12.1 DC Circuit Analysis & Network Theorems

12.1 DC Circuit Analysis & Network Theorems

Direct current (DC) circuit theory is the cornerstone of electrical engineering practice and a heavily weighted topic on the PRC Registered Electrical Engineer (REE) Licensure Examination. Mastering fundamental circuit laws, systematic formulation methods (nodal and mesh analysis), and network reduction theorems enables candidates to solve complex resistive networks efficiently under board exam conditions.

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1. Fundamental Circuit Laws & Energy Relations

Electric circuit analysis relies on charge conservation and energy conservation laws.

Ohm's Law & Power Dissipation

For a linear bilateral resistor of resistance $R$ (Ohms, $\Omega$), the voltage $V$ (Volts, $\text{V}$) across its terminals is directly proportional to the current $I$ (Amperes, $\text{A}$) passing through it:

$V = I R \quad \implies \quad I = \frac{V}{R} \quad \text{and} \quad R = \frac{V}{I}$

Power $P$ (Watts, $\text{W}$) dissipated as heat in a resistor is expressed by Joule's Law:

$P = V I = I^2 R = \frac{V^2}{R}$

$\text{Energy (Joules, J)} = W = \int_{0}^{t} P(\tau) d\tau = P \cdot t$

Kirchhoff's Current Law (KCL)

KCL (Conservation of Charge): The algebraic sum of currents entering any node (junction) in an electrical circuit is identically equal to zero:

$\sum_{k=1}^{n} I_k = 0 \quad \implies \quad \sum I_{\text{entering}} = \sum I_{\text{leaving}}$

Kirchhoff's Voltage Law (KVL)

KVL (Conservation of Energy): The algebraic sum of all potential differences (voltage sources and voltage drops) around any closed loop in a circuit is identically equal to zero:

$\sum_{k=1}^{m} V_k = 0 \quad \implies \quad \sum V_{\text{rises}} = \sum V_{\text{drops}}$

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2. Nodal Voltage & Mesh Current Analysis Methods

Systematic formulation techniques transform complex multi-loop circuits into sets of simultaneous linear algebraic equations solvable via matrix inversion or Cramer's Rule.

Nodal Voltage Analysis

Nodal analysis uses KCL at ungrounded essential nodes with node voltages as variables:

1. Select a reference node (ground, $0\ \text{V}$).
2. Assign node voltage variables ($V_1, V_2, \dots, V_N$) relative to reference.
3. Apply KCL at each non-reference node using branch currents expressed via Ohm's Law: $I_{ab} = \frac{V_a - V_b}{R_{ab}}$.

$\begin{bmatrix} G_{11} & -G_{12} & \dots & -G_{1N} \\ -G_{21} & G_{22} & \dots & -G_{2N} \\ \vdots & \vdots & \ddots & \vdots \\ -G_{N1} & -G_{N2} & \dots & G_{NN} \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \\ \vdots \\ V_N \end{bmatrix} = \begin{bmatrix} I_{s1} \\ I_{s2} \\ \vdots \\ I_{sN} \end{bmatrix}$

Supernode Technique: When an independent or dependent voltage source is connected directly between two non-reference nodes without series resistance, enclose the source and its nodes inside a supernode surface and write a combined KCL equation.

Mesh Current Analysis

Mesh analysis applies KVL around planar mesh loops using loop currents ($i_1, i_2, \dots, i_M$):

$\begin{bmatrix} R_{11} & -R_{12} & \dots & -R_{1M} \\ -R_{21} & R_{22} & \dots & -R_{2M} \\ \vdots & \vdots & \ddots & \vdots \\ -R_{M1} & -R_{M2} & \dots & R_{MM} \end{bmatrix} \begin{bmatrix} i_1 \\ i_2 \\ \vdots \\ i_M \end{bmatrix} = \begin{bmatrix} v_{s1} \\ v_{s2} \\ \vdots \\ v_{sM} \end{bmatrix}$

Supermesh Technique: When a current source is shared between two adjacent meshes, create a supermesh by temporarily removing the current source branch and applying KVL around the outer perimeter loop.

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3. Fundamental Network Reduction Theorems

Thévenin's Theorem

Any linear two-terminal DC network containing independent or dependent sources and linear resistors can be replaced across terminals $a\text{--}b$ by an equivalent circuit consisting of an open-circuit voltage source $V_{\text{th}}$ in series with a Thévenin equivalent resistance $R_{\text{th}}$.

