S1 CSE Electrical and Electronics Engineering · Module 1

Basic circuit elements: Resistors, Inductors and Capacitors

Syllabus: Elementary Concepts of DC Electric Circuits

Basics: Capacitors & Inductors: V-I relations and Energy stored. Current and Voltage Division Rule - Relative potential
Ohm's Law and Kirchhoff's laws - numerical problems.
Star-delta conversion (resistive networks only - derivation not required) - numerical problems.

Resistors

An illustrated cutaway of a carbon composition resistor showing internal structure and color code bands

Resistor

A resistor is a passive two-terminal electrical component that restricts or regulates the flow of electric current in a circuit. Resistance opposes current flow through atomic collisions.

Resistance

Resistance is measured in ohms ( $\Omega$ ).

Ohm's Law

Ohm's law defines a resistor's behavior through the formula $R = V/I$ , where $R$ is resistance, $V$ is voltage, and $I$ is current.

Diagram comparing pure copper crystal structure with brass alloy structure containing copper and zinc atoms

V-I Characteristics of a Resistor

V-I characteristics graph showing a linear relationship between voltage and current

Capacitor

A capacitor consists of two conductors separated by an insulating material (the dielectric).

Capacitance

Capacitance is the ability of a circuit element to store electric charge. If $Q$ is the stored charge in coulombs and $V$ is the applied voltage:

Q=CV,\qquad C=\frac{Q}{V}.

The unit of capacitance is the farad (F).

Energy Stored in a Capacitor

For a capacitor charged from zero to $V$ , the average charging voltage is $V/2$ . The stored energy is:

E=\frac{1}{2}QV=\frac{1}{2}CV^2.

V-I Relation of a Capacitor

V-I relation of capacitor during charging and discharging

Inductors

Types of inductors: air core, iron core, ferrite core, and variable core

Inductance

Inductance is the property of an electrical circuit that resists changes in current. The SI unit of inductance is the henry (H). It stores energy in the form of a magnetic field when current flows through it.

The $v - i$ relationship in an inductance is given by:

v = L \frac{di}{dt}

V-I Characteristics and Switching Behavior of an Inductor

V-I characteristic and inductor response graph with circuit diagrams

An inductor acts as a short circuit to DC current in steady state.

During switching transitions, the inductor operates as follows:

Switching ON ( $t = 0$ ):
- The inductor acts like an open circuit.
- Voltage ( $V$ ): Maximum, equal to the source voltage.
- Current ( $I$ ): Exactly zero.
Steady State:
- Current ( $I$ ): Reaches its maximum value.
- Voltage: Across an ideal inductor drops to 0 V.
De-energizing (Load on):
- Current ( $I$ ): Gradually reduces to zero.
- Voltage ( $V$ ): Maximum reversed (negative polarity).

Energy Stored in an Inductor

Inductor

An inductor stores energy in a magnetic field created by current flowing through a coil of wire.

The energy stored in an inductor's magnetic field is given by:

W_L = \frac{1}{2}Li^2

Independent and dependent voltage and current sources

Constant Voltage Source

A constant voltage source is a device that maintains a constant voltage across its terminals, with the output current changing depending on the connected load resistance (such as a battery).

Internal Resistance: Has a very low (ideally zero) internal resistance.
Behavior on Short Circuit: Tries to supply infinite current, which can damage the device or blow a fuse.
Behavior on Open Circuit: Outputs its rated voltage with zero current flow.

Behavior Under Load

For a connected load resistor ( $R$ ), the output current ( $I$ ) varies while terminal voltage ( $V$ ) remains constant.

Circuit diagram showing a resistor connected across a constant voltage source battery

Circuit Conditions for Voltage Sources

Behavior on Short Circuit

Circuit diagram showing a short circuit with a battery, fuse, and switch, with a lightning bolt pointing to the fuse.

Behavior on Open Circuit

Circuit diagram showing a battery, a fuse, and two switches connected in series

Voltage Source Characteristics

Ideal Voltage Source

An ideal voltage source maintains an exact constant voltage output ( $E$ ) independent of load current demand.

Two alternative circuit symbols for an ideal voltage source between terminals A and B

Or $E$ :

Circuit diagram and V-I or V-time characteristic curve for an ideal voltage source

V-I Characteristic

Practical Voltage Source and Its Graph:

Circuit diagram of a practical voltage source with internal resistance R_int and a V-I characteristic graph showing ideal and practical curves.

Constant Current Source

A constant current source is a device that maintains a constant current; the output voltage changes depending on the connected load resistance.

Internal Resistance: Has a very high (ideally infinite) internal resistance.
Behavior on Short Circuit: Safely delivers its rated current at near-zero voltage.
Behavior on Open Circuit: Tries to drive voltage to its maximum limit to force current.

Behavior on Open Circuit

Circuit diagram showing a solar panel or battery connected in series with a fuse, an open switch, and another open switch.

V-I Characteristic

Circuit diagram of a current source in parallel with a resistor

Graph showing the V-I characteristic of a practical current source

Circuit diagram with a resistor connected to a DC power source or solar panel

Circuit fundamentals: SI units, definitions and Ohm's law

SI units of electrical quantities

Quantity	Unit	Symbol
Current	Ampere	A
Voltage	Volt	V
Power	Watt	W
Energy	Joule	J
Resistance	Ohm	Ω

Series and parallel connections

Component	Series Connection	Parallel Connection
Resistors ( $R$ )	$R_1 + R_2 \text{ ...}$ (Add)	$\frac{1}{R_1} + \frac{1}{R_2} \text{ ...}$ (Reciprocals)
Capacitors ( $C$ )	$\frac{1}{C_1} + \frac{1}{C_2} \text{ ...}$ (Reciprocals)	$C_1 + C_2 \text{ ...}$ (Add)
Inductors ( $L$ )	$L_1 + L_2 \text{ ...}$ (Add)	$\frac{1}{L_1} + \frac{1}{L_2} \text{ ...}$ (Reciprocals)

Network

A network is the interconnection of two or more circuit elements (resistor, inductor, capacitor, voltage source, current source) connected in any manner.

Electric Circuit

An electric circuit is a network that contains at least one closed path. Every circuit is a network, but not all networks are circuits.

Figure 1.2: Basic electric circuit showing a voltage source and load resistor in a closed loop

Figure 1.2: Basic electric circuit

Ohm's Law

Ohm's law states that the current flowing through a conductor is directly proportional to the voltage across it and inversely proportional to its resistance, provided the temperature remains constant.

Ohm's Law Triangle diagram showing the relationship between Voltage V, Current I, and Resistance R

Georg Simon Ohm

Georg Simon Ohm (16 March 1789 – 6 July 1854) was a German physicist and mathematician who formulated the relationship between voltage, current, and resistance.

Portrait of Georg Simon Ohm

Joseph Henry

Joseph Henry was an American scientist whose work on electromagnetism led to the characterization of electromagnetic induction.

Portrait of American scientist Joseph Henry

Henry

One henry is defined as the inductance of a circuit where an induced voltage of one volt is produced by a current changing at a rate of one ampere per second.

Series and parallel resistor combinations

Resistances in Series

Series Circuit

A series circuit is a circuit in which resistances are connected end to end so that there is only one path for current flow.

In a series circuit:

The same current flows through all resistances.
There will be a voltage drop across each resistance given by Ohm's law.
The sum of the voltage drops is equal to the applied voltage.

Series circuit diagram with voltage source Vs, current I, and three series resistances R1, R2, R3 with voltage drops V1, V2, V3 respectively.

3rd point:

Series resistor circuit diagram with voltage source V_s, current I, resistors R1, R2, R3, and voltage drops V1, V2, V3

From Figure 1.7, $V_S = V_1 + V_2 + V_3$

\begin{aligned} &= IR_1 + IR_2 + IR_3 \\ &= I(R_1 + R_2 + R_3) \\ &= IR \quad \text{where } R \text{ is the equivalent resistance.} \end{aligned}

Then the equivalent resistance, $R = R_1 + R_2 + R_3$

Resistances in Parallel

Parallel Circuit

A parallel circuit is a circuit in which one end of each resistance is joined to a common point and the other end of each resistance is joined to another common point so that there are many paths for current flow (refer to Figure 1.8).

In a parallel circuit:

The voltage across each resistance of a parallel combination is the same.
The total current is the sum of individual branch currents.

Circuit diagram showing three resistors in parallel connected to a voltage source

Figure 1.8: Resistances in parallel

From the figure, $I = I_1 + I_2 + I_3$

= \frac{V_S}{R_1} + \frac{V_S}{R_2} + \frac{V_S}{R_3}

Let $R$ be the equivalent resistance of the circuit. Then $I = \frac{V_S}{R}$

So, $\frac{V_S}{R} = V_S \left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\right)$

\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}

If resistances $R_1, R_2, \dots R_n$ are connected in parallel, then

G \quad \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}

R = R_1R_2/R_1+R_2

Voltage divider and current divider rules

Applications of Ohm's Law

Voltage divider and current divider rules
Source transformation rule

If current is not given in a circuit, the voltage division rule can be used. If voltage is not given in a circuit, the current division rule can be used.

Voltage Divider Rule

The voltage across any resistor in a series network is equal to the ratio of its individual resistance to the total resistance of the series network, multiplied by the total voltage in the circuit.

Circuit diagram for voltage divider rule showing a voltage source V and two series resistors R1 and R2 with voltage drops V1 and V2

Figure 2 - Voltage Division Rule

$v_1$ = voltage across first resistor
$V$ = total voltage in the circuit
$R_1 + R_2$ = total resistance

V_1 = \frac{VR_1}{R_1 + R_2}

V_2 = \frac{VR_2}{R_1 + R_2}

General Formula ( $N$ Resistors)

For a series connection of $N$ resistors, the voltage across any target resistor $R_x$ is:

V_x = V_{\text{total}} \cdot \left(\frac{R_x}{R_{\text{total}}}\right)

where $R_{\text{total}} = R_1 + R_2 + \dots + R_N$ .

Derivation of Voltage Divider Rule

According to Ohm's law,

V_1 = IR_1 \text{ and } V_2 = IR_2

Let $R$ is the total resistance of the circuit, and it is given by,

R = R_1 + R_2 \quad \cdots (1)

Also, from the circuit, we have,

V = IR = I(R_1 + R_2) \quad \cdots (2)

But

I = \frac{V_1}{R_1} = \frac{V_2}{R_2} \quad \cdots (3)

Therefore, from equations (2) & (3), we finally get,

V = \frac{V_1}{R_1}(R_1 + R_2) \implies V \cdot R_1 = V_1 \cdot (R_1 + R_2) \implies \frac{V \cdot R_1}{R_1 + R_2} = V_1

\therefore V_1 = \frac{VR_1}{R_1 + R_2} \quad \cdots (4)

Similarly,

V = \frac{V_2}{R_2}(R_1 + R_2)

\therefore V_2 = \frac{VR_2}{R_1 + R_2} \quad \cdots (5)

Current Divider Rule

The current in any branch is equal to the ratio of opposite branch resistance to the total resistance of the circuit multiplied by the total current in the circuit.

Current divider circuit diagram with two parallel resistors R1 and R2

I_1 = \frac{IR_2}{R_1 + R_2}

I_2 = \frac{IR_1}{R_1 + R_2}

Derivation of Current Divider Rule

The current $I$ has been divided into $I_1$ and $I_2$ into two parallel branches:

I_1 = \frac{V}{R_1} \hspace{5cm} I_2 = \frac{V}{R_2}

R_T = \frac{R_1 R_2}{R_1 + R_2}

Blue arrow pointing to the right

\text{Parallel equivalent: } R_T = \frac{R_1 R_2}{R_1 + R_2}

I = \frac{V}{R_T}

= \frac{V(R_1 + R_2)}{R_1 R_2}

But $V = I_1 R_1$ . Substituting the value of $V$ in the above equation we get:

Current flow through $R_1$ :

I = \frac{I_1 R_1 (R_1 + R_2)}{R_1 R_2} = \frac{I_1 (R_1 + R_2)}{R_2}

Blue arrow pointing to the right

I_1 = \frac{I R_2}{R_1 + R_2}

Current flow through $R_2$ :

Blue arrow pointing to the right

I_2 = \frac{I R_1}{R_1 + R_2}

Important Notes

If current is not given in a circuit, the voltage division rule can be used.
If voltage is not given in a circuit, the current division rule can be used.
If voltage and current are available in a circuit, use Ohm's law.
When specifically asked in the question, apply the divider rules.

