations. Through multiple worked examples, students learn how to set up mesh equations, handle current sources within meshes, and compute specific circuit parameters like voltage across components and branch currents.

🗂️ Topics Covered

The lecture covers six complete examples applying loop analysis techniques. The first example calculates voltage across specific nodes using a single loop circuit. The second example finds voltage between two nodes by first computing voltage across a resistor. The third example calculates current through a shared branch in a two-mesh circuit with a known current source. The fourth example solves for both unknown current and voltage in a two-mesh configuration. The final two examples demonstrate loop analysis in circuits containing independent current sources, showing how to handle mesh currents that are directly determined by source values.

📝 Lecture Summary

Example: Calculate the voltage Vac

We want to calculate the voltage across nodes a and c. To solve this problem, the circuit is first redrawn to clearly show the loop.

Let the current I₁ flow through the circuit. Applying KVL: 10kI₁ + 20kI₁ + 30kI₁ - 6 = 0 10kI₁ + 20kI₁ + 30kI₁ = 6 60kI₁ = 6 I₁ = 6/60k I₁ = 0.1mA

Now Vac will be equal to the voltage across the 10kΩ and 20kΩ resistors combined: Vac = (10k + 20k)I₁ Vac = 30k(0.1mA) Vac = 3 Volts

Example: Calculate the voltage Vbd

We want to calculate the voltage between nodes b and d. First, we calculate the voltage across the 40kΩ resistor.

Applying KVL to the single loop: 10kI₁ + 9 + 40kI₁ + 10kI₁ - 6 = 0 60kI₁ = -3 I₁ = -3/60k I₁ = -0.05 mA

So, voltage across the 40kΩ resistor: V₄₀ₖ = (-0.05m)(40k) V₄₀ₖ = -2 volts — (A)

To calculate Vbd, we redraw the circuit and take the path b-d-c-b. Applying KVL along this path: Vbd - V₄₀ₖ - 9 = 0 Substituting the value of V₄₀ₖ from equation (A): Vbd - (-2) - 9 = 0 Vbd + 2 - 9 = 0 Vbd = 7 Volts

Example: Calculate the current I₀

We want to calculate the current through the 4kΩ resistor (I₀). The circuit is redrawn to identify two meshes.

For Mesh 1: I₁ = 120 mA (determined by the current source)

Applying KVL to Mesh 2: 8kI₂ + 4kI₂ + 4k(I₂ - I₁) = 0 16kI₂ - 4kI₁ = 0 16kI₂ - 4k(120m) = 0 16kI₂ = 480 I₂ = 480/16k I₂ = 30 mA

So I₀ will be the current through the 4kΩ resistor shared by both meshes: I₀ = I₁ - I₂ I₀ = 120 - 30 I₀ = 90 mA

Example: Calculate the current I₀ and the voltage V₀

We want to calculate both the unknown current and voltage. The circuit is redrawn to show two meshes.

Applying KVL to Mesh 1: 2kI₁ + 6k(I₁ - I₂) = 12 8kI₁ - 6kI₂ = 12 — (Equation 1)

Applying KVL to Mesh 2: 8kI₂ + 4kI₂ + 6k(I₂ - I₁) = 0 18kI₂ - 6kI₁ = 0 — (Equation 2)

Solving equations of mesh 1 and mesh 2 by multiplying Equation 1 by 3: 24kI₁ - 18kI₂ = 36 — (Equation 1 × 3) -24kI₁ + 72kI₂ = 0 — (Equation 2 × 4) Adding: 54kI₂ = 36 I₂ = 36/54k I₂ = 0.67mA — (A)

Now V₀ will be equal to the voltage across the 4kΩ resistor: V₀ = 4k × I₂ Substituting the value of I₂ from equation (A): V₀ = 4k × 0.67m V₀ = 2.66 Volts

Example: Calculate the current I₀

We want to calculate the current I₀. The circuit contains two independent current sources.

Here: I₁ = -2 mA (negative because the assumed direction of I₁ is opposite to the independent current source direction) I₃ = 4 mA (determined by the current source)

Applying KVL to Mesh 2: 4kI₂ + 6k(I₂ - I₃) + 2k(I₂ - I₁) = 12 4kI₂ + 6kI₂ - 6k(4m) + 2kI₂ - 2k(-2m) = 12 12kI₂ - 24 + 4 = 12 12kI₂ - 20 = 12 12kI₂ = 32 I₂ = 32/12k I₂ = 8/3 mA I₂ = 2.66 mA

I₀ is the same as mesh current I₂: I₀ = 2.66 mA

🔑 Definition — Mesh current: The current that flows around a complete loop (mesh) in a circuit, used as the variable in loop analysis.

Example: Calculate the current I₀

We want to calculate the current I₀ in this circuit configuration. The circuit is redrawn to identify three meshes.

Here: I₁ = -2 mA (opposite direction to current source) I₃ = 4 mA (determined by current source)

Applying KVL to Mesh 2: 2kI₂ + 1k(I₂ - I₁) = 12 2kI₂ + 1kI₂ - 1k(-2m) = 12 3kI₂ + 2 = 12 3kI₂ = 10 I₂ = 10/3 mA I₂ = 3.33 mA

I₀ is the current through the shared branch between meshes 1 and 2: I₀ = I₁ - I₂ I₀ = -2 - 3.33 I₀ = -5.33 mA

💡 Why this matters: The negative sign indicates that the actual direction of I₀ is opposite to the assumed direction. The magnitude tells us the current value, while the sign tells us the direction.

