Network Laws and Theorems

LessonAbout 30 min

Network laws and theorems provide systematic techniques to analyze complex multi-loop electric circuits.

1. Kirchhoff’s Laws and Network Analysis

1.1 Kirchhoff’s Voltage Law

Kirchhoff’s voltage law states that the algebraic sum of electric potential changes around any closed circuit loop is zero.

Loop rule: algebraic sum of voltages around closed path equals zero
Source sign: positive exiting positive terminal, negative exiting negative terminal
Drop sign: negative in current direction, positive opposite current direction

Kirchhoff’s Voltage Law

\[ \sum V = 0 \qquad\qquad \sum E = \sum I R \]

EE Board Exam August 1977

Example. In a DC network, a branch has resistor \(R_2 = 1\ \Omega\) in series with battery \(V_B = 120\ \text{V}\) carrying \(I_2 = 2\ \text{A}\), in parallel with resistor \(R_3 = 3\ \Omega\). If generator branch has \(R_1 = 1\ \Omega\), find generated EMF \(E_g\).

Solution.

\[ V = V_B + I_2 R_2 = 120 + 2(1) = 122\ \text{V} \]

\[ I_1 = I_2 + \dfrac{V}{R_3} = 2 + \dfrac{122}{3} = 42.67\ \text{A} \]

\[ E_g = V + I_1 R_1 = 122 + (42.67)(1) = 164.67\ \text{V} \]

Answer: \(164.67\ \text{V}\)

1.2 Kirchhoff’s Current Law

Kirchhoff’s current law states that the algebraic sum of currents entering any circuit node is zero.

Junction rule: total current entering node equals total current leaving node
Sign convention: currents towards node are positive, currents away are negative
Charge conservation: electric charge cannot accumulate at a circuit node

Kirchhoff’s Current Law

\[ \sum I_{\text{entering}} = \sum I_{\text{leaving}} \qquad\qquad \sum I = 0 \]

Example. A battery charger of \(130\ \text{V}\) with \(3\text{-}\Omega\) internal resistance is connected in parallel with a \(120\text{-}\text{V}\) battery of \(2\text{-}\Omega\) internal resistance, supplying a \(40\text{-}\Omega\) load. Solve for battery current \(I_b\).

Solution.

\[ \dfrac{130 - V_L}{3} + \dfrac{120 - V_L}{2} = \dfrac{V_L}{40} \implies V_L = 120.43\ \text{V} \]

\[ I_b = \dfrac{120 - V_L}{2} = \dfrac{120 - 120.43}{2} = -0.215\ \text{A} \]

Answer: \(-0.215\ \text{A}\)

1.3 Mesh Method

Mesh method analyzes planar networks by assigning independent loop currents to each circuit mesh.

Planar circuit: network drawn on plane without crossing conductors
Mesh current: fictitious circulating current assumed around independent loop
System order: equation count equals number of independent meshes

Mesh Current Equations

\[ R_{11} I_A - R_{12} I_B = E_1 \]

\[ -R_{21} I_A + R_{22} I_B = E_2 \]

Where: \(R_{11}\) = total resistance around mesh 1, \(R_{12}\) = mutual resistance

Example. In a two-mesh network with \(6\text{-}\text{A}\) and \(3\text{-}\text{A}\) current sources, resistors \(2\ \Omega\), \(8\ \Omega\), and \(2\ \Omega\) are connected around mesh B. Find voltage drop \(V\) across the \(8\text{-}\Omega\) resistor using mesh analysis.

Solution.

\[ (6 - I_B)(2) - I_B(8) - (I_B + 3)(2) = 0 \]

\[ 12 - 2 I_B - 8 I_B - 2 I_B - 6 = 0 \implies I_B = 0.5\ \text{A} \]

\[ V = I_B R = (0.5\ \text{A})(8\ \Omega) = 4\ \text{V} \]

Answer: \(4.0\ \text{V}\)

1.4 Nodal Method

Nodal method determines essential node voltages relative to a chosen reference node using KCL.