Worked Examples

Example 1

Find voltage drop across each resistance.

Circuit diagram with a 6V voltage source and two series resistors of 200 ohm and 100 ohm.

Applying the voltage division rule, voltage across $200\,\Omega$ resistance we get:

E_1 = \frac{E}{r_1+r_2} r_1

E_1 = \frac{6}{200+100} \text{ X } 200 = 4\text{V}

Voltage across $100\,\Omega$ resistance we get:

E_2 = \frac{E}{r_1+r_2} r_2

E_2 = \frac{6}{200+100} \text{ X } 100 = 2\text{V}

Example 2

Consider two resistors $20\,\Omega$ and $40\,\Omega$ are connected in a parallel with a current source of $20\text{ A}$ . Find out the current flowing through each resistor in the parallel circuit.

Circuit diagram showing a parallel circuit with a current source of 20A and two resistors R1 (20 ohms) and R2 (40 ohms)

Applying the current division rule voltage across $20\,\Omega$ resistance we get:

I_1 = \frac{I}{r_1+r_2} r_2

I_1 = \frac{20}{20+40} \times 40 = 13.33\text{A}

Voltage across $40\,\Omega$ resistance we get:

I_2 = \frac{I}{r_1+r_2} r_1

I_2 = \frac{20}{20+40} \times 20 = 6.67\text{A}

Example 1.11.1

A resistor of $5\;\Omega$ is connected in series with a parallel combination of $6\;\Omega$ and $3\;\Omega$ . Find the supply current taken from a $35\;\text{V}$ dc source.

Solution

Let $R_P$ be the equivalent resistance of the parallel combination of $3\;\Omega$ and $6\;\Omega$ .

Then $\dfrac{1}{R_P} = \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{3}{6}$

Circuit diagram showing a parallel combination of 3 ohm and 6 ohm resistors in series with a 5 ohm resistor, connected across a 35 V source.

Therefore, $R_P = 2\;\Omega$

Total resistance in the circuit, $R = R_P + 5 = 2 + 5 = 7\;\Omega$

Supply current, $I = \dfrac{V}{R} = \dfrac{35}{7} = 5\;\text{A}$

Example 1.11.2

Find the total resistance of the circuit across the terminals X and Y. Also find the power consumed by the circuit if a $12\text{ V}$ battery is connected across XY.

Circuit diagram with resistors connected between terminals X and Y

Figure 1.13

Solution

\begin{aligned} R_{XZ} &= \frac{2 \times 2}{2 + 2} = 1\text{ }\Omega\\ \frac{1}{R_{ZY}} &= \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{4}{4} = 1 \end{aligned}

So, $R_{ZY} = 1\text{ }\Omega$

Simplified circuit diagram

R_{XZ} + R_{ZY} = 1 + 1 = 2\, \Omega

\frac{1}{R_{XY}} = \frac{1}{4} + \frac{1}{2} + \frac{1}{4} = 1

Therefore, equivalent resistance, $R_{XY} = 1\, \Omega$

Total current in the circuit, $I = \frac{V}{R_{XY}} = \frac{12}{1} = 12\text{ A}$

Power consumed $= VI = 12 \times 12 = 144\text{ W}$

Example 1.11.3

Three resistors $10\ \Omega$ , $20\ \Omega$ and $30\ \Omega$ are connected in series across a $100\text{ V}$ supply. Find the voltage across each resistor.

Solution

Circuit diagram of three resistors connected in series with a voltage source V = 100 V

Figure 1.15

Let $R_S$ be the equivalent series resistance:

R_S = 10 + 20 + 30 = 60\ \Omega

I = \frac{V}{R_S} = \frac{100}{60} = 1.666\text{ A}

Voltage drop across $10\ \Omega$ , $V_1 = 1.666 \times 10 = 16.66\text{ V}$

Voltage drop across $20\ \Omega$ , $V_2 = 1.666 \times 20 = 33.33\text{ V}$

Voltage drop across $30\ \Omega$ , $V_3 = 1.666 \times 30 = 50\text{ V}$

Example 2.6.2

Find the voltage drop and current through all the resistances in the circuit shown in Figure 2.20.

Circuit diagram with a 50V DC source, a 6 ohm series resistor, and two parallel branches with 12 ohm and 6 ohm resistors.

Figure 2.20

Solution

\text{Equivalent resistance} = 6 + \frac{12 \times 6}{12 + 6} = 10\;\Omega

\text{Source current}, I = \frac{\text{Source voltage}}{\text{Equivalent resistance}} = \frac{50\;\text{V}}{10\;\Omega} = 5\;\text{A}

Voltage drop across $6\;\Omega$ resistor:

= \text{Current through } 6\;\Omega \times \text{Resistance}

= 5 \times 6 = 30\;\text{V}

Using current divider rule, current through $12\ \Omega$ resistor $= I_1$ :

\quad = 5 \times \frac{6}{12 + 6} = \frac{5}{3}\text{ A}

By Kirchhoff's current law, $I = I_1 + I_2$

Therefore, $I_2 = I - I_1 = 5 - \frac{5}{3} = \frac{10}{3}\text{ A}$

Voltage drop across $12\ \Omega = I_1 \times 12 = \frac{5}{3} \times 12 = 20\text{ V}$

Voltage drop across $6\ \Omega = I_2 \times 6 = \frac{10}{3} \times 6 = 20\text{ V}$

The voltage across $6\ \Omega$ and $12\ \Omega$ resistors are the same because they are connected in parallel.

Source transformation and relative potential

Source transformation

Ohm's law

Voltage source to current source

Diagram showing voltage source to current source transformation

Figure 2.5: Voltage source to current source transformation

A voltage source $V_s$ in series with resistance $R$ can be transformed into a current source $I_s$ in parallel with resistance $R$ where $I_s = \frac{V_s}{R}$ .

Current source to voltage source

Current source to voltage source transformation diagram

Figure 2.6: Current source to voltage source transformation

Similarly, a current source $I_S$ in parallel with a resistance $R$ can be transformed into a voltage $V_S$ series with a resistance $R$ where $V_S = I_S R$ .

Network topology definitions

Node

A node is a junction where two or more network elements are connected together.

Node diagram

Branch

A branch is an element or number of elements connected between two nodes.

Branch diagram

Junction

A junction is a point where three or more circuit components meet.

Junction diagram

Electrical Network

An electrical network is an interconnection of electrical elements.

Key network components include:

Node
Junction
Branch
Loop
Mesh
Electric network

Circuit diagram illustrating nodes

Circuit diagram illustrating branch

Circuit versus network

Circuit vs Network comparison diagram showing a simple circuit with a battery, resistor, and capacitor, alongside a complex network of resistors.

Electrical Circuit

An electrical circuit is an electrical network with at least one closed path. All circuits are networks, but all networks are not circuits.

Loop

A loop is any closed part of the circuit.

Mesh

A mesh is the most elementary form of a loop and cannot be further divided into other loops. All meshes are loops, but all loops are not meshes.

Two-mesh circuit diagram illustrating loop currents i1 and i2.

Single mesh circuit labeled Mesh 1 with current i1, voltage source V1, and resistor R1.

Potential and reference levels

Absolute (Node) Potential ( $V_A$ )

Absolute potential is the voltage at Node A measured relative to the reference node (Ground, which is assumed to be $0\text{ V}$ ).

Relative Potential ( $V_{AB}$ )

Relative potential is the voltage at Node A relative to Node B.

The formula:

V_{AB} = V_A - V_B

Relative Potential

Relative potential (or potential difference) is the voltage of one point in a circuit measured with respect to another point, rather than to ground.

Circuit diagram illustrating absolute and relative potential with nodes A, B, C, D, resistors, a voltage source, and ground.

Circuit diagram illustrating electrical concepts including Nodes, Branches, and Loops

Branch

A branch is a single path containing a circuit component connected between two nodes.

Circuit diagram illustrating branches with resistors and a voltage source

Kirchhoff's laws: KVL, KCL and network loops

Finding Loops in a Circuit

Circuit diagram illustrating loops with resistors, a voltage source, and mesh currents i1 and i2

Mesh

A mesh is a closed loop that does not contain any other loop within it. In other words, any path within a circuit that does not include any other path is considered a mesh.

Circuit diagram illustrating Mesh 1 and Mesh 2

The overall circuit depicted in Figure 1 comprises 6 Nodes, 7 Branches, 3 Loops, and 2 Meshes, as illustrated in Figure 6.

Circuit diagram illustrating nodes, branches, loops, and meshes

Robert Kirchhoff (1824–1887) was a German physicist and mathematician who contributed significantly to the fundamental understanding of electrical circuits.

Portrait illustration of Robert Kirchhoff

Kirchhoff's Voltage Law (KVL)

Kirchhoff's voltage law states that the algebraic sum of all voltages in a closed path in a circuit is always zero at all instants of time. Alternatively, the sum of potential drops is equal to the sum of potential rises.

Circuit diagram illustrating Kirchhoff's Voltage Law

Voltage Polarities and Sign Conventions

Circuit diagram showing a closed loop with a voltage source V and four resistors R1, R2, R3, R4 with indicated polarities and current I

Potential rise and potential drops:

$V + V_1 + V_2 + V_3 + V_4 = 0$

V across the resistance $R_1 = IR_1$

Single loop electrical circuit with resistors and voltage sources

Potential rise icon showing transition from minus to plus

Potential rise

Potential drop icon showing transition from plus to minus

Potential drop

Key conventions for analysis:

Current direction: Shared branches take $(i_1 - i_2)$ for the first loop and $(i_2 - i_1)$ for the adjacent loop when mesh currents flow in the same direction, or $(i_1 + i_2)$ and $(i_2 + i_1)$ when in opposing directions.
Voltage polarity: Always start with the voltage source when setting up mesh equations.

Sign Convention for KVL Equations

When traversing a loop:

If the positive terminal is encountered before the negative terminal, the voltage is interpreted as positive in the KVL equation.
If the negative terminal is encountered first, the voltage is interpreted as negative in the KVL equation.

Formulating KVL Equations

Circuit diagram illustrating KVL with a voltage source V and resistors R1, R2, R3, and R4 in a single closed loop with clockwise current I.

Polarity convention inset box showing plus to minus gives negative (-ve) and minus to plus gives positive (+ve).

Traversing the loop in the clockwise direction yields:

+V - IR_1 - IR_2 - IR_3 - IR_4 = 0

Single-Mesh Analysis

Example 2.6.1 Find the current in the circuit shown in Figure 2.

Circuit diagram with a single loop, three resistors, and two voltage sources

Solution

Let the current $I$ flow in the clockwise direction. Marking the polarities of voltages across resistors according to the sign convention as shown in Figure 2.19:

Circuit diagram with clockwise current I and voltage polarities across components

Figure 2.19

Blue arrow pointing right

Applying KVL, we get:

\begin{aligned} E_1 - IR_1 - IR_2 - E_2 - IR_3 &= 0 \\ 12 - 2 \times I - 3 \times I - 3 - 4 \times I &= 0 \\ 9 - 9I &= 0 \\ \therefore \text{Current}, I &= 1\text{ A} \end{aligned}

Example 2.6.1 (Variant) Find the current in the circuit shown in Figure 2.18.

Circuit diagram with two voltage sources E1 (12 V) and E2 (3 V) and three resistors R1 (2 ohm), R2 (3 ohm), and R3 (4 ohm).

Figure 2.18

Solution

Let the current $I$ flow in the clockwise direction. Marking the polarities of voltages across resistors according to the sign convention as shown in Figure 2.19:

Circuit diagram for Example 2.6.1 showing assumed clockwise loop current I and voltage polarities across all components.