⭐ Key Takeaways

When using loop analysis with independent current sources, mesh currents may be directly determined by these sources, significantly simplifying the problem by reducing the number of unknown variables. Remember that when a current source is shared between two meshes, you must account for the current flow direction carefully — a negative mesh current simply means the actual current flows opposite to your assumed direction. For multi-mesh circuits, always set up KVL equations systematically, paying special attention to voltage drops across resistors shared by adjacent meshes (these involve current differences). When calculating voltage between two nodes, you can take any convenient path between those nodes and apply KVL, as long as all component voltages along that path are known. Finally, in single-loop circuits, once you find the loop current, you can directly compute any voltage using Ohm's law across the relevant resistors.

🧠 Quick Revision Questions

1. In the voltage Vbd example, why did the voltage across the 40kΩ resistor come out negative, and how did this affect the final calculation of Vbd?

2. When a mesh contains an independent current source, how do you determine that mesh current, and what special consideration is needed for the direction?

3. In the last example, why was I₀ calculated as I₁ - I₂ rather than I₂ - I₁, and what does the negative result tell you?

4. For the example calculating both I₀ and V₀, explain the steps to solve the two simultaneous equations derived from KVL.

5. In the current I₀ example with 120mA source, how was the current through the 4kΩ resistor determined from the two mesh currents?

📘 Lecture 17 — Applications of Loop Analysis

📖 Overview: This lecture demonstrates the practical application of mesh (loop) analysis to solve circuits containing dependent sources and multiple meshes. Through three detailed examples, students learn how to set up and solve mesh equations systematically to find unknown voltages and currents.

🗂️ Topics Covered

Three example problems are solved in detail: the first example calculates voltage V₀ using two mesh equations solved simultaneously, the second example involves a dependent source requiring substitution of the controlling variable, and the third example features a dependent voltage source where mesh currents simplify significantly. Each example reinforces the mesh analysis methodology.

📝 Lecture Summary

Example: Calculate the voltage V₀

The circuit has three meshes. Mesh 1 has a known current source of 2mA directly assigned. For mesh 2, the KVL equation includes the 6V source and resistors 4kΩ and 2kΩ: 4k(I₂ – I₃) + 2k(I₂ – I₁) = 6. With I₁ = 2mA, this simplifies to 6kI₂ – 4kI₃ = 10.

For mesh 3, all resistors (6kΩ, 2kΩ, and 4kΩ) contribute: 6kI₃ + 2k(I₃ – I₁) + 4k(I₃ – I₂) = 0. Substituting I₁ = 2mA gives 12kI₃ – 4kI₂ = 4.

Solving the two equations simultaneously: multiply the mesh 2 equation by 4 and mesh 3 equation by 6, then subtract to eliminate I₂. This yields 56kI₃ = 64, so I₃ = 1.142mA.

The output voltage V₀ is across the 6kΩ resistor: V₀ = 6k × I₃ = 6k × 1.142mA.

📌 Example: V₀ = 6.85 volts

Example: Calculate the voltage V₀

This circuit contains a dependent voltage source (4Va) in mesh 2. The circuit is redrawn to identify meshes clearly. For mesh 1: Va + 4k(I₁ – I₂) = 12. For mesh 2: 4Va + 6kI₂ + 4k(I₂ – I₁) = 0.

The controlling variable Va is the voltage across the 2kΩ resistor in mesh 1: Va = 2kI₁.

Substituting Va = 2kI₁ into the mesh 1 equation: 2kI₁ + 4kI₁ – 4kI₂ = 12 → 6kI₁ – 4kI₂ = 12. Substituting into mesh 2: 8kI₁ + 10kI₂ – 4kI₁ = 0 → 4kI₁ + 10kI₂ = 0.

Solving: multiply the first equation by 1.5 → 9kI₁ – 6kI₂ = 18. Subtract the second equation (4kI₁ + 10kI₂ = 0) from this. The system is solved by elimination: from 4kI₁ + 10kI₂ = 0, I₁ = -2.5I₂. Substituting into 6k(-2.5I₂) – 4kI₂ = 12 gives -19kI₂ = 12.

🔑 Definition — I₂: I₂ = -12/19 mA = -0.6315mA

The output voltage V₀ is across the 6kΩ resistor: V₀ = 6k × I₂ = 6k × 0.6315mA.

📌 Example: V₀ = 3.78 volts

💡 Why this matters: When dependent sources appear, the controlling variable must be expressed in terms of mesh currents and substituted into the equations — this creates a system that can still be solved with standard linear algebra.

Example: Calculate the voltage V₀

This circuit has a dependent voltage source (2kIₓ) in series with a 2kΩ resistor. The circuit is redrawn to show two meshes. For mesh 1: 2kI₁ + 4k(I₁ – Iₓ) – 2kIₓ = 0. Note the dependent source subtracts because of polarity.

Simplifying mesh 1: 6kI₁ – 4kIₓ – 2kIₓ = 0 → 6kI₁ – 6kIₓ = 0 → I₁ = Iₓ.

For mesh 2: 2kIₓ + 4k(Iₓ – I₁) – 6 = 0. Since I₁ = Iₓ, this simplifies to 2kIₓ + 4k(Iₓ – Iₓ) – 6 = 0 → 2kIₓ – 6 = 0.

Solving: 2kIₓ = 6 → Iₓ = 3mA.

The output voltage V₀ is across the 2kΩ resistor in mesh 2: V₀ = 2k × Iₓ = 2k × 3mA.

📌 Example: V₀ = 6 Volts

💡 Why this matters: When dependent sources cause mesh currents to become equal, the circuit simplifies dramatically — always check for such relationships before solving the full system.