Reference node: zero-potential datum node chosen for circuit analysis
Node voltage: potential of essential node measured relative to reference
Equation count: requires \((n - 1)\) KCL equations for \(n\) nodes

Nodal Voltage Equations

\[ \dfrac{V_a - E_1}{R_1} + \dfrac{V_a}{R_2} + \dfrac{V_a - V_b}{R_3} = 0 \]

Example. Find voltage \(V\) across a \(4\text{-}\Omega\) resistor connected at a node linked to a \(12\text{-}\text{V}\) source via \(4\ \Omega\), a \(4\text{-}\text{A}\) current source, and a dependent source \(4I\).

Solution.

\[ I = \dfrac{12 - V}{4} \qquad\qquad \text{and} \qquad\qquad I_1 = \dfrac{V}{4} \]

\[ I + 4 = I_1 + 4I \implies 4 = \dfrac{V}{4} + 3\left(\dfrac{12 - V}{4}\right) \]

\[ 4 = 0.25 V + 9 - 0.75 V \implies 0.5 V = 5 \implies V = 10\ \text{V} \]

Answer: \(10\ \text{V}\)

2. Network Theorems

2.1 Superposition Theorem

Superposition theorem states that response in a linear circuit equals the algebraic sum of responses from each independent source acting alone.

Linear network: circuit composed of constant parameter linear elements
Voltage source inactive: deactivated independent voltage source becomes short circuit
Current source inactive: deactivated independent current source becomes open circuit

Superposition Response

\[ I = I' + I'' + \dots + I^{(n)} \]

\[ V = V' + V'' + \dots + V^{(n)} \]

Example. A network contains a \(6\text{-}\text{V}\) voltage source, a \(3\text{-}\text{A}\) current source, and a dependent source \(3V_1\). Find current in the \(6\text{-}\Omega\) resistor using superposition.

Solution.

\[ I' \text{ (voltage source alone)}: 6 + I'(1) - 6 I' + 3 I' = 0 \implies I' = 3\ \text{A} \]

\[ I'' \text{ (current source alone)}: 3 = \dfrac{V_1}{1} + \dfrac{V_1 + 3V_1}{6} \implies V_1 = 1.8\ \text{V} \implies I'' = 1.2\ \text{A} \]

\[ I = I' + I'' = 3\ \text{A} + 1.2\ \text{A} = 4.2\ \text{A} \]

Answer: \(4.2\ \text{A}\)

1.2 Thevenin’s Equivalent Circuit

Thevenin’s theorem reduces any two-terminal linear network to an equivalent voltage source in series with an equivalent resistance.

Thevenin voltage (\(E_o\)): open-circuit voltage across load terminals
Thevenin resistance (\(R_o\)): resistance looking back with independent sources deactivated
Equivalent current: load current determined by single-loop series equivalent

Thevenin Equivalent Circuit

\[ I_L = \dfrac{E_o}{R_o + R_L} \]

Where: \(E_o\) = open-circuit voltage, \(R_o\) = equivalent resistance looking back

EE Board Exam April 1980

Example. A \(12\text{-}\text{V}\) DC source energizes a bridge having arms of \(20\ \Omega\), \(10\ \Omega\), \(40\ \Omega\), and \(30\ \Omega\). Determine current through a \(50\text{-}\Omega\) detector using Thevenin’s theorem.

Solution.

\[ R_o = \dfrac{(20)(40)}{20 + 40} + \dfrac{(10)(30)}{10 + 30} = 13.333 + 7.5 = 20.833\ \Omega \]

\[ E_o = \left(\dfrac{12}{40}\right)(10) - \left(\dfrac{12}{60}\right)(20) = 3 - 4 = -1\ \text{V} \]

\[ I = \dfrac{|E_o|}{R_o + R_L} = \dfrac{1\ \text{V}}{20.833\ \Omega + 50\ \Omega} = 0.014\ \text{A} \]

Answer: \(0.014\ \text{A}\)

1.3 Norton’s Equivalent Circuit

Norton’s theorem reduces any two-terminal linear network to an equivalent current source in parallel with an equivalent resistance.