Figure 2.19

Applying KVL, we get:

\begin{aligned} E_1 - IR_1 - IR_2 - E_2 - IR_3 &= 0 \\ 12 - 2 \times I - 3 \times I - 3 - 4 \times I &= 0 \\ 15 - 9I &= 0 \\ 9I &= 15 \\ I = \frac{15}{9} = \frac{5}{3} \text{ A} \approx 1.67 \text{ A} \end{aligned}

Two-Mesh Analysis

Example 2.6.5 Find the current through $4 \; \Omega$ in the circuit shown in Figure 2.23.

Circuit diagram with two meshes, resistors, voltage sources, and mesh currents I1 and I2

When all mesh currents are drawn clockwise by default, shared branches always carry opposing currents.

Electrical circuit diagram with two meshes

Solution

From loop 1: $15 - 10I_1 - 5(I_1 - I_2) = 0$

15I_1 - 5I_2 = 15

3I_1 - I_2 = 3 \quad (2.4)

From loop 2: $-5(I_2 - I_1) - 4I_2 - 9 = 0$

9I_2 - 5I_1 = -9 \quad (2.5)

From expressions 2.4 and 2.5, we get: $I_1 = 0.82 \text{ A}$

I_2 = -0.54 \text{ A}

Note: The current flowing through the $4\,\Omega$ resistor is the mesh current $I_2$ :

I_{4\Omega} = I_2 = - \frac{6}{11} \text{ A} \approx -0.545 \text{ A}

Solving the System of Simultaneous Equations

From Equation 1, express $I_2$ in terms of $I_1$ :

I_2 = 3I_1 - 3

From Equation 1, express $I_2$ in terms of $I_1$ :

I_2 = 3I_1 - 3

Substitute $I_2$ into Equation 2:

\begin{aligned} 5I_1 - 9(3I_1 - 3) &= 9 \\ 5I_1 - 27I_1 + 27 &= 9 \\ -22I_1 &= 9 - 27 \\ -22I_1 &= -18 \\ I_1 = \frac{-18}{-22} &= \frac{9}{11}\text{ A} \approx 0.818\text{ A} \end{aligned}

Now, calculate $I_2$ :

\begin{aligned} I_2 &= 3\left(\frac{9}{11}\right) - 3 \\ I_2 &= \frac{27}{11} - \frac{33}{11} = -\frac{6}{11}\text{ A} \approx -0.545\text{ A} \end{aligned}

Three-Mesh Analysis

Example 2.6.3 Find the loop currents in the electric circuit shown in Figure 2.21.

Electric circuit diagram with 3 meshes and 3 mesh currents I1, I2, and I3

Solution

From loop 1: $5 - 5I_1 - 1(I_1 - I_2) = 0$

6I_1 - I_2 = 5 \tag{2.1}

From loop 2: $-2I_2 - 2(I_2 - I_3) - 1(I_2 - I_1) = 0$

5I_2 - I_1 - 2I_3 = 0 \tag{2.2}

Circuit diagram with three loops and multiple resistors and voltage sources.

From loop 3: $-2(I_3 - I_2) - 3I_3 + 10 = 0$

5I_3 - 2I_2 = 10 \tag{2.3}

Solving Equations 2.1, 2.2, and 2.3, we get:

I_1 = 1.03\text{ A}

I_2 = 1.198\text{ A}

I_3 = 2.48\text{ A}

From (Eq. 1): $I_2 = 6I_1 - 5$

From (Eq. 3): $5I_3 = 10 + 2I_2 = 10 + 2(6I_1 - 5) = 10 + 12I_1 - 10 = 12I_1 \implies I_3 = 2.4I_1$

Substitute $I_2$ and $I_3$ into (Eq. 2):

\begin{aligned} I_1 - 5(6I_1 - 5) + 2(2.4I_1) &= 0 \\ I_1 - 30I_1 + 25 + 4.8I_1 &= 0 \\ -24.2I_1 &= -25 \end{aligned}

I_1 = \frac{25}{24.2} = \frac{125}{121} \approx 1.033\text{ A}

Now, calculate $I_2$ and $I_3$ :

I_2 = 6 \cdot \left(\frac{125}{121}\right) - 5 = \frac{750}{121} - \frac{605}{121} = \frac{145}{121} \approx 1.198\text{ A}

I_3 = 2.4 \cdot \left(\frac{125}{121}\right) = \frac{12}{5} \cdot \frac{125}{121} = \frac{300}{121} \approx 2.479\text{ A}

Supermesh analysis and numerical problems

Supermesh

A supermesh is created in circuit analysis when a current source (independent or dependent) is located on the boundary shared by two adjacent meshes.

Supermesh Analysis Procedure

To analyze a circuit containing a supermesh:

Identify shared sources: Determine when a current source is located on the boundary shared by two meshes.
Form the supermesh: Temporarily remove the shared current source and any element connected strictly in series with it, leaving an open circuit to create a larger combined loop.
Apply KVL: Apply Kirchhoff's Voltage Law around the newly formed supermesh loop.
Write the constraint equation: Formulate an auxiliary equation relating the mesh currents flowing through the shared current source branch.
Unshared current sources: If a current source is located on the outside of the circuit (not shared between meshes), a supermesh is not needed because the mesh current is determined directly.

Worked Numerical Problems

A current source is located on the boundary shared by two meshes.

Example 1. Determine the current in the $5\text{ }\Omega$ resistor in the network given in Fig.

Electrical network with meshes, resistors, a voltage source, and a current source

Electrical circuit diagram with three mesh currents, voltage source, current source, and resistors

50 - 10(I_1 - I_2) - 5(I_1 - I_3) = 0

(1)

15I_1 - 10I_2 - 5I_3 = 50

Step B: Constraint Equation for the Supermesh

The $2\text{ A}$ current source is in a branch shared by Mesh 2 and Mesh 3.

Current $I_2$ flows in the same direction as the source (right to left).
Current $I_3$ flows in the opposite direction (left to right).

Circuit diagram showing a supermesh with a 50V source, 10 ohm, 5 ohm, 2 ohm, 3 ohm, and 1 ohm resistors, and a 2A current source between mesh 2 and mesh 3.

I_2 - I_3 = 2 \implies I_2 = I_3 + 2 - (\text{Eq. 2})

Electrical circuit diagram with three mesh currents I1, I2, I3, resistors of 10 ohm, 2 ohm, 5 ohm, 1 ohm, a 50V voltage source, and a 2A current source with a blue overlay box.

-2I_2 - I_3 - 5(I_3 - I_1) - 10(I_2 - I_1) = 0

\ -15I_1 + 12I_2 + 6I_3 = 0 \quad (3)

Step C: Solving the System of Equations

Substitute $I_2 = I_3 + 2$ from (Eq. 2) into (Eq. 1) and (Eq. 3):

From (Eq. 1):

3I_1 - 2(I_3 + 2) - I_3 = 10 \implies 3I_1 - 3I_3 = 14 \implies I_1 - I_3 = \frac{14}{3}\text{ A}

Interestingly, we do not need to go further because the question asks for the current in the $5\text{ }\Omega$ resistor. Looking at the diagram, the net downward current in the $5\text{ }\Omega$ resistor is exactly $(I_1 - I_3)$ .

I_{5\Omega} = I_1 - I_3 = \frac{14}{3}\text{ A} \approx 4.67\text{ A}

If a current source is on the outside of the circuit (not shared), you don't need a supermesh.

Example 2. Write the mesh equations for the circuit shown in Fig. and determine the currents $I_1$ , $I_2$ and $I_3$ .

Circuit diagram with three meshes containing a 10A current source, resistors of 3 ohms, 2 ohms, and 1 ohm, and a 10V voltage source.

Summing the voltage drops to zero:

3(I_2 - I_1) + 10 + 2(I_2 - I_3) = 0

Substitute $I_1 = 10$ :

3(I_2 - 10) + 10 + 2(I_2 - I_3) = 0

3I_2 - 30 + 10 + 2I_2 - 2I_3 = 0

5I_2 - 2I_3 = 20 \quad \text{--- (Equation 1)}

1 \cdot I_3 + 2(I_3 - I_2) - 10 = 0

-2I_2 + 3I_3 = 10 \quad \text{--- (Equation 2)}

Circuit diagram showing a mesh analysis problem with mesh currents I1, I2, I3, a 10A current source, a 3 ohm resistor, a 10V voltage source, a 2 ohm resistor, and a 1 ohm resistor.

Solving the System of Equations

From equation 2:

3I_3=2I_2+10,\qquad I_3=\frac{2I_2+10}{3}.

Substitute into equation 1:

5I_2-2\left(\frac{2I_2+10}{3}\right)=20.

Multiplying by 3 and simplifying:

15I_2-4I_2-20=60,\qquad 11I_2=80.

I_2=\frac{80}{11}\,\mathrm{A}\approx7.27\,\mathrm{A}.

Then:

I_3=\frac{2(80/11)+10}{3}=\frac{270}{33}=\frac{90}{11}\,\mathrm{A}\approx8.18\,\mathrm{A}.

Final Currents

I_1=10\,\mathrm{A},\qquad I_2=\frac{80}{11}\,\mathrm{A},\qquad I_3=\frac{90}{11}\,\mathrm{A}.

Mesh current method with matrix representation

Matrix Method of Analysis

An $n$ -mesh network can be represented by $n$ simultaneous equations. Refer to the three-mesh network shown in Figure 2.25.

Three-mesh electrical network with resistors R1, R2, R3, R4, R5, voltage sources V1, V2, and mesh currents I1, I2, I3

Figure 2.25

Circuit diagram with three loops containing resistors and voltage sources

The matrix equation for the network is

$R_{11} = 1^{\text{st}} \text{ loop total resistance}$

$R_{12} = \text{common resistance in between}$ $\text{loop } 1 \text{ and } 2$

$R_{12} = R_{21}$ $R_{23} = R_{32}$ $R_{13} = R_{31}$

$V_1 \text{ is } - \text{ to } + \text{ then } v_1$ $V_2 \text{ is } + \text{ to } - \text{ then } -v_2$

\begin{bmatrix} R_{11} & -R_{12} & -R_{13} \\ -R_{21} & R_{22} & -R_{23} \\ -R_{31} & -R_{32} & R_{33} \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \\ I_3 \end{bmatrix} = \begin{bmatrix} V_1 \\ V_2 \\ V_3 \end{bmatrix}

Matrix Formulation Rules

Diagonal Elements ( $R_{11}, R_{22}, R_{33}$ ): Always positive ( $+$ ). Represents the sum of all resistances in that specific loop.
Off-Diagonal Elements ( $R_{12}, R_{23}, R_{31}$ , etc.): Usually negative ( $-$ $-$ ). Represents the sum of resistances shared between two loops.
- Sign Rule: If both loop currents pass through the shared resistor in opposite directions, the entry is negative ( $-$ ). If they flow in the same direction, the entry is positive ( $+$ ). When all mesh currents are chosen clockwise by convention, currents in shared branches oppose each other, resulting in negative off-diagonal terms.

Voltage Source Summation

If two voltage sources lie in the same loop, sources aiding one another are added, whereas sources opposing one another are subtracted.

Voltage Matrix Polarity

For entries in the voltage matrix, polarity depends on the assumed mesh current direction.

Applying KVL to each mesh yields the following equations:

Circuit diagram with three meshes and mesh currents I1, I2, I3

\begin{bmatrix} R_{11} & R_{12} & R_{13} \\ R_{21} & R_{22} & R_{23} \\ R_{31} & R_{32} & R_{33} \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \\ I_3 \end{bmatrix} = \begin{bmatrix} V_1 \\ V_2 \\ V_3 \end{bmatrix}

Rearranging the equations gives:

\begin{aligned} (R_1 + R_2)I_1 &- R_2 I_2 & &= V_1 \\ -R_2 I_1 &+ (R_2 + R_3 + R_4)I_2 &- R_4 I_3 &= 0 \\ &-R_4 I_2 &+ (R_4 + R_5)I_3 &= -V_2 \end{aligned}

These equations are expressed in matrix form as:

\begin{bmatrix} (R_1 + R_2) & -R_2 & 0 \\ -R_2 & R_2 + R_3 + R_4 & -R_4 \\ 0 & -R_4 & R_4 + R_5 \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \\ I_3 \end{bmatrix} = \begin{bmatrix} V_1 \\ 0 \\ -V_2 \end{bmatrix}

Example 2.6.8

Determine the mesh currents in the network shown in Figure 2.26.