⭐ Key Takeaways

Mesh (loop) analysis is a systematic method for solving circuits with multiple loops. When a current source is present in a mesh, its value can be directly assigned as the mesh current. For dependent sources, the controlling variable must be expressed in terms of mesh currents and substituted into the KVL equations. Solving two simultaneous equations can be done by elimination or substitution. The output voltage is always found by multiplying the appropriate mesh current by the resistance across which it is measured. Always verify current directions and source polarities before writing KVL equations.

🧠 Quick Revision Questions

1. In the first example, why was I₁ directly set to 2mA without writing a KVL equation for mesh 1?
2. In the second example, how was the dependent source voltage 4Va expressed in terms of mesh currents?
3. In the third example, what relationship between I₁ and Iₓ simplified the circuit solution?
4. What is the general procedure when a dependent source appears in a mesh analysis problem?
5. If the mesh equations 6kI₁ – 4kI₂ = 12 and 4kI₁ + 10kI₂ = 0 are given, solve for I₂ using elimination.

📘 Lecture 18 — Applications of Loop Analysis-Super Mesh Technique

📖 Overview: This lecture focuses on applying loop analysis and the super mesh technique to solve complex circuits. It demonstrates how to handle circuits with dependent sources, current sources, and multiple meshes, showing that loop analysis often simplifies calculations compared to nodal analysis.

🗂️ Topics Covered

The lecture covers five detailed examples demonstrating loop analysis and super mesh technique applications. Examples include calculating voltage across resistors using mesh currents with dependent sources, solving circuits with voltage-controlled voltage sources, handling circuits with multiple current sources using super mesh approach, and determining currents through specific branches using simplified mesh analysis.

📝 Lecture Summary

Example: Calculate the voltage V₀

The circuit contains a dependent voltage source Vx and requires solving for output voltage V₀ across a 6kΩ resistor.

First, the circuit is redrawn to identify three meshes. The relationships between mesh currents are established using the dependent source constraints.

🔑 Definition — Vx: The voltage across the 6kΩ resistor between meshes 2 and 3, given by Vx = (I₂ - I₃)6kΩ.

From the dependent source connection: I₁ = Vx/2kΩ. Substituting Vx gives I₁ = 3(I₂ - I₃).

Given that I₂ = 2mA, this simplifies to I₁ = 6mA - 3I₃.

📐 Formula — KVL for Mesh 3: 6kΩ I₃ + 6kΩ(I₃ - I₂) + 2kΩ(I₃ - I₁) = 0

Substituting values: 6kI₃ + 6kI₃ - 12 + 2kI₃ - 2k(6mA - 3I₃) = 0

Simplifying: 14kI₃ - 12 - 12 + 6kI₃ = 0 → 20kI₃ = 24

📌 Example: I₃ = 24/20k = 1.2mA, therefore V₀ = 6kΩ × 1.2mA = 7.2 volts

Example: Calculate the voltage Vₐ

This circuit contains a voltage-controlled voltage source 2Vₐ. The problem notes that while super node technique could solve this, loop analysis is more efficient.

The circuit has four meshes with I₄ = 1A given directly.

For mesh 1: 1Ω(I₁ - I₂) + 3Ω(I₁ - I₃) - 5 = 0 → 4I₁ - 3I₃ - I₂ = 5 ... Equation (A)

For mesh 2: 2I₂ + 3I₂ - 2Vₐ + I₂ - I₁ = 0 → 6I₂ - I₁ = 2Vₐ ... Equation (B)

🔑 Definition — Vₐ: The voltage across the 3Ω resistor in mesh 2, given by Vₐ = 3I₂

Substituting Vₐ into equation (B): 6I₂ - I₁ = 2(3I₂) → I₁ = 0

For mesh 3: 2Vₐ + 5(I₃ - I₄) + 3(I₃ - I₁) = 0

Substituting I₁ = 0, I₄ = 1A, and Vₐ = 3I₂: 6I₂ + 3I₃ + 5I₃ - 5 = 0 → 8I₃ + 6I₂ = 5

Solving equations (A) and mesh 3 equation simultaneously:

• From (A) with I₁ = 0: -3I₃ - I₂ = 5 → multiply by 6: -18I₃ - 6I₂ = 30
• From mesh 3: 8I₃ + 6I₂ = 5

Adding: -10I₃ = 35 → I₃ = -3.5A

📌 Example: Substituting I₃ into equation (A): 10.5 - I₂ = 5 → I₂ = 5.5A

Therefore Vₐ = 3 × 5.5 = 16.5 volts

Example: Find Current through all meshes

This example demonstrates the super mesh technique for a circuit containing a current source between two meshes.

The instructor first shows an incorrect approach and then the correct super mesh method.

The circuit has four meshes with the following known currents from current sources: I₂ = -5A and I₃ = +5A.

🔑 Definition — Super Mesh: When a current source exists between two meshes, we combine those meshes into a single "super mesh" and apply KVL around the combined loop, while using the current source to relate the mesh currents.

KVL for mesh 1: -5 + I₁ - I₃ + 2I₁ = 0 ... Equation (1)

KVL for super mesh (combining meshes 2, 3, and 4): 1Ω(I₃ - I₁) + 4Ω I₂ + 3Ω I₂ + 2 + 4Ω I₃ - 4Ω I₄ + 6Ω I₃ = 0 ... Equation (2)

KVL for mesh 4: 4Ω I₄ - 4Ω I₃ + 2Ω I₄ - 3 = 0 ... Equation (3)

Simplifying:

• Equation (1): 3I₁ - I₃ = 5
• Equation (2): -I₁ + 7I₂ + 11I₃ - 4I₄ = -2
• Equation (3): -4I₃ + 6I₄ = 3
• From current source: -I₂ + I₃ = 5

📌 Example: Solving the system yields: I₁ = 2.481A, I₂ = -2.556A, I₃ = 2.444A, I₄ = 2.130A

Example: Calculate current I₀

The circuit contains two current sources that directly set two mesh currents. This simplifies the analysis significantly.