Norton current (\(I_{sc}\)): short-circuit current between output terminals
Norton resistance (\(R_o\)): resistance looking back into deactivated network terminals
Dual relation: \(E_o = I_{sc} R_o\) relates Thevenin and Norton equivalents

Norton Equivalent Circuit

\[ I_L = I_{sc} \dfrac{R_o}{R_o + R_L} \]

Example. A circuit has a \(4\text{-}\text{A}\) current source, a \(4\text{-}\Omega\) resistor in series with \(12\ \text{V}\), and a \(1\text{-}\Omega\) load in series with \(20\ \text{V}\). Determine current in the \(1\text{-}\Omega\) load using Norton’s theorem.

Solution.

\[ R_o = 4\ \Omega \]

\[ I_{sc} = 4 - \dfrac{20 - 12}{4} = 4 - 2 = 2\ \text{A} \]

\[ I_L = I_{sc} \dfrac{R_o}{R_o + R_L} = 2\ \text{A} \left(\dfrac{4\ \Omega}{4\ \Omega + 1\ \Omega}\right) = 1.6\ \text{A} \]

Answer: \(1.6\ \text{A}\)

1.4 Maximum Power Transfer Theorem

Maximum power transfer theorem states that maximum power is transferred to a load when load resistance equals internal resistance looking back.

Matching condition: load resistance \(R_L\) equals source resistance \(R_o\)
Efficiency at maximum power: power transfer efficiency equals \(50\%\)
Maximum power value: depends directly on square of open-circuit voltage

Maximum Power Transfer

\[ R_L = R_o \]

\[ P_{\max} = \dfrac{E_o^2}{4 R_o} \]

Example. A \(120\text{-}\text{V}\) DC generator has internal resistance of \(0.5\ \Omega\) and supplies load through line conductors of total resistance \(9.5\ \Omega\). Determine load resistance \(R_L\) for maximum power transfer and maximum power delivered.

Solution.

\[ R_L = R_o = r + R_{\text{line}} = 0.5\ \Omega + 9.5\ \Omega = 10\ \Omega \]

\[ I_L = \dfrac{E_o}{R_o + R_L} = \dfrac{120\ \text{V}}{10\ \Omega + 10\ \Omega} = 6\ \text{A} \]

\[ P_{\max} = I_L^2 R_L = (6\ \text{A})^2 (10\ \Omega) = 360\ \text{W} \]

Answer: \(360\ \text{watts}\)

Formula Summary

Formula Name	Formula
Kirchhoff’s Voltage Law	\(\sum V = 0 \quad \sum E = \sum I R\)
Kirchhoff’s Current Law	\(\sum I_{\text{entering}} = \sum I_{\text{leaving}} \quad \sum I = 0\)
Mesh Current Equations	\(R_{11} I_A - R_{12} I_B = E_1 \quad -R_{21} I_A + R_{22} I_B = E_2\)
Nodal Voltage Equations	\(\dfrac{V_a - E_1}{R_1} + \dfrac{V_a}{R_2} + \dfrac{V_a - V_b}{R_3} = 0\)
Superposition Response	\(I = I' + I'' + \dots + I^{(n)} \quad V = V' + V'' + \dots + V^{(n)}\)
Thevenin Equivalent Circuit	\(I_L = \dfrac{E_o}{R_o + R_L}\)
Norton Equivalent Circuit	\(I_L = I_{sc} \dfrac{R_o}{R_o + R_L}\)
Maximum Power Transfer	\(R_L = R_o \quad P_{\max} = \dfrac{E_o^2}{4 R_o}\)