Circuit diagram with three meshes and mesh currents I1, I2, I3

Figure 2.26

Solution

The matrix equation for the network is:

\begin{bmatrix}8&-3&0\\-3&9&2\\0&2&10\end{bmatrix}\begin{bmatrix}I_1\\I_2\\I_3\end{bmatrix}=\begin{bmatrix}5\\10\\35\end{bmatrix}\tag{2.19}

\Delta_R = \begin{vmatrix} 8 & -3 & 0 \\ -3 & 9 & 2 \\ 0 & 2 & 10 \end{vmatrix} = 8(90 - 4) - 3(+30) = 598

\Delta_1 = \begin{vmatrix} 5 & -3 & 0 \\ 10 & 9 & 2 \\ 35 & 2 & 10 \end{vmatrix} = 5(90 - 4) + 3(100 - 70) = 520

\Delta_2 = \begin{vmatrix} 8 & 5 & 0 \\ -3 & 10 & 2 \\ 0 & 35 & 10 \end{vmatrix} = 8(100 - 70) - 5(-30) = 390

\Delta_3 = \begin{vmatrix} 8 & -3 & 5 \\ -3 & 9 & 10 \\ 0 & 2 & 35 \end{vmatrix} = 8(315 - 20) + 3(-105) + 5(-6) = 2015

\therefore I_1 = \frac{\Delta_1}{\Delta_R} = \frac{520}{598} = 0.87\text{ A}

I_2 = \frac{\Delta_2}{\Delta_R} = \frac{390}{598} = 0.65\text{ A}

I_3 = \frac{\Delta_3}{\Delta_R} = \frac{2015}{598} = 3.37\text{ A}

Example 2.6.9

Write the resistance matrix of the network shown in Figure 2.27 and determine the current drawn from the voltage source.

Electrical network diagram with three mesh currents I1, I2, and I3, resistors of 3 ohm, 4 ohm, 2 ohm, 3 ohm, 2 ohm, and 4 ohm, and a 10 V DC voltage source.

Figure 2.27

Solution

The matrix equation for the network is:

\begin{bmatrix} R_{11} & R_{12} & R_{13} \\ R_{21} & R_{22} & R_{23} \\ R_{31} & R_{32} & R_{33} \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \\ I_3 \end{bmatrix} = \begin{bmatrix} V_1 \\ V_2 \\ V_3 \end{bmatrix}

\begin{bmatrix} 10 & -4 & -3 \\ -4 & 8 & -2 \\ -3 & -2 & 9 \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \\ I_3 \end{bmatrix} = \begin{bmatrix} 10 \\ 0 \\ 0 \end{bmatrix} \tag{2.20}

\Delta_R = \begin{vmatrix} 10 & -4 & -3 \\ -4 & 8 & -2 \\ -3 & -2 & 9 \end{vmatrix} = 10(72 - 4) + 4(-36 - 6) - 3(8 + 24) = 416

\Delta_1 = \begin{vmatrix} 10 & -4 & -3 \\ 0 & 8 & -2 \\ 0 & -2 & 9 \end{vmatrix} = 10(72 - 4) + 4(0 - 0) - 3(0 - 0) = 680

\Delta_2 = \begin{vmatrix} 10 & 10 & -3 \\ -4 & 0 & -2 \\ -3 & 0 & 9 \end{vmatrix} = 10(0 - 0) - 10(-36 - 6) - 3(0 - 0) = 420

\Delta_3 = \begin{vmatrix} 10 & -4 & 10 \\ -4 & 8 & 0 \\ -3 & -2 & 0 \end{vmatrix} = 10(0 - 0) + 4(0 - 0) + 10(8 + 24) = 320

\therefore I_1 = \frac{\Delta_1}{\Delta_R} = \frac{680}{416} = 1.635\text{ A}

I_2 = \frac{\Delta_2}{\Delta_R} = \frac{420}{416} = 1.01\text{ A}

I_3 = \frac{\Delta_3}{\Delta_R} = \frac{320}{416} = 0.77\text{ A}

Circuit diagram with mesh currents I1, I2, and I3

Figure 2.27

Node voltage method and numerical problems

Homework Problem

(a) Determine the current $I_L$ in the circuit shown in the given figure by mesh analysis using matrix form.

Circuit diagram for mesh analysis with multiple voltage sources and resistors

(10)

Kirchhoff's Current Law (KCL)

Kirchhoff's Current Law

Kirchhoff's current law states that any current flowing into a node must simultaneously equal the current flowing out of it at any given moment.

If entering currents are assigned a positive sign, leaving currents are given a negative sign.

Circuit diagram illustrating Kirchhoff's Current Law with a voltage source connected in parallel to three resistor branches, showing a node A with currents I_T entering and I_1, I_2, I_3 leaving.

I_T = I_1 + I_2 + I_3

Alternative Formulation

The algebraic sum of all currents meeting at a node is equal to zero.

Node

A node is a junction where two or more network elements are connected together.

Algebraic sum: $\sum I = 0$

Current meeting at a node diagram

I_1 + I_5 - I_2 - I_3 - I_4 = 0

I_1 + I_5 = I_2 + I_3 + I_4

Node Voltage Method

Node Voltage Analysis

In this method, one of the independent nodes is chosen as the reference node ( $0\text{v}$ ) and the voltages of the remaining independent nodes are defined with respect to it.

Circuit diagram with a voltmeter and nodes A, B, and C

Consider the circuit above containing 3 independent nodes: A, B, and C.

Let C be the reference node (ground).
Let $V_A$ and $V_B$ be the potentials of nodes A and B with respect to the reference node.
In node voltage analysis, at any identified node point, currents must enter and leave.

Circuit diagram with nodes A, B, C, resistors R1 to R5, and voltage sources V1 and V2

Procedural Steps

Find the node voltages first; branch currents flowing through circuits and components can then be calculated.

Circuit diagram for node voltage method analysis showing two voltage sources and five resistors.

The network has two independent nodes (A and B), one relative potential, and two absolute potential differences.

Circuit diagram with two voltage sources V1 and V2, five resistors R1 through R5, two nodes A and B, a ground node C, and labeled currents corresponding to each branch.

Three branches connect to node A. Assuming all currents leave node A, the current expression is $I_1+I_2+I_3=0$ :

Left Branch (towards $V_1$ ): Current flows from potential $V_A$ to potential $V_1$ across resistor $R_1$ .

I_{\text{left}} = \frac{V_A - V_1}{R_1}

Middle-Down Branch (towards Node C): Current flows from $V_A$ to ground ( $0\text{ V}$ ) through $R_2$ .

I_{\text{down}} = \frac{V_A - 0}{R_2} = \frac{V_A}{R_2}

Right Branch (towards Node B): Current flows from $V_A$ to $V_B$ across $R_3$ .

I_{\text{right}} = \frac{V_A - V_B}{R_3}

Circuit diagram with two nodes A and B, resistors R1 to R5, voltage sources V1 and V2, and a ground reference C.

Applying KCL at node A ( $\rightarrow I_1 + I_2 + I_3 = 0 \rightarrow I = V/R$ ):

\frac{V_A - V_1}{R_1} + \frac{V_A}{R_2} + \frac{V_A - V_B}{R_3} = 0

Applying KCL at node B:

\frac{V_B - V_2}{R_5} + \frac{V_B}{R_4} + \frac{V_B - V_A}{R_3} = 0

Solving the two simultaneous equations yields the node voltages $V_A$ and $V_B$ .

Numerical Examples

Example 1

Using nodal analysis, find the current through the $2\Omega$ resistance.

Circuit diagram with nodes A, B, and C, resistors of 6 ohms, 1 ohm, 2 ohms, 3 ohms, and 10 ohms, and DC voltage sources of 10V and 20V.

First determine the node voltages to calculate the branch current.

Applying KCL at node A:

\frac{V_A - 10}{6} + \frac{V_A}{1} + \frac{V_A - V_B}{2} = 0

V_A \left( \frac{1}{6} + 1 + \frac{1}{2} \right) - \frac{V_B}{2} = \frac{10}{6}

1.667 V_A - 0.5 V_B = 1.667 \quad ---\text{(1)}

Applying KCL at node B:

\frac{V_B - 20}{10} + \frac{V_B}{3} + \frac{V_B - V_A}{2} = 0

-\frac{V_A}{2} + V_B \left( \frac{1}{10} + \frac{1}{3} + \frac{1}{2} \right) = \frac{20}{10}

-0.5 V_A + 0.933 V_B = 2 \quad ---\text{(2)}

Solving the equations:

V_A = 1.957\text{V}, \quad V_B = 3.192\text{V}

I = \frac{V_A - V_B}{2} = \frac{1.957 - 3.192}{2} = -0.617\text{A}

Example 2.7.1

Find the current through resistor $20\ \text{ }\ \Omega$ in the circuit shown in Figure 2.32 using nodal analysis.

Circuit diagram with resistors and voltage sources

Figure 2.32: Circuit for Example 2.7.1

Determine the node voltages to find the required current.

At node $V_1$ :

\frac{V_1 - 8}{10} + \frac{V_1}{20} + \frac{V_1 - 2}{30} = 0

Electrical circuit diagram showing an 8V source, 10 ohm resistor, 20 ohm resistor, 30 ohm resistor, and a 2V source connected in a loop with a central node labeled node V1.

Step 2: Solve for Node Voltage $V_1$

Multiply the entire equation by the common denominator $60$ to clear fractions:

6(V_1 - 8) + 3V_1 + 2(V_1 - 2) = 0

6V_1 - 48 + 3V_1 + 2V_1 - 4 = 0

11V_1 - 52 = 0

11V_1 = 52

Step 3: Calculate Current through the $20\ \text{}\Omega$ Resistor

The current flowing downward through the $20\ \text{}\Omega$ resistor is:

I_{20\ \text{}\Omega} = \frac{V_1}{20} = \frac{4.727}{20} \approx 0.236\ \text{A} \quad \left(\text{or}\ \frac{13}{55}\ \text{A}\right)

The current through the $20\ \text{}\Omega$ resistor is $0.236\ \text{A}$ (flowing downwards).

Solution Verification

The circuit is redrawn with node voltages labelled, resistors designated, and reference node assigned as shown in Figure 2.33.

Electrical circuit diagram with a node voltage V_b and reference node V_d = 0

\frac{V_b - 8}{10} + \frac{V_b - 0}{20} + \frac{V_b - 2}{30} = 0

V_b \left(\frac{1}{10} + \frac{1}{20} + \frac{1}{30}\right) - \left(\frac{8}{10} + \frac{2}{30}\right) = 0

\therefore V_b = 4.7\text{ V}

\text{Current through } 20\text{ }\Omega,\ i_2 = \frac{V_b}{20} = \frac{4.7}{20} = 0.24\text{ A}

Example 2.7.2

Find the current through the resistor $R_1$ for the network shown in Figure 2.34.

Circuit diagram with a voltage source, a current source, and two resistors

Figure 2.34: Circuit for Example 2.7.2

First find the node voltages, then calculate the branch current.

Circuit diagram with a voltage source, two resistors, and a current source

\frac{V_1 - 10}{10} + \frac{V_1}{5} - 1 = 0

V_b \left(\frac{1}{10} + \frac{1}{5}\right) = 2

Solving for $V_b$ , we get $V_b = 6.67\text{ V}$ .

Current through resistor $R_1$ , $i_1 = \frac{V_a - V_b}{R_1}\text{ A}$ :

= \frac{10 - 6.67}{10} = 0.333\text{ A}

Example 2.7.4

Determine the node voltages for the network shown in Figure 2.38.

Circuit diagram with resistors R1=10 ohms, R2=20 ohms, R3=30 ohms, R4=40 ohms, R5=50 ohms, and a DC voltage source Vs=100V.

Figure 2.38: Circuit for Example 2.7.4

First find the node voltages to calculate branch currents.

Figure 2.39 displays the chosen reference node and current directions.