I₁ = -1mA and I₂ = -2mA are given directly from the current sources.

KVL for loop 3: 1kΩ(I₃ + I₁) + 2kΩ I₃ + 1kΩ(I₃ + I₂) = 2 + 4

Expanding: 1kI₃ + 1kI₁ + 2kI₃ + 1kI₃ + 1kI₂ = 6

Simplifying: 4kI₃ - 1 - 2 = 6 → 4kI₃ - 3 = 6 → 4kI₃ = 9

📌 Example: I₃ = 9/4 mA = 2.25mA. Since I₀ = I₃, the answer is I₀ = 2.25mA

Example: Calculate current I₀

Similar to the previous example, current sources directly define two mesh currents.

I₁ = 2mA and I₂ = -4mA from the current sources.

KVL for loop 3: 2kΩ I₃ + 2kΩ(I₃ + I₂) = 12

Expanding: 2kI₃ + 2kI₃ + 2kI₂ = 12 → 4kI₃ - 8 = 12 → 4kI₃ = 20

📌 Example: I₃ = 5mA. Note that I₀ = I₃ + I₁ = 5mA + 2mA = 7mA

💡 Why this matters: This example shows that the desired current may be a combination of mesh currents, not just a single mesh current.

⭐ Key Takeaways

The super mesh technique is essential when a current source lies between two meshes, as it allows us to create a combined loop for KVL while using the current source relation between meshes. Dependent sources require expressing the controlling variable in terms of mesh currents before solving the system of equations. Loop analysis often simplifies circuits with many voltage sources compared to nodal analysis, especially when current sources directly define some mesh currents. When writing mesh equations, pay careful attention to the sign conventions and ensure all terms are collected properly. The desired output current or voltage may be a combination of multiple mesh currents, not simply one mesh current.

🧠 Quick Revision Questions

1. How do you handle a current source that exists between two meshes in loop analysis?
2. In the Vₐ example, why did I₁ become zero, and how did this simplify the solution?
3. What is the relationship between I₁ and the dependent source Vx in the first example?
4. In the last example, why was I₀ equal to I₃ + I₁ instead of just I₃?
5. When using the super mesh technique, how many KVL equations do you write for the combined super mesh?

📘 Lecture 19 — (Examples of Loop Analysis)

📖 Overview: This lecture demonstrates the practical application of loop analysis (mesh analysis) to solve for unknown currents and voltages in circuits containing multiple current sources. It shows how to systematically assign loop currents, apply Kirchhoff’s Voltage Law (KVL), and solve simultaneous equations to find the desired quantities, reinforcing the method through three worked examples.

🗂️ Topics Covered

This lecture covers three detailed examples of loop analysis. The first example calculates the current (I_o) in a circuit with three loops and two independent current sources. The second example calculates the voltage (V_o) across a resistor in a four-loop circuit. The third example calculates the voltage (V_o) in a more complex four-loop circuit with three current sources. Each example demonstrates the step-by-step process of redrawing the circuit, identifying known mesh currents from current sources, applying KVL to the unknown loop, and solving for the required value.

📝 Lecture Summary

Example: Calculate the current (I_o).

The objective is to find the current (I_o) through a specific branch. The circuit is redrawn to clearly define loop currents. In this configuration, two current sources directly determine two loop currents: (I_1 = 2,mA) and (I_2 = -4,mA). The negative sign indicates the direction of the source is opposite to the assumed loop current direction.

The Kirchhoff’s Voltage Law (KVL) equation is written for the third loop (loop 3), which does not have a current source. The sum of voltage drops around loop 3 is set equal to the voltage source (+12V). The equation is: [ 2k\Omega I_3 + 1k\Omega (I_3 - I_2) + 2k\Omega (I_3 + I_1 - I_2) = 12V ]

🔑 Definition — Loop Analysis: A method that uses loop (mesh) currents as unknowns and applies KVL to each independent loop to form a set of simultaneous equations.

📐 Formula: The KVL equation for loop 3 is simplified by combining like terms: (5k\Omega I_3 - 3k\Omega I_2 + 2k\Omega I_1 = 12V)

📌 Example: Substitute (I_1 = 2,mA) and (I_2 = -4,mA) into the simplified equation: (5k\Omega I_3 - 3k\Omega(-4m) + 2k\Omega(2m) = 12V) → (5k\Omega I_3 + 12 + 4 = 12) → (5k\Omega I_3 = -4) → (I_3 = -0.8,mA) The required current (I_o) is the sum of loop currents flowing through the output branch: (I_o = I_3 + I_1 - I_2 = -0.8 + 2 - (-4)) → (I_o = 5.2,mA).

💡 Why this matters: This example shows how to handle independent current sources in loop analysis. They directly set the loop current, simplifying the system of equations.

Example: Calculate the voltage (V_o).

The goal is to find the voltage (V_o) across a 1k(\Omega) resistor. The circuit is redrawn to show four clear loop currents. Three current sources define the first three loop currents directly: (I_1 = -2,mA), (I_2 = -4,mA), and (I_3 = 4,mA).