From figure, $V_b = V_s = 100$

Circuit diagram with nodes labeled V_a, V_b, V_c, V_d=0, resistors R1 through R5, and a voltage source V_s of 100 V.

From figure, $V_b = 100$

V_a \left(\frac{1}{10} - \frac{1}{20} + \frac{1}{30}\right) - V_c \left(\frac{1}{30}\right) = \frac{100}{20}

0.183V_a - 0.033V_c = 5

\frac{V_a - V_d}{R_1} + \frac{V_a - V_b}{R_2} + \frac{V_a - V_c}{R_3} = 0

\frac{V_a}{R_1} + \frac{V_a - 100}{R_2} + \frac{V_a - V_c}{R_3} = 0

KCL at Node $V_c$ :

$\frac{V_c - V_a}{R_3} + \frac{V_c - V_b}{R_4} + \frac{V_c - V_d}{R_5} = 0$

Substituting $V_d = 0\text{ V}$ and $V_b = 100\text{ V}$ gives:

\frac{V_c - V_a}{R_3} + \frac{V_c - 100}{R_4} + \frac{V_c}{R_5} = 0

V_a \left(\frac{1}{30}\right) - V_c \left(\frac{1}{30} + \frac{1}{40} + \frac{1}{50}\right) = - \frac{100}{40}

0.033V_a - 0.078V_c = -2.5

Solving for $V_a$ and $V_c$ , we get $V_a = 35.84\text{ V}$ and $V_c = 47.21\text{ V}$ .

An ideal voltage source cannot be expressed using Ohm's Law directly because an ideal voltage source has zero internal resistance.

Two-Node Circuit Formulation

KCL at Node 1 ( $V_1$ ):

The sum of currents leaving Node 1 equals zero:

\frac{V_1 - 0}{10} + \frac{V_1 - 100}{20} + \frac{V_1 - V_2}{30} = 0

Multiply the entire equation by $60$ to clear denominators:

6V_1 + 3(V_1 - 100) + 2(V_1 - V_2) = 0

6V_1 + 3V_1 - 300 + 2V_1 - 2V_2 = 0

11V_1 - 2V_2 = 300 \quad - (\text{Equation } 1)

KCL at Node 2 ( $V_2$ ):

The sum of currents leaving Node 2 equals zero:

\frac{V_2 - 0}{50} + \frac{V_2 - 100}{40} + \frac{V_2 - V_1}{30} = 0

Multiply the entire equation by $600$ to clear denominators:

12V_2 + 15(V_2 - 100) + 20(V_2 - V_1) = 0

12V_2 + 15V_2 - 1500 + 20V_2 - 20V_1 = 0

-20V_1 + 47V_2 = 1500 \quad - (\text{Equation } 2)

Circuit diagram with resistors and a DC voltage source

Step 3: Solve the System of Equations

From Equation 1, express $V_2$ in terms of $V_1$ :

2V_2 = 11V_1 - 300 \implies V_2 = 5.5V_1 - 150

Substitute $V_2$ into Equation 2:

-20V_1 + 47(5.5V_1 - 150) = 1500

-20V_1 + 258.5V_1 - 7050 = 1500

-20V_1 + 258.5V_1 - 7050 = 1500

238.5V_1 = 8550

V_1 = \frac{8550}{238.5} \approx 35.85 \text{ V}

Substitute $V_1$ back to find $V_2$ :

V_2 = 5.5(35.85) - 150 \approx 47.17 \text{ V}

Node voltage method with matrix representation

Nodal Voltage Matrix Formulation

In general, for any circuit, the node voltage equation can be written as

V/R=I

G \times V = I_S

where

G = \begin{bmatrix} G_{11} & -G_{12} & \dots & -G_{1n} \\ -G_{21} & G_{22} & \dots & -G_{nn} \\ \dots & \dots & \dots & \dots \\ G_{n1} & -G_{n2} & \dots & G_{nn} \end{bmatrix}

Matrix $G$ is called the conductance matrix.

Conductance

Conductance, $G = \frac{1}{R}$ , is defined as the reciprocal of resistance:

V = \begin{bmatrix} V_1 \\ V_2 \\ \dots \\ V_n \end{bmatrix}

I_S = \begin{bmatrix} I_{S1} \\ I_{S2} \\ \dots \\ I_{Sn} \end{bmatrix}

Comparison of Mesh Current and Node Voltage Matrix Methods

Feature	Mesh Current Matrix Method $[R][I] = [V]$	Node Voltage Matrix Method $[G][V] = [I]$
Governing Law	Kirchhoff's Voltage Law (KVL) around loops	Kirchhoff's Current Law (KCL) at nodes
Unknown Vector $(X)$	Mesh currents $[I_1, I_2, \dots, I_m]^T$	Non-reference node voltages $[V_1, V_2, \dots, V_n]^T$
Coefficient Matrix $(A)$	Resistance Matrix $[R]$ (Units: Ohms, $\Omega$ )	Conductance Matrix $[G]$ (Units: Siemens, $S$ )
Diagonal Elements $(A_{ii})$	Sum of resistances in Mesh $i$ (Always positive)	Sum of conductances connected to Node $i$ (Always positive)
Off-Diagonal Elements ( $A_{ij}$ )	Negative sum of resistances shared between Mesh $i$ and Mesh $j$	Negative sum of conductances connected directly between Node $i$ and Node $j$
Source Vector $(B)$	Net driving voltage in each mesh $[V]$	Net injected current into each node $[I]$

Matrix Formats

Mesh Current Matrix Equation

\begin{bmatrix} R_{11} & R_{12} & \dots & R_{1m} \ R_{21} & R_{22} & \dots & R_{2m} \\\vdots & \vdots & \ddots & \vdots \ R_{m1} & R_{m2} & \dots & R_{mm} \end{bmatrix} \begin{bmatrix} I_1 \ I_2 \ \vdots \ I_m \end{bmatrix} = \begin{bmatrix} V_{s1} \ V_{s2} \ \vdots \ V_{sm} \end{bmatrix}

$R_{ii}$ : Sum of all branch resistances bounding mesh $i$ .
$R_{ij}$ ( $i \neq j$ ): $-(\text{resistance shared by mesh } i \text{ and mesh } j)$ when both mesh currents are defined clockwise.
$V_{si}$ : Sum of independent voltage source rises in the direction of mesh current $i$ .

Node Voltage Matrix Equation

\begin{bmatrix} G_{11} & G_{12} & \dots & G_{1n} \ G_{21} & G_{22} & \dots & G_{2n} \ \vdots & \vdots & \ddots & \vdots \ G_{n1} & G_{n2} & \dots & G_{nn} \end{bmatrix} \begin{bmatrix} V_1 \ V_2 \ \vdots \ V_n \end{bmatrix} = \begin{bmatrix} I_{s1} \ I_{s2} \ \vdots \ I_{sn} \end{bmatrix}

$G_{ii}$ : Sum of conductances ( $1/R$ ) connected to node $i$ .
$G_{ij}$ ( $i \neq j$ ): $-(\text{conductance connected directly between node } i \text{ and node } j)$ .
$I_{si}$ : Net sum of independent source currents entering node $i$ .

Side-by-Side Network Comparison

Mesh Analysis

Mesh analysis circuit diagram showing two loops with currents i1 and i2, resistors R1, R2, R3, and a voltage source E1

\text{KVL: } R_1 i_1 + R_2(i_1 - i_2) = E_1

R_3 i_2 + R_2(i_2 - i_1) = 0

Nodal Analysis

Nodal analysis circuit diagram showing node voltage V, reference ground at 0V, resistors R1, R2, R3, voltage source E1, and branch current direction arrows

\text{KCL: } \frac{V - E_1}{R_1} + \frac{V}{R_2} + \frac{V}{R_3} = 0

Example 2.7.5

Consider the electric network shown in Figure 2.41. Using the conductance matrix method, solve for node voltages $V_1$ and $V_2$ .

Circuit diagram for Example 2.7.5 showing a network with two current sources, three resistors, and labeled node voltages V1, V2, and V3=0.

Figure 2.41: Circuit for Example 2.7.5

Node Voltage Matrix Equation:

\begin{bmatrix} G_{11} & G_{12} & \dots & G_{1n} \\ G_{21} & G_{22} & \dots & G_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ G_{n1} & G_{n2} & \dots & G_{nn} \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \\ \vdots \\ V_n \end{bmatrix} = \begin{bmatrix} I_{s1} \\ I_{s2} \\ \vdots \\ I_{sn} \end{bmatrix}

Solution

In this network, the number of nodes is $3$ .

Therefore, the order of the conductance matrix is $n = 3 - 1 = 2$ .

Therefore, we can write the conductance matrix as described above:

G = \begin{bmatrix} G_{11} & -G_{12} \\ -G_{21} & G_{22} \end{bmatrix}

$G_{11}$ is the sum of conductance connected to node 1, i.e., $G_{11} = \left(\frac{1}{R_1} + \frac{1}{R_2}\right) = \left(1 + \frac{1}{4}\right) = \frac{5}{4}$ .
$G_{12}$ and $G_{21}$ represent the conductance connected to node 1 and node 2, i.e., $G_{12} = G_{21} = \frac{1}{R_2} = \frac{1}{4}$ .
$G_{22}$ is the sum of conductance connected to node 2, i.e., $G_{22} = \left(\frac{1}{R_2} + \frac{1}{R_3}\right) = \left(\frac{1}{4} + \frac{1}{6}\right) = \frac{10}{24} = \frac{5}{12}$ .

Therefore, the conductance matrix is:

Electrical circuit diagram with current sources and resistors

G = \begin{bmatrix} G_{11} & -G_{12} \\ -G_{21} & G_{22} \end{bmatrix} = \begin{bmatrix} \frac{5}{4} & -\frac{1}{4} \\ -\frac{1}{4} & \frac{5}{12} \end{bmatrix}

We can write node voltage equations in matrix form as:

G.V = I_s

Where

G = \begin{bmatrix} \frac{5}{4} & -\frac{1}{4} \\ -\frac{1}{4} & \frac{5}{12} \end{bmatrix}, \quad V = \begin{bmatrix} V_1 \\ V_2 \end{bmatrix}, \quad \text{and } I_s = \begin{bmatrix} I_{S1} \\ I_{S2} \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \end{bmatrix}

Therefore,

\begin{bmatrix} \frac{5}{4} & -\frac{1}{4} \\ -\frac{1}{4} & \frac{5}{12} \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \end{bmatrix}

We can solve for node voltages $V_1$ and $V_2$ using Cramer's rule as discussed in section 2.6.1.

We get, $V_1 = 3.82\text{ V}$ and $V_2 = 7.09\text{ V}$ .