KVL is applied to loop 4, which does not contain a current source. The sum of voltage drops across the resistors in loop 4 equals zero (no voltage source in this loop). The equation is: [ 1k\Omega (I_4 - I_1) + 1k\Omega (I_4 - I_2) + 1k\Omega I_3 + 2k\Omega I_4 = 0 ]

🔑 Definition — Voltage across a resistor: Given by Ohm’s Law, (V = IR), where (I) is the net current flowing through the resistor.

📐 Formula: Simplifying the KVL equation for loop 4: (4k\Omega I_4 + 1k\Omega I_3 - 1k\Omega I_1 - 1k\Omega I_2 = 0)

📌 Example: Substitute the known currents: (4k\Omega I_4 + 1k(4m) - 1k(-2m) - 1k(-4m) = 0) → (4k\Omega I_4 + 4 + 2 + 4 = 0) → (4k\Omega I_4 = -10) → (I_4 = -2.5,mA). The current through the 1k(\Omega) resistor where (V_o) is measured is (I_{1k} = I_4 - I_1 = -2.5 - (-2) = -0.5,mA). Therefore, (V_o = 1k\Omega \times 0.5,mA = 0.5,V). Note: The magnitude is used for the voltage value.

💡 Why this matters: This example demonstrates how to calculate a specific voltage by first finding its branch current using loop analysis, and then applying Ohm's Law.

Example: Calculate the voltage (V_o).

The circuit is redrawn to identify loop currents. Three current sources directly set three loop currents: (I_1 = 4,mA), (I_2 = -2,mA), and (I_3 = 1,mA).

KVL is applied to loop 4. The resistors in this loop have voltages that sum to zero (no independent voltage source). The equation is: [ 1k\Omega (I_4 - I_1) + 1k\Omega (I_4 - I_2 + I_3) + 2k\Omega (I_4 + I_3) + 1k\Omega I_4 = 0 ]

🔑 Definition — Supermesh: When a current source is common to two loops, a supermesh is formed by combining the two loops and applying KVL. This example uses the standard approach where the source sets one current.

📐 Formula: Simplifying the KVL equation for loop 4: (5k\Omega I_4 - 1k\Omega I_1 - 1k\Omega I_2 + 1k\Omega I_3 + 2k\Omega I_3 = 0) → (5k\Omega I_4 - 4V + 2V + 1V + 2V = 0)

📌 Example: Substitute known values: (5k\Omega I_4 - 4 + 2 + 1 + 2 = 0) → (5k\Omega I_4 = -1) → (I_4 = -0.2,mA). The voltage (V_o) is across the 2k(\Omega) resistor, and the net current through it is (I_4 + I_3). So, (V_o = 2k\Omega , (I_4 + I_3) = 2k\Omega (-0.2m + 1m)) → (V_o = 2k\Omega \times 0.8,mA = 1.6,V).

💡 Why this matters: This example illustrates how loop currents can be combined to find the net current through a specific component for voltage calculation.

⭐ Key Takeaways

The core skill reinforced in this lecture is the systematic application of loop analysis to circuits with multiple current sources. First, identify all independent loops. Second, note that a current source that is alone in a loop directly determines that loop's current, often with a sign change. Third, for loops without a source, write a KVL equation summing all voltage drops to zero or the source voltage. Fourth, carefully account for all current contributions from neighboring loops when writing the voltage drop across a shared resistor. Finally, to find a specific branch current or voltage, combine the relevant loop currents with proper signs and apply Ohm’s Law.

🧠 Quick Revision Questions

1. In loop analysis, what is the first step when a current source is present in a single loop?
2. In Example 1, why is the equation (I_o = I_3 + I_1 - I_2) used to find the final current?
3. In Example 2, what is the value of (V_o) and how is it calculated after finding (I_4) and (I_1)?
4. In Example 3, why is the current through the 2k(\Omega) resistor (I_4 + I_3) and not just (I_4)?
5. State the general rule for determining the sign of a loop current when a current source points opposite to the assumed loop direction.

📘 Lecture 20 — (Super Mesh Technique - Coupling equation)

📖 Overview: This lecture introduces the super mesh technique, a powerful method for analyzing circuits that contain current sources shared between two meshes. It also covers the coupling equation used to relate mesh currents when dependent sources are present. Multiple worked examples demonstrate how to apply these techniques to find unknown voltages and currents.

🗂️ Topics Covered

The lecture covers several complete circuit analysis examples using mesh analysis. It demonstrates how to handle circuits with independent current sources (using super mesh when sources are common to two meshes) and dependent sources (using coupling equations). Each example shows the step-by-step process of setting up and solving simultaneous equations to find unknown voltages and currents.

📝 Lecture Summary

Example: Calculate the voltage Vo

We want to calculate the voltage V₀. The circuit is redrawn with mesh currents labeled. It contains a 2mA current source that establishes I₁.

Here, I₁ = -2mA (the negative sign indicates the assumed direction is opposite to the source's actual direction).

KVL equation for loop 2: 2kΩ·I₂ + 4kΩ·I₂ + 2kΩ·(I₂ + I₁) = 12 6kΩ·I₂ + 2kΩ·I₂ + 2kΩ·I₁ = 12 8kΩ·I₂ – 4 = 12 8kΩ·I₂ = 16 I₂ = 2mA

V₀ = 2kΩ·(I₂) = 2kΩ·(2mA) = 4 volts

📌 Example: This example shows a simple two-mesh circuit where one current is set by a source. Finding I₂ allows direct calculation of output voltage V₀.

Example: Calculate the voltage Vo

We want to find the voltage V₀. The circuit is redrawn with mesh currents. It contains two independent current sources that set I₁ and I₂.

Here: I₁ = 3mA I₂ = 1mA

KVL for loop 3: 4kΩ·I₃ + 2kΩ·(I₃ + I₂) + 4kΩ·(I₃ – I₁) + 2kΩ·(I₃ – I₁) = 6 4kΩ·I₃ + 2kΩ·I₃ + 2kΩ·I₂ + 4kΩ·I₃ – 4kΩ·I₁ + 2kΩ·I₃ – 2kΩ·I₁ = 6 12kΩ·I₃ + 2 – 12 – 6 = 12 12kΩ·I₃ = 22 I₃ = 11/6 mA = 1.833mA

V₀ = 4kΩ·I₃ = 4kΩ·1.833mA = 7.33 volts

📌 Example: With two currents set by sources, only one unknown mesh current remains. Solving the single KVL equation gives V₀.

Example: Calculate the current Io

We want to find the current I₀. The circuit is redrawn with mesh currents. It contains two independent current sources and requires solving a system of simultaneous equations.

Here: I₁ = -2mA I₃ = 1mA

KVL for mesh 2: 4kΩ·(I₂ + I₄) + 2kΩ·(I₂ – I₃) + 2kΩ·(I₂ – I₁) = 0 8kΩ·I₂ + 4kΩ·I₄ – 2 + 4 = 0 4kΩ·I₂ + 2kΩ·I₄ = -1

KVL for loop 4: 4kΩ·(I₄ + I₂) + 2kΩ·(I₄ + I₃) + 6kΩ·I₄ – 6 = 0 12kΩ·I₄ + 4kΩ·I₂ + 2 – 6 = 0 12kΩ·I₄ + 4kΩ·I₂ – 4 = 0 3kΩ·I₄ + 1kΩ·I₂ = 1

Multiplying the mesh 2 equation by 3 and the loop 4 equation by 2, then subtracting: 12kΩ·I₂ + 6kΩ·I₄ = -3 -2kΩ·I₂ – 6kΩ·I₄ = -2 10kΩ·I₂ = -5 I₂ = -0.5mA

I₀ = I₁ – I₂ = -2mA – (-0.5mA) = -1.5mA

📌 Example: When two unknowns remain (I₂ and I₄), two simultaneous equations are needed. The solution requires algebraic manipulation to eliminate one variable.

Example: Find the currents I₁, I₂ and I₃

We want to find currents I₁, I₂, and I₃. The circuit contains a dependent voltage source, requiring a super mesh and a coupling equation.

The circuit is redrawn. For mesh 1: 4I₁ - 4I₂ + 1 = 0 ... equation (1)

For the super mesh (mesh 2 and mesh 3 combined): 3I₂ + 2I₃ + 4I₂ - 4I₁ = 0 ... equation (2)

Coupling Equation: I₃ – I₂ = 2Vx

Here, Vx = 3I₂, so: I₃ – I₂ = 2(3I₂) = 6I₂ I₃ – I₂ = 6I₂ -7I₂ + I₃ = 0 ... equation (3)

Solving equation (1): 4I₁ – 4I₂ = -1

Solving equation (2): -4I₁ + 7I₂ + 2I₃ = -1

Solving equations (1), (2), and (3): I₁ = -308.8 mA I₂ = -58.82 mA I₃ = -411.8 mA

🔑 Definition — Super Mesh: When a current source (independent or dependent) is present between two meshes, a "super mesh" is formed by combining those two meshes into one larger loop, excluding the current source and any elements in series with it.

🔑 Definition — Coupling Equation: An additional equation derived from the relationship between the current source (especially dependent sources) and the mesh currents, used alongside super mesh analysis to solve for unknown currents.

💡 Why this matters: Circuits with dependent sources cannot be solved using standard mesh analysis alone. The super mesh technique combined with coupling equations provides the complete set of equations needed for analysis.

⭐ Key Takeaways

The super mesh technique is essential when a current source is shared between two meshes — the two meshes are combined into one larger loop for KVL, while the coupling equation from the current source provides the missing relationship between mesh currents. For dependent sources, the coupling equation expresses the controlling variable in terms of mesh currents. Multiple examples demonstrate that systematic equation setup and simultaneous solving yields all unknown mesh currents, from which branch currents and voltages can be calculated. The sign convention must be carefully maintained throughout, and current sources directly set their respective mesh currents when they appear in only one mesh.

🧠 Quick Revision Questions

1. When must a super mesh be used instead of regular mesh analysis?
2. What is a coupling equation, and how is it derived for a dependent voltage source?
3. In the first example, why is I₁ negative (-2mA) rather than positive?
4. How do you handle a circuit with two independent current sources and two remaining unknown mesh currents?
5. What algebraic technique was used to solve the system of two equations in the I₀ example?

📘 Lecture 21 — Examples of Loop Analysis by using dependent sources - Coupling equation

📖 Overview: This lecture demonstrates the application of loop analysis techniques in circuits containing dependent sources. It shows how to set up KVL equations for loops and supermeshes, and introduces the coupling equation method for solving circuits with dependent sources efficiently. These techniques are essential for analyzing more complex circuits where standard mesh analysis becomes cumbersome.

🗂️ Topics Covered

The lecture covers three detailed examples of loop analysis with dependent sources. The first example calculates voltage across a resistor using KVL and the relationship between a dependent current source and its controlling voltage. The second example involves a more complex circuit with multiple dependent sources and loops. The third example demonstrates the use of the coupling equation technique to find loop currents in a circuit with a current source common to two meshes, forming a supermesh.

📝 Lecture Summary

Example: Calculate the voltage V₀.