Example: Matrix Node Analysis with Branch Currents

Electrical circuit diagram with current sources, resistors, and node voltages

Conductance Terms ( $G_{ij}$ ):

$G_{11} = \text{Sum of conductances connected to Node } 1 = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{4} + \frac{1}{4} = 1.25\text{ S}$
$G_{22} = \text{Sum of conductances connected to Node } 2 = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{4} + \frac{1}{6} = \frac{5}{12}\text{ S} \approx 0.4167\text{ S}$
$G_{12} = G_{21} = -\text{Conductance between Node } 1 \text{ and Node } 2 = -\frac{1}{R_2} = -\frac{1}{4} = -0.25\text{ S}$

Source Currents ( $I_{si}$ ):

$I_{s1} = \text{Net current entering Node } 1 = 3\text{ A } (\text{from } I_{S1})$
$I_{s2} = \text{Net current entering Node } 2 = -2\text{ A } (\text{since } I_{S2} = 2\text{ A leaves Node } 2)$

The matrix equation:

\begin{bmatrix} 1.25 & -0.25 \\ -0.25 & \frac{5}{12} \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \end{bmatrix} = \begin{bmatrix} 3 \\ -2 \end{bmatrix}

Using Cramer's Rule or matrix inversion:

\Delta = \det(G) = (1.25)\left(\frac{5}{12}\right) - (-0.25)(-0.25) = \frac{5}{4} \cdot \frac{5}{12} - \frac{1}{16} = \frac{25}{48} - \frac{3}{48} = \frac{22}{48}

$\circ$ Solve $V_1$ :

\Delta_1 = \det \begin{bmatrix} 3 & -0.25 \\ -2 & \frac{5}{12} \end{bmatrix} = (3)\left(\frac{5}{12}\right) - (-2)(-0.25) = \frac{5}{4} - \frac{1}{2} = \frac{3}{4}

V_1 = \frac{\Delta_1}{\Delta} = \frac{3/4}{11/24} = \frac{3}{4} \times \frac{24}{11} = \frac{\mathbf{18}}{\mathbf{11}} \text{ V} \approx \mathbf{1.636}\text{ V}

Solve $V_2$ :

\Delta_2 = \det \begin{bmatrix} 1.25 & 3 \\ -0.25 & -2 \end{bmatrix} = (1.25)(-2) - (-0.25)(3) = -2.5 + 0.75 = -1.75

V_2 = \frac{\Delta_2}{\Delta} = \frac{-7/4}{11/24} = -\frac{7}{4} \times \frac{24}{11} = -\frac{\mathbf{42}}{\mathbf{11}} \text{ V} \approx \mathbf{-3.818}\text{ V}

Step 3: Calculate Branch Currents ( $i_1, i_2, i_3$ )

Using the calculated node voltages:

Current $i_1$ (downward through $R_1$ ):

i_1 = \frac{V_1 - 0}{R_1} = \frac{18/11}{1} = \frac{18}{11} \text{ A} \approx \mathbf{1.636} \text{ A}

Current $i_2$ (left-to-right through $R_2$ ):

i_2 = \frac{V_1 - V_2}{R_2} = \frac{\frac{18}{11} - \left(-\frac{42}{11}\right)}{4} = \frac{\frac{60}{11}}{4} = \frac{\mathbf{15}}{\mathbf{11}} \text{ A} \approx \mathbf{1.364} \text{ A}

Current $i_3$ (downward through $R_3$ ):

i_3 = \frac{V_2 - 0}{R_3} = \frac{-42/11}{6} = -\frac{\mathbf{7}}{\mathbf{11}} \text{ A} \approx \mathbf{-0.636} \text{ A}

Supernode analysis and procedure

Supernode Analysis Procedure

Supernode

A supernode is formed when a voltage source (independent or dependent) is connected between two non-reference nodes.

To analyze a circuit containing a supernode, use the following procedure:

Identify the supernode containing the voltage source connected between two non-reference nodes.
Formulate the constraint equation relating the voltages of the two connected nodes.
Enclose the voltage source and apply Kirchhoff's Current Law (KCL) ( $1^{\text{st}}$ statement: the current entering the node is equal to the current leaving the node) across the combined supernode boundary.
Solve the resulting system of equations for the node voltages.

Apply the rule: current entering = current leaving (supernode case).

Circuit diagram illustrating a super node case

Figure 1

Apply the rule: current entering = current leaving (supernode case).

Circuit diagram showing a supernode with two voltage nodes V1 and V2, a 2V independent voltage source between them, resistors of 2 ohms and 1 ohm, current sources of 3A (upward) and 2A (downward), and branch currents I1, I2, I3.

\text{Currents Entering} = \text{Currents Leaving}

3 = \frac{V_1}{2} + \frac{V_2}{1} + 2

Example 1

Step 1: Constraint Equation

The independent voltage source is connected directly between $N_1$ and $N_2$ (positive terminal at $N_1$ ):

V_1 - V_2 = 2 \implies V_1 = V_2 + 2 - (\text{Equation 1})

\begin{aligned} -3 &= \frac{V_1}{2} + \frac{V_2}{1} + 2\\ -3 + \frac{V_1}{2} + \frac{V_2}{1} + 2 &= 0 \end{aligned}

Step 2: KCL at the Supernode ( $N_1 - N_2$ )

Summing all currents leaving the supernode boundary to equal zero:

\frac{V_1}{2} + V_2 - 1 = 0

Multiply the entire equation by 2:

V_1 + 2V_2 = 2 - (\text{Equation 2})

Step 3: Solve for Node Voltages

Substitute $V_1 = V_2 + 2$ from Equation 1 into Equation 2: $(V_2 + 2) + 2V_2 = 2$

3V_2 + 2 = 2 \quad 3V_2 = 0 \implies \mathbf{V_2 = 0\text{ V}}

Substitute $V_2 = 0\text{ V}$ back into Equation 1: $\mathbf{V_1 = 2}$

Step 4: Branch Currents

Using the calculated node voltages:

$I_1$ (through $2\text{ }\Omega$ resistor):

Circuit diagram with nodes N1 and N2, a 3A current source, a 2 ohm resistor, a 1 ohm resistor, and a 2A current source.

Example 2

Apply the rule: current entering = current leaving (supernode case).

Circuit diagram illustrating a super node with voltage and current sources

Figure 1

Step 1: Constraint Equation

v_1 - v_2 = 1 \\[6pt] v_1 = v_2 + 1 \quad - (\text{Equation 1})

Step 2: KCL Equation at the Supernode $(1 - 2)$

Applying KCL (sum of all currents leaving the supernode boundary = 0):

\underbrace{2}_{\text{from } 2\text{A source}} + \underbrace{\frac{v_1}{1}}_{\text{down } 1\Omega \text{ resistor}} + \underbrace{\frac{v_2}{1}}_{\text{down middle } 1\Omega \text{ resistor}} + \underbrace{\frac{v_2 - 2}{1}}_{\text{down right } 1\Omega \text{ branch}} = 0

Step 3: Solve the System of Equations

Substitute Equation 1 ( $v_1 = v_2 + 1$ ) into Equation 2:

(v_2 + 1) + 2v_2 = 0

3v_2 + 1 = 0

3v_2 = -1 \implies \mathbf{v_2 = -\frac{1}{3} \text{ V} \approx -0.333 \text{ V}}

Substitute $v_2 = -\frac{1}{3} \text{ V}$ back into Equation 1:

v_1 = -\frac{1}{3} + 1 \implies \mathbf{v_1 = \frac{2}{3} \text{ V} \approx 0.667 \text{ V}}

Example 3

Apply the rule: current entering = current leaving (supernode case).

Determine the current in the $5\,\Omega$ resistor for the circuit shown in fig.

Circuit diagram with nodes V1, V2, V3, a 10A current source, a 20V independent voltage source, and resistors of 2Ω, 3Ω, 1Ω, 5Ω, and 2Ω.

Step 1: Constraint Equation

V_2 - V_3 = 20 \implies V_2 = V_3 + 20

Step 2: KCL at Node 1 ( $V_1$ )

Summing currents leaving node $V_1$ :

-10 + \frac{V_1}{3} + \frac{V_1 - V_2}{2} = 0

Step 3: KCL at the Supernode ( $2-3$ )

Summing all currents leaving the combined supernode $2-3$ :

\frac{V_2 - V_1}{2} + \frac{V_2}{1} + \frac{V_3 - 10}{5} + \frac{V_3}{2} = 0

Step 4: Calculate Current in the $5\,\Omega$ Resistor

The $5\,\Omega$ resistor is connected between node $V_3$ and the $10\text{ V}$ source. The current $I_{5\,\Omega}$ flowing downwards through the resistor is given by:

I_{5\,\Omega} = \frac{V_3 - 10}{5}

Substitute $V_3 = -8.421\text{ V}$ :

I_{5\,\Omega} = \frac{-8.421 - 10}{5} = \frac{-18.421}{5} = -3.684\text{ A}

Star-delta transformation and bridge circuits

Star and Delta Configurations

Star (Wye, Y) and Delta ( $\Delta$ ) are the two main ways to connect three-phase electrical loads or impedance networks.

T-network circuit diagram

Y-network (Star) circuit diagram

T-network circuit diagram with resistors R_AB, R_AC, R_BC and terminals A, B, C

Delta network circuit diagram with resistors R_AC, R_BC, R_AB and terminals A, B, C forming a triangle

Y-Shaped to Delta-Shaped resistor conversion diagram

In complex bridge circuits, it is often difficult to identify the total equivalent resistance directly using series and parallel reduction rules alone.

Electric circuit diagram showing delta-wye or triangle configuration with nodes A, B, C and resistors RA, RB, RC, RAB, RBC, RCA.

Delta network circuit diagram

(a) Delta network

Star network circuit diagram

(b) Star network

Star-to-Delta and Delta-to-Star Transformations

Star and Delta resistive network transformation diagram and formulas

Delta-to-Star Transformation

R_A = \frac{R_{AB} R_{CA}}{R_{AB} + R_{BC} + R_{CA}}

R_B = \frac{R_{BC} R_{AB}}{R_{AB} + R_{BC} + R_{CA}}

R_C = \frac{R_{CA} R_{BC}}{R_{AB} + R_{BC} + R_{CA}}

Star-to-Delta Transformation

R_{AB} = \frac{R_A R_B + R_B R_C + R_C R_A}{R_C}

R_{BC} = \frac{R_A R_B + R_B R_C + R_C R_A}{R_A}

R_{CA} = \frac{R_A R_B + R_B R_C + R_C R_A}{R_B}

Star and Delta network interconnected with nodes A, B, C and resistors R_A, R_B, R_C, R_AB, R_BC, R_CA

Transformation Summary

DELTA to STAR TRANSFORMATION	STAR to DELTA TRANSFORMATION
$R_A = \frac{R_{AB} R_{CA}}{R_{AB} + R_{BC} + R_{CA}}$ <br><br> $R_B = \frac{R_{BC} R_{AB}}{R_{AB} + R_{BC} + R_{CA}}$ <br><br> $R_C = \frac{R_{CA} R_{BC}}{R_{AB} + R_{BC} + R_{CA}}$	$R_{AB} = \frac{R_A R_B + R_B R_C + R_C R_A}{R_C}$ <br><br> $R_{BC} = \frac{R_A R_B + R_B R_C + R_C R_A}{R_A}$ <br><br> $R_{CA} = \frac{R_A R_B + R_B R_C + R_C R_A}{R_B}$

Worked Examples

Example 1

Use star-delta transformation to find the equivalent resistance between C and D. Also, find the current from the source.

Bridge circuit with resistors and a 50V voltage source

Electrical circuit diagram with a bridge configuration connected to a 50V voltage source

R_A = \frac{R_{AB}R_{CA}}{R_{AB} + R_{BC} + R_{CA}} = \frac{5 \times 3}{5 + 3 + 2} = 1.5\,\Omega

R_B = \frac{R_{BC}R_{AB}}{R_{AB} + R_{BC} + R_{CA}} = \frac{3 \times 2}{5 + 3 + 2} = 0.6\,\Omega

R_C = \frac{R_{CA}R_{BC}}{R_{AB} + R_{BC} + R_{CA}} = \frac{5 \times 2}{5 + 3 + 2} = 1\,\Omega

Electrical bridge circuit diagram with resistors and a DC voltage source

Circuit diagram with a 1 ohm resistor in series with a parallel combination of 6.5 ohm and 8.6 ohm resistors, connected to a 50V DC source.

$R_{eq} = 1 + \frac{6.5 \times 8.6}{6.5 + 8.6} = 4.7\Omega$

\text{Current drawn from the supply } I = \frac{50}{4.7} = 10.64\text{ A}

Example 2

For the network shown in figure, determine the equivalent delta network using delta-star and star-delta transformations and hence find currents $I_1$ , $I_2$ and $I_3$ .

Electrical circuit diagram with three voltage sources, three series resistors, and a star-connected load network.

An electrical circuit diagram with two 100V DC sources, three line resistors of 2Γ each, current indicators I1, I2, I3, and a delta-connected load of three 24Γ resistors between nodes A, B, and C.

Using delta to star transformation:

Circuit diagram showing three-wire supply with 100V sources, 2 ohm resistors in each line, and load resistors of 8 ohms arranged in a bridge network.

Transformed circuit diagram showing a star-connected load with 30 ohm resistors and corresponding equations for currents I1, I2, and I3.

\begin{aligned} I_1 &= \frac{10}{3} + \frac{20}{3} = 10\text{ A} \\ I_2 &= \frac{10}{3} - \frac{10}{3} = 0 \\ I_3 &= -\frac{10}{3} - \frac{20}{3} = -10\text{ A} \end{aligned}

Example 3

Using delta-star transformation, find the voltage $V_{BD}$ .