We are given a circuit with a 6V source, resistors of 4kΩ, 2kΩ, and 2kΩ, and a dependent current source controlled by voltage V₀. The circuit is redrawn for clarity. The voltage V₀ appears across a 2kΩ resistor. The solution begins by defining loop currents I₁ and I₂.

First, the output voltage is expressed as: V₀ = 2k(I₁ + I₂).

Next, we apply KVL for loop 1, which includes the dependent source. The dependent current source I₁ is defined as 2V₀/1k. Substituting the expression for V₀ gives I₁ = 4(I₁ + I₂), which simplifies to -3I₁ = 4I₂, leading to the relationship I₁ = -4/3 I₂.

We then write KVL for loop 2: 4kI₂ + 4k(I₁ + I₂) + 2k(I₁ + I₂) = 6. This simplifies to 10kI₂ + 6kI₁ = 6, or 5kI₂ + 3kI₁ = 3. Substituting I₁ = -4/3 I₂ gives 5kI₂ – 4kI₂ = 3, so I₂ = 3mA.

Consequently, I₁ = -4mA. Substituting these into the expression for V₀: V₀ = 2k(-4m + 3m) = -2 volts.

🔑 Definition — Loop Analysis: A circuit analysis method that uses Kirchhoff's Voltage Law (KVL) to determine unknown currents in independent loops of a circuit. 📐 Formula: V₀ = 2k(I₁ + I₂) → The voltage across a resistor is the product of its resistance and the sum of currents flowing through it. 📌 Example: With I₁ = -4mA and I₂ = 3mA, V₀ = 2kΩ * (-4mA + 3mA) = 2kΩ * (-1mA) = -2V.

Example: Calculate the voltage V₀.

This circuit is more complex. It contains a 2V source, a 1V source, and resistors of 1kΩ, plus dependent sources. We define loop currents I₁, I₂, I₃, and I₄.

First, from the circuit we find I₂ = -4mA and I₃ = -1mA.

The voltage Vx is defined across two branches. We use the expression Vx = -(I₃ + I₄)1k.

The dependent source I₁ is given by 2Vx/1k. Substituting Vx gives I₁ = -2(I₃ + I₄). Since I₃ = -1mA, this becomes I₁ = 2 - 2I₄.

Now we apply KVL for loop 4. This is a complex loop equation: -1k(I₄ + I₃) + 1k(I₄ + I₃) + 1k(I₄ + I₃ – I₂) + 1k(I₄ - I₁) + 2kI₄ = 0. Simplifying and substituting known values (I₂ = -4mA, I₃ = -1mA, I₁ = 2 - 2I₄) yields: 2kI₄ – 2 -1kI₄ +1 +4 +2kI₄ +1k(I₄ -2 +2I₄) = 0, which simplifies to 6kI₄ + 1 = 0.

Solving gives I₄ = -1/6k A.

Finally, V₀ = 1k(I₄ + I₃) = 1k(-1/6k – 1m) = 1k(-0.166m – 1m) = -1.166 volts.

📌 Example: With I₄ = -0.166mA and I₃ = -1mA, V₀ = 1kΩ * (-0.166mA – 1mA) = 1kΩ * (-1.166mA) = -1.166V.

Example: Find loop currents.

This example introduces the coupling equation technique. The circuit has a 7V source, a 1A current source, and resistors of 1Ω, 2Ω, and 3Ω.

Because the 1A current source is shared between mesh 1 and mesh 3, they form a supermesh. The KVL equation for this supermesh is: -7 + 1(I₁ – I₂) + 3(I₃ – I₂) + 1I₃ = 0, which simplifies to I₁ - 4I₂ + 4I₃ = 7 (Equation 1).

The KVL equation for mesh 2 is: 1(I₂ – I₁) + 2I₂ + 3(I₂ – I₃) = 0, which simplifies to -I₁ + 6I₂ - 3I₃ = 0 (Equation 2).

The coupling equation defines the relationship between the loop currents at the shared current source. Since the 1A source is oriented from mesh 1 to mesh 3, the coupling equation is I₃ – I₁ = 1 (Equation 3), but the lecture uses it in the form I₁ – I₃ = -1 which when rearranged gives I₁ – I₃ = 7... Wait, the lecture text shows a different step. Let's follow the lecture's solution.

The lecture states the coupling equation as I₁ – I₃ = 7 (Equation 3, based on a different source value). Then, to solve, we manipulate Equations 1 and 2. First, multiply Equation 1 by 6 and Equation 2 by 4: 6I₁ - 24I₂ + 24I₃ = 42 (Equation A) -4I₁ + 24I₂ - 12I₃ = 0 (Equation B)

Adding Equations A and B gives 2I₁ + 12I₃ = 42 (Equation 4).

Now we have two equations: I₁ – I₃ = 7 (Equation 3) and 2I₁ + 12I₃ = 42 (Equation 4). To solve, multiply Equation 3 by 2: 2I₁ – 2I₃ = 14. Subtract this from Equation 4: (2I₁ + 12I₃) – (2I₁ – 2I₃) = 42 – 14 → 14I₃ = 28, so I₃ = 2A.

Substituting I₃ = 2A into Equation 3: I₁ – 2 = 7, so I₁ = 9A.

Finally, substitute I₁ and I₃ into Equation 1 to find I₂: 9 – 4I₂ + 8 = 7 → -4I₂ = -10, so I₂ = 2.5A.

🔑 Definition — Supermesh: A larger loop created when a current source is common to two meshes, allowing KVL to be applied around the outer boundary that excludes the current source. 🔑 Definition — Coupling Equation: An equation that expresses the relationship between mesh currents at a shared current source, typically current = I₁ – I₂ (or I₂ – I₁), depending on direction. 📌 Example: For a 1A source between meshes 1 and 3, the coupling equation is I₃ – I₁ = 1 (if the source direction matches the current direction from mesh 1 to mesh 3).

⭐ Key Takeaways

The critical skills from this lecture are: (1) how to express a dependent source's value in terms of loop currents and then substitute into KVL equations; (2) how to handle supermeshes by writing KVL around the outer loop that bypasses the shared current source; (3) how to write the coupling equation to relate the loop currents at a shared current source; and (4) how to systematically solve the resulting system of linear equations for all loop currents. These techniques are essential for analyzing any circuit with multiple independent and dependent sources.

🧠 Quick Revision Questions

1. In the first example, what was the relationship derived between I₁ and I₂ from the dependent source definition?
2. In the second example, what was the value of I₄, and how was V₀ calculated from it?
3. In the third example, why was a supermesh necessary, and what was the resulting KVL equation?
4. What is the general form of a coupling equation for a current source of value 'I' shared between mesh X (with current Iₓ) and mesh Y (with current Iᵧ), with the source arrow pointing from mesh X to mesh Y?
5. If you have solved for I₁ and I₃ in a supermesh problem, how do you find I₂, the current in the branch that contains the dependent and independent elements?

📘 Lecture 22 — (Matrices and determinants)

📖 Overview: This lecture introduces matrices and determinants as mathematical tools for solving simultaneous equations in circuit analysis. It demonstrates how matrix methods simplify the solution of node voltages and branch currents in complex circuits using KVL and KCL, with practical examples showing the complete solution process.

🗂️ Topics Covered

The lecture covers the fundamental concepts of matrices and determinants, including matrix notation, characteristic matrices, and determinant calculation. It then applies these concepts to solve circuit equations through matrix inversion, demonstrating the process with two complete examples involving node voltage analysis and current calculations using given circuit parameters.

📝 Lecture Summary

Matrices and Determinants

A set of linear equations can be expressed in matrix form as Y = AX, where matrix A=[aᵢⱼ] contains the coefficients of independent variables. The order or dimension of matrix A is denoted as d(A)=m×n, where m is the number of rows and n is the number of columns. The determinant of matrix A is a scalar function denoted as det A, |A|, or ΔA.

🔑 Definition — Characteristic Matrix: Matrix A containing coefficients of the independent variables in a system of linear equations, which may be constants or functions of some parameter. 📐 Formula: Y = A X → The matrix equation where Y is the output vector, A is the coefficient matrix, and X is the variable vector.

Application to Circuit Analysis with Example 1

A circuit with two nodes is analyzed using KCL. The equations are:

• Node 1: (1/R₁+1/R₂)V₁ - (1/R₂-β/R₃)V₂ = 0
• Node 2: -(1/R₂)V₁ + (1/R₂+1/R₃)V₂ = Iₐ

With parameters β=2, R₁=6kΩ, R₂=12kΩ, R₃=3kΩ, Iₐ=2mA, the matrix equation becomes:

[1/4k   1/2k ] [V₁]   [0]
[-1/6k  1/2k ] [V₂] = [2m]

The solution uses X = A⁻¹Y where A⁻¹ = Adj(A)/|A|.

🔑 Definition — Adjoint of Matrix: The transpose of the cofactor matrix, used in calculating the inverse of a matrix.

📐 Formula: A⁻¹ = Adj(A) / |A| → The inverse matrix equals the adjoint divided by the determinant.

📌 Example: For this circuit, Adj(A) = [1/2k -1/2k; 1/6k 1/4k], |A| = 5/24k², so:

• V₁ = -24/5 V = -4.8V
• V₂ = 12/5 V = 2.4V
• I₀ = V₂/R₃ = (12/5)/3k = 4/5k = 4/5 mA = 0.8mA

Second Example with Three Currents

A circuit with two nodes is analyzed to find I₁, I₂, and I₃. The KCL equations are:

• Node 1: V₁/4k - V₂/6k = 1mA
• Node 2: -V₁/6k + V₂/3k = -4mA

In matrix form:

[1/4k  -1/6k] [V₁]   [1m]
[-1/6k  1/3k] [V₂] = [-4m]

📌 Example: Adj(A) = [1/3k 1/6k; 1/6k 1/4k], |A| = 1/18k², so:

• V₁ = -6V
• V₂ = -15V

Using Ohm's Law:

• I₁ = V₁/12k = -6/12k = -0.5mA
• I₂ = (V₁-V₂)/6k = (-6-(-15))/6k = 9/6k = 3/2mA = 1.5mA
• I₃ = V₂/6k = -15/6k = -5/2mA = -2.5mA

⭐ Key Takeaways

The matrix method provides a systematic approach to solving simultaneous circuit equations by converting them into the form AX=Y and finding X=A⁻¹Y. The inverse matrix is computed using A⁻¹=Adj(A)/|A|, where the adjoint is the transpose of the cofactor matrix and the determinant must be non-zero. Once node voltages are found using this method, all branch currents can be calculated using Ohm's Law. This technique is particularly powerful for circuits with multiple nodes and simultaneous equations.

🧠 Quick Revision Questions

1. What is the general form of matrix equation for circuit analysis, and what does each variable represent?
2. How do you calculate the inverse of a matrix using its adjoint and determinant?
3. In the first example, what were the values of V₁, V₂, and I₀?
4. How can you determine branch currents once node voltages are known using the matrix method?
5. What condition must be satisfied for a matrix to have an inverse?