Circuit diagram containing a bridge network and a DC voltage source

R_A = \frac{R_{AB} R_{CA}}{R_{AB} + R_{BC} + R_{CA}} = \frac{1 \times 2}{1 + 2 + 5} = 0.25\Omega

R_B = \frac{R_{BC} R_{AB}}{R_{AB} + R_{BC} + R_{CA}} = \frac{1 \times 5}{1 + 2 + 5} = 0.625\Omega

R_C = \frac{R_{CA} R_{BC}}{R_{AB} + R_{BC} + R_{CA}} = \frac{5 \times 2}{1 + 2 + 5} = 1.25\Omega

Electrical circuit diagram with a 4V DC source and multiple resistors

Current drawn from source, $I = \dfrac{4}{3 + 0.25 + \dfrac{4.625 \times 7.25}{4.625 + 7.25}} = 0.6586\text{ A}$

Current through $4\,\Omega$ resistance, $I_{4\,\Omega} = \dfrac{7.25}{4.625 + 7.25} \times 0.6586 = 0.402\text{ A}$

$V_{BD} = 4 \times I_{4\,\Omega} = 4 \times 0.402 = 1.608\text{ V}$

Example 2.8.1

Find the total resistance of the circuit between the points A and B of Figure 2.43.

Circuit diagram with resistors connected in series and parallel between points A and B

Figure 2.43

Solution:

The delta CDE of Figure 2.43 is converted into equivalent star as shown in Figure 2.44(a).

R_{CS} = \frac{8 \times 4}{18} = \frac{16}{9} \ \Omega

R_{ES} = \frac{8 \times 6}{18} = \frac{24}{9} \ \Omega

R_{DS} = \frac{6 \times 4}{18} = \frac{12}{9} \ \Omega

Figure 2.44 showing delta to star conversion steps (a) and (b)

Figure 2.44

Figure 2.43 original bridge circuit

Figure 2.43

Blue arrow pointing right

Two parallel resistances between S and B can be simplified to a single resistance of $\frac{35}{9} \ \Omega$ as:

R_{SB} = \left(\frac{12}{9} + 8\right) \parallel \left(\frac{24}{9} + 4\right) = \frac{35}{9} \ \Omega

\therefore \text{Total resistance, } R_{AB} = 4 + \frac{16}{9} + \frac{35}{9} = \frac{87}{9} \ \Omega

Example 2.8.2

Find the resistance between the points A and B in the network shown in Figure 2.45.

Resistor network diagram (Figure 2.45)

Figure 2.45

Circuit diagram of a bridge network with outer delta ABC and inner delta DEF

Circuit diagram for Figure 2.47

Solution:

The inner delta $DEF$ is first transformed to its equivalent star network as given below:

R_D = \frac{3 \times 1}{1 + 1 + 3} = 0.6\;\Omega

R_E = \frac{1 \times 1}{1 + 1 + 3} = 0.2\;\Omega

R_F = \frac{3 \times 1}{1 + 1 + 3} = 0.6\;\Omega

Circuit diagram with star transformation, Figure 2.46

Arrow pointing upward

R_{FE} = 1.6 + 2.2 + \frac{1.6 \times 2.2}{1.6} = 6\;\Omega

R_{DE} = 1.6 + 2.2 + \frac{1.6 \times 2.2}{1.6} = 6\;\Omega

R_{FD} = 1.6 + 1.6 + \frac{1.6 \times 1.6}{2} = 4.36\;\Omega

Resistances $R_{AB}$ , $R_{BC}$ and $R_{AC}$ are given by:

R_{AB} = \frac{6 \times 1}{6 + 1} = 0.86\;\Omega

R_{BC} = \frac{6 \times 2}{6 + 2} = 1.5\;\Omega

R_{AC} = \frac{3 \times 4.36}{3 + 4.36} = 1.78\;\Omega

\therefore \text{ Resistance across } AB = (R_{AC} + R_{BC}) \parallel R_{AB}

= \frac{(1.78 + 1.5) \times 0.8}{(1.78 + 1.5) + 0.8} = 0.64\;\Omega

Example 2.8.3

Determine the equivalent resistance of the network shown in Figure 2.48.

Circuit diagram for Example 2.8.3 with resistors and terminals A and B

Figure 2.48

Solution:

The star network CBE is transformed to its equivalent delta network as shown in Figure 2.49.

Circuit diagram Figure 2.49 showing a bridge network with resistors

Figure 2.49

\begin{aligned} R_{CE} &= 2 + 3 + \frac{2 \times 3}{4} = 6.5 \; \Omega \\ R_{EB} &= 3 + 4 + \frac{4 \times 3}{2} = 13 \; \Omega \\ R_{BC} &= 2 + 4 + \frac{2 \times 4}{3} = 8.66 \; \Omega \end{aligned}

The delta network ACE is transformed to its equivalent star network as shown in Figure 2.50.

Circuit diagram Figure 2.50 showing star-delta transformation

Figure 2.50

\begin{aligned} R_1 &= \frac{4 \times 4}{4 + 2 + 6.5} = 0.64 \; \Omega \\ R_2 &= \frac{6.5 \times 2}{4 + 2 + 6.5} = 1.04 \; \Omega \\ R_3 &= \frac{4 \times 6.5}{4 + 2 + 6.5} = 2.08 \; \Omega \end{aligned}

Two pairs of parallel branches $3 \; \Omega \parallel 8.66 \; \Omega$ and $13 \; \Omega \parallel 1 \; \Omega$ are simplified as shown in Figure 2.51:

3 \; \Omega \parallel 8.66 \; \Omega = \frac{3 \times 8.66}{3 + 8.66} = 2.23 \; \Omega

and $13 \; \Omega \parallel 1 \; \Omega = \frac{13 \times 1}{13 + 1} = 0.93 \; \Omega$

Circuit diagram Figure 2.51 showing simplified equivalent circuit

Figure 2.51

Equivalent resistance between the terminals A and B is:

0.64 + (2.23 + 2.08) \parallel (1.04 + 0.93) = 0.64 + \frac{4.31 \times 1.97}{4.31 + 1.97} = 1.99 \; \Omega

Example 2.8.4

Find the equivalent resistance across the battery terminals

Bridge circuit with resistors and a battery

in the bridge network shown in Figure 2.52.

Bridge network with resistors connected in a diamond shape and a DC voltage source at the bottom.

Figure 2.52

Solution:

The delta network DAC has been transformed to its equivalent star as shown in Figure 2.53:

R_1 = \frac{2 \times 1}{2 + 1 + 1} = 0.5\ \Omega

R_2 = \frac{1 \times 1}{2 + 1 + 1} = 0.25\ \Omega

R_3 = \frac{2 \times 1}{2 + 1 + 1} = 0.5\ \Omega

Equivalent circuit diagram after delta-to-star transformation.

Figure 2.53

Now the total resistance across the points D and B is:

0.5 + (0.25 + 1) \parallel (0.5 + 1) = 0.5 + \frac{1.25 \times 1.5}{1.25 + 1.5} = \frac{13}{11}\ \Omega

Battery symbol

Figure 2.52

Example 2.8.5

Obtain a simplified delta network from the network shown in Figure 2.54.

Circuit diagram of a bridged delta-star network with nodes A, B, C, and D, and resistors of various values.

Figure 2.54

Solution:

The star in the Figure 2.54 is converted to delta and combined with the existing delta as shown in Figure 2.55(a).

Circuit conversion from star-delta to combined delta and simplified delta network

\begin{aligned} R_{AB} &= 6 + 4 + \frac{6 \times 4}{3} = 18\text{ }\Omega\\ R_{AC} &= 6 + 3 + \frac{6 \times 3}{4} = 13.5\text{ }\Omega\\ R_{BC} &= 4 + 3 + \frac{4 \times 3}{6} = 9\text{ }\Omega \end{aligned}

The simplified delta network is shown in figure 2.55(b).

Magnetic circuits: basic terminology and laws

Elementary Concepts of Magnetic Circuits ( $2^{\text{nd}}$ half)

Magnetic circuits basic terminology includes MMF, field strength, flux density, and reluctance, alongside comparisons between electric and magnetic circuits and series/parallel magnetic circuits with composite materials.

Magnetism

A magnet is a material that possesses magnetic properties, attracting pieces of iron and aligning North–South when freely suspended. The phenomenon of attraction is termed magnetism. Magnets are broadly classified into permanent magnets and electromagnets.

Understanding magnetic circuit principles is essential for the design and analysis of electromagnetic devices such as transformers, generators, and motors.

A magnetic field surrounds any magnet. The lines of magnetic field from a bar magnet form continuous, closed loops directed outward from the North Pole and into the South Pole.

Magnetic field around a bar magnet showing closed lines of force originating from the North (N) pole and entering the South (S) pole.

Figure 3.1: Magnetic field around a bar magnet

Magnetic Field Around a Current-Carrying Conductor

A magnetic field is likewise established surrounding any conductor carrying an electric current.

Diagram of a straight vertical conductor carrying current I, surrounded by concentric circular magnetic field lines.

(b)

Magnetic Flux

Magnetic flux is the total number of magnetic lines of force existing in a magnetic field. It is denoted by $\Phi$ and measured in Webers (Wb).

Properties of magnetic flux:

Magnetic flux lines are imaginary lines.
Each line of magnetic flux forms a closed loop.
Magnetic flux lines never intersect each other.
They behave like stretched elastic threads, tending to shorten along their length.
Parallel flux lines acting in the same direction exert a repulsive force on each other.

Magnetic Circuit

A magnetic circuit is the closed path followed by magnetic flux lines.

Magnetic Flux Density

Magnetic flux density is the flux per unit area perpendicular to the direction of the flux. If $\Phi$ is the flux and $A$ is the cross-sectional area, magnetic flux density $B$ is defined as:

B = \frac{\Phi}{A}

The unit of magnetic flux density is $\text{Wb/m}^2$ or Tesla (T).

Magnetomotive Force (MMF)

Magnetomotive force (MMF) is the driving force that establishes magnetic flux in a magnetic circuit. When current passes through an excitation coil, magnetic flux is produced. The product of current ( $I$ ) and the number of turns ( $N$ ) of the coil gives the mmf.

MMF, $F$ is expressed as:

F = NI

The unit of MMF is Ampere-Turns (AT).

Magnetic Field Intensity

Magnetic field intensity is the magnetomotive force per unit length of the magnetic circuit. Denoted by $H$ , its unit is Ampere-Turns per meter (AT/m). It is also known as magnetic field strength or magnetizing force.

H = \frac{\text{mmf}}{l} = \frac{F}{l}

H = \frac{NI}{l}

Or, $F = NI = Hl$ \dots(3.1)

Permeability

Permeability is the ability of a material to conduct magnetic flux through it. Represented by $\mu$ , its unit is Henry per meter (H/m). Flux density $B$ is directly proportional to magnetizing force $H$ :

B \propto H

B = \mu H \tag{3.2}

\mu = \frac{B}{H} \tag{3.3}

where $\mu$ is the permeability of the material. The permeability of free space is denoted by $\mu_o$ , equal to $\mu_o = 4\pi \times 10^{-7}$ H/m.

Relative Permeability

Relative permeability is the ratio of the flux density produced in a given material to the flux density produced in vacuum by the same magnetizing force. When an mmf is applied to ferromagnetic materials, the resulting flux is significantly greater than in air, vacuum, or nonmagnetic materials. It is denoted by $\mu_r$ :

\text{i.e., } \mu_r = \frac{\text{Flux density produced in a material}}{\text{Flux density in vacuum}}

Also:

\mu_r = \frac{\mu}{\mu_0}

This yields the relation $B = \mu_0 \mu_r H$ .

Reluctance

Reluctance is the opposition offered by a material to the creation of magnetic flux. Denoted by $S$ , reluctance is directly proportional to the path length and inversely proportional to the cross-sectional area of the magnetic path, analogous to resistance in an electric circuit:

S \propto \frac{l}{A}

\boxed{S = \frac{l}{\mu_0 \mu_r A}} \tag{3.4}

where:

$l = \text{mean length of magnetic circuit in meter}$
$A = \text{area of cross-section of magnetic circuit in } m^2$
$\mu_0 = \text{permeability of free space} = 4\pi \times 10^{-7} \text{ Henry/meter}$
$\mu_r = \text{relative permeability of magnetic material}$

Thus, reluctance depends directly on the physical dimensions and core material of the magnetic circuit.

Reluctance can also be expressed through circuit variables:

\text{Reluctance} = \frac{\text{mmf}}{\text{Flux}}

i.e.,

$S = \frac{F}{\Phi}$

From expressions 3.1 and 3.2:

F = Hl = \frac{B}{\mu_0 \mu_r} l = \frac{\Phi}{A \mu_0 \mu_r} l \quad \text{since } B = \frac{\Phi}{A}

From equation 3.4, $F = S\Phi$ :

i.e., $\text{mmf} = \text{Reluctance} \times \text{Flux} \text{ (3.5)}$

\boxed{\text{Reluctance, } S = \frac{F}{\Phi} = \frac{NI}{\Phi} = \frac{Hl}{\Phi}}

The unit of reluctance is Ampere-Turns per Weber (AT/Wb).

Comparison of Electric and Magnetic Circuits

Electric and magnetic circuits exhibit direct mathematical analogies as well as key operational differences.

Electric circuit diagram showing a closed loop with a DC voltage source E, a resistor R, and current I flowing clockwise.

Magnetic circuit diagram showing a magnetic core wrapped with a coil carrying current I, producing flux Phi.

Sl. No.	Electric circuit	Magnetic circuit
1	The closed path for electric current is called an electric circuit.	The closed path for magnetic flux is called a magnetic circuit.
2	The cause of current flow is electromotive force (emf).	The cause of generation of flux is magnetomotive force (mmf).
3	Opposition to current flow is offered by resistance $R$ .	Opposition to flux is offered by reluctance $S$ .
4	Current $I = \frac{\text{emf}}{\text{Resistance}}$ is expressed in amperes (A).	Flux $\Phi = \frac{\text{mmf}}{\text{Reluctance}}$ is expressed in webers (Wb).
5	Electromotive force is expressed in volts (V).	Magnetomotive force is expressed in Ampere-Turns (AT).
6	Current density $J = \frac{I}{A}$ is expressed in $A/m^2$ .	Flux density $B = \frac{\Phi}{A}$ is expressed in $Wb/m^2$ .
7	Resistance $R = \frac{l}{\sigma A}$ , where $\sigma$ is conductivity.	Reluctance $S = \frac{l}{\mu A}$ , where $\mu$ is permeability.

Differences Between Electric and Magnetic Circuits

No.	Electric circuit	Magnetic circuit
1	Electric current actually flows via physical charge carrier movement.	Magnetic flux is a field quantity; it does not physically flow like electric charge.
2	Energy is continuously expended as long as current flows in a resistive circuit.	Energy is required only to establish the magnetic field; maintaining a steady field idealizes zero continuous energy loss, though coil resistance consumes power.
3	At constant temperature, resistance is independent of current for an ohmic conductor.	Reluctance depends nonlinearly on flux density because material permeability varies with saturation.
4	Good electrical insulators can virtually eliminate current leakage outside the conductor path.	Magnetic flux cannot be entirely insulated or confined; some leakage flux always exists in surrounding air.

Series and parallel composite magnetic circuits

Series Magnetic Circuit

Consider a simple magnetic circuit shown in Figure 3.7. It consists of a coil of $N$ turns wound over a single core material having uniform cross-sectional area $A$ and mean length of flux path $l$ . mmf is indicated by $NI$ which is equal to $Hl$ . It is represented by $\mathcal{F}$ . Reluctance offered by the circuit to the flow of flux is denoted as $S$ in its electrical equivalent representation. This is useful for simplifying analysis of magnetic circuit even though there is no physical connection between winding and core.

Diagram showing a physical magnetic core with a coil and its electrical equivalent circuit

Figure 3.7: Simple magnetic circuit

Polarity of mmf is determined based on the direction of flux. Here it is in clockwise direction as indicated in figure. mmf source is expressed by $NI$ and mmf drops are expressed by $Hl$ , where $H$ is the magnetizing force on a section of a magnetic circuit and $l$ is the length of the section.

Magnetic circuit with air gap and its equivalent electrical/magnetic circuit diagram

Consider a composite magnetic circuit with an air gap of length $l_g$ . Thus there are two mediums in the magnetic circuit namely (a) iron and (b) air. Magnetic flux is same through these mediums and thus it is a series magnetic circuit. Also total reluctance of the circuit is the sum of reluctances of iron and air.

Parameters for this circuit are defined as follows:

Number of turns: $= N$
Current through the coil (exciting current): $= I$ (A)
Mean length of the flux path through iron: $= l_i$ (m)
Length of the flux path through air gap: $= l_g$ (m)
Cross-sectional area: $= A$ ( $\text{m}^2$ )
Magnetic flux density: $= B$ ( $\text{Wb}/\text{m}^2$ )
Permeability of iron: $= \mu_i$ ( $\text{Henry}/\text{m}^2$ )
Reluctance of iron: $= \mathcal{S}_i$ ( $\text{AT}/\text{Wb}$ )
Reluctance of air gap: $= \mathcal{S}_g$ ( $\text{AT}/\text{Wb}$ )

We can develop equations for this magnetic circuit as:

Total mmf in the circuit,

NI = H_i l_i + H_g l_g

Magnetic field intensity required for iron, $H_i = \dfrac{B}{\mu_0 \mu_i}$

Magnetic field intensity required for air, $H_g = \dfrac{B}{\mu_0}$

Therefore,

NI = H_i l_i + H_g l_g

Derivation steps for the total mmf and flux in a magnetic circuit with an air gap

Role of the Air Gap

An air gap ( $l_g$ ) is intentional in magnetic cores (like inductors, flyback transformers, and DC-biased systems) to control magnetic flux and prevent core saturation.

Prevents Core Saturation: High magnetic permeability in iron cores means they saturate quickly under high current or DC bias. The air gap acts as high magnetic reluctance, flattening the B-H curve so the core can handle higher currents without saturating.
Stores Magnetic Energy: Ferromagnetic core materials hold very little energy, whereas air stores significant magnetic energy. The air gap serves as the primary energy storage region in devices like inductors and flyback transformers ( $W = \frac{1}{2} \text{ }\text{·}\text{ } \frac{B^2}{\mu_0} \text{ }\text{·}\text{ } V_{\text{gap}}$ ).
Stabilizes Inductance: An air gap makes total circuit reluctance dominated by the gap length rather than fluctuations in core permeability ( $\mu_r$ ), keeping inductance ( $L$ ) stable across temperature changes and varying operating currents.
Linearizes the Magnetic Circuit: Air has a completely linear B-H response, reducing harmonic distortion in signal-processing inductors and audio transformers.

Series-Parallel Magnetic Circuits

Now consider a parallel magnetic circuit in which a coil of $N$ turns is wound on an iron core as shown in Figure 3.12.

For analyzing it, we draw its equivalent electrical representation as indicated in Figure 3.13.

Figure 3.12: A series-parallel magnetic circuit showing an iron core with a coil of N turns and magnetic flux paths.

Figure 3.12: A series-parallel magnetic circuit

Figure 3.13: Electrical equivalent circuit of series-parallel magnetic circuit.

Figure 3.13: Electrical equivalent circuit of series-parallel magnetic circuit

We usually refer vertical links of core as limbs and horizontal links as yoke. In the figure $AF$ , $BE$ and $CD$ are the limbs whereas $AB$ , $BC$ , $FE$ and $ED$ are the yokes.

When the coil carries a current $I$ in the direction shown, a magnetic flux $\Phi_1$ will be created in the first limb ( $AF$ ) which will be in the upward direction. Flux $\Phi_1$ will be same along the yoke $AB$ . At point $B$ , there are two parallel paths for flux. Thus $\Phi_1$ will be split into two components $\Phi_2$ and $\Phi_3$ such that $\Phi_1 = \Phi_2 + \Phi_3$ , $\Phi_2$ along the central limb $BE$ and $\Phi_3$ along the yoke $BC$ . Values of $\Phi_1$ and $\Phi_2$ depend on respective reluctances of the paths. $\Phi_2$ and $\Phi_3$ again recombine at point $E$ and completes the path.

\text{Total flux, } \Phi_1 = \Phi_2 + \Phi_3

\text{Total mmf required} = \text{mmf for path EFAB} + \text{mmf for path BE or path BCDE}

NI = \Phi_1 S_1 + \Phi_2 S_2 = \Phi_1 S_1 + \Phi_3 S_3

Some other forms of series-parallel magnetic circuits and their electric equivalent representations are shown below.

Magnetic core (a) with Gap a and Gap b and its magnetic circuit equivalent (b)

Figure 3.14: A series-parallel magnetic circuit and its electric-equivalent representation

Magnetic core with two windows and two windings (a)

Magnetic core with center gap and side gap (a) and its equivalent magnetic circuit (b)

Magnetic circuit equivalent with two mmf sources and two loops

Series-parallel magnetic circuit and its electric-equivalent

Basic circuit elements: Resistors, Inductors and Capacitors

Syllabus: Elementary Concepts of DC Electric Circuits

Resistors

V-I Characteristics of a Resistor

V-I Relation of a Capacitor

Inductors

V-I Characteristics and Switching Behavior of an Inductor

Energy Stored in an Inductor

Independent and dependent voltage and current sources

Behavior Under Load

Circuit Conditions for Voltage Sources

Behavior on Short Circuit

Behavior on Open Circuit

Voltage Source Characteristics

V-I Characteristic

Behavior on Open Circuit

V-I Characteristic

Circuit fundamentals: SI units, definitions and Ohm's law

SI units of electrical quantities

Series and parallel connections

Series and parallel resistor combinations

Resistances in Series

Resistances in Parallel

Voltage divider and current divider rules

Applications of Ohm's Law

General Formula (NNN Resistors)

Derivation of Voltage Divider Rule

Derivation of Current Divider Rule

Important Notes

Worked Examples

Source transformation and relative potential

Source transformation

Voltage source to current source

Current source to voltage source

Network topology definitions

Circuit versus network

Potential and reference levels

Kirchhoff's laws: KVL, KCL and network loops

Finding Loops in a Circuit

Voltage Polarities and Sign Conventions

Formulating KVL Equations

Single-Mesh Analysis

Two-Mesh Analysis

Solving the System of Simultaneous Equations

Three-Mesh Analysis

Supermesh analysis and numerical problems

Supermesh Analysis Procedure

Worked Numerical Problems

Step B: Constraint Equation for the Supermesh

Step C: Solving the System of Equations

Solving the System of Equations

Final Currents

Mesh current method with matrix representation

Matrix Method of Analysis

Matrix Formulation Rules

Node voltage method and numerical problems

Homework Problem

Kirchhoff's Current Law (KCL)

Node Voltage Method

Procedural Steps

Numerical Examples

Example 1

Example 2.7.1

Example 2.7.2

Example 2.7.4

Two-Node Circuit Formulation

Node voltage method with matrix representation

Nodal Voltage Matrix Formulation

Comparison of Mesh Current and Node Voltage Matrix Methods

Matrix Formats

Side-by-Side Network Comparison

Mesh Analysis

Nodal Analysis

Example 2.7.5

Example: Matrix Node Analysis with Branch Currents

Step 3: Calculate Branch Currents (i1,i2,i3i_1, i_2, i_3i1​,i2​,i3​)

Supernode analysis and procedure

Supernode Analysis Procedure

Example 1

Step 1: Constraint Equation

General Formula ( $N$ Resistors)

Step 3: Calculate Branch Currents ( $i_1, i_2, i_3$ )

Step 2: KCL at the Supernode ( $N_1 - N_2$ )

Step 2: KCL Equation at the Supernode $(1 - 2)$

Step 2: KCL at Node 1 ( $V_1$ )

Step 3: KCL at the Supernode ( $2-3$ )

Step 4: Calculate Current in the $5\,\Omega$ Resistor

Elementary Concepts of Magnetic Circuits ( $2^{\text{nd}}$ half)