• $V_{\text{th}}$ (Thévenin Voltage): The open-circuit voltage measured across terminals $a\text{--}b$ ($V_{\text{th}} = V_{ab,\text{oc}}$).
• $R_{\text{th}}$ (Thévenin Resistance): The input resistance looking into terminals $a\text{--}b$ with all independent voltage sources replaced by short circuits ($0\ \text{V}$) and all independent current sources replaced by open circuits ($0\ \text{A}$).

$R_{\text{th}} = \frac{V_{ab,\text{oc}}}{I_{ab,\text{sc}}}$

Norton's Theorem

Any linear two-terminal DC network can be replaced across terminals $a\text{--}b$ by an equivalent ideal current source $I_N$ connected in parallel with Norton resistance $R_N$.

• $I_N$ (Norton Current): The short-circuit current flowing from terminal $a$ to terminal $b$ when terminals are shorted ($I_N = I_{ab,\text{sc}}$).
• $R_N$ (Norton Resistance): Equal to Thévenin resistance ($R_N = R_{\text{th}}$).

$\text{Source Transformation: } V_{\text{th}} = I_N \cdot R_{\text{th}} \quad \iff \quad I_N = \frac{V_{\text{th}}}{R_{\text{th}}}$

Maximum Power Transfer Theorem

Maximum power is delivered from a linear network with Thévenin equivalent $V_{\text{th}}$ and $R_{\text{th}}$ to an adjustable load resistor $R_L$ when the load resistance equals the Thévenin internal resistance:

$R_L = R_{\text{th}}$

Under maximum power transfer, the maximum power absorbed by $R_L$ is:

$P_{\max} = I_L^2 R_L = \left( \frac{V_{\text{th}}}{R_{\text{th}} + R_L} \right)^2 R_L = \left( \frac{V_{\text{th}}}{2 R_{\text{th}}} \right)^2 R_{\text{th}} = \frac{V_{\text{th}}^2}{4 R_{\text{th}}}$

$\text{Efficiency at Max Power: } \eta = \frac{P_{\text{load}}}{P_{\text{total}}} = \frac{I_L^2 R_L}{I_L^2 (R_{\text{th}} + R_L)} = \frac{R_{\text{th}}}{2 R_{\text{th}}} = 50\%$

Superposition Theorem

In any linear network containing multiple independent sources, the response (voltage or current) across any branch equals the algebraic sum of the individual responses caused by each independent source acting alone, with all other independent voltage sources shorted and current sources opened.

Caution: Superposition applies strictly to linear variables (voltage and current), not to power ($P = I^2 R$), which is a quadratic non-linear relationship.

Millman's Theorem

For $N$ parallel voltage sources ($V_1, V_2, \dots, V_N$) with internal series resistances ($R_1, R_2, \dots, R_N$), the common terminal voltage $V_m$ is given directly by:

$V_m = \frac{\sum_{i=1}^{N} \frac{V_i}{R_i}}{\sum_{i=1}^{N} \frac{1}{R_i}} = \frac{\sum_{i=1}^{N} I_{si}}{\sum_{i=1}^{N} G_i}$

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4. Delta-Wye ($\Delta\text{--}Y$) & Wye-Delta ($Y\text{--}\Delta$) Transformations

When resistors form bridge networks that are neither in series nor parallel, Delta-Wye transformation formulas reduce the circuit.

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Solved Board Exam Examples

Example 1: Thévenin Equivalent & Maximum Power Calculation

Problem: A DC active network connected across terminals $a\text{--}b$ consists of a $60\ \text{V}$ independent voltage source in series with a $10\ \Omega$ resistor, parallel-connected across a branch containing a $20\ \Omega$ resistor and a $30\ \Omega$ resistor. Determine: (a) Thévenin voltage $V_{\text{th}}$, (b) Thévenin resistance $R_{\text{th}}$, and (c) maximum power $P_{\max}$ delivered to an adjustable load resistor $R_L$.

Solution:

1. Calculate open-circuit voltage $V_{\text{th}} = V_{ab,\text{oc}}$ across the $30\ \Omega$ resistor using voltage division: $V_{\text{th}} = 60\ \text{V} \times \frac{30\ \Omega}{10\ \Omega + 20\ \Omega + 30\ \Omega} = 60 \times \frac{30}{60} = 30\ \text{V}$
2. Calculate $R_{\text{th}}$ looking into $a\text{--}b$ with the $60\ \text{V}$ source shorted. Resistors $(10 + 20) = 30\ \Omega$ are in parallel with the $30\ \Omega$ resistor: $R_{\text{th}} = (10\ \Omega + 20\ \Omega) \parallel 30\ \Omega = \frac{30 \times 30}{30 + 30} = 15\ \Omega$
3. Set $R_L = R_{\text{th}} = 15\ \Omega$ for maximum power transfer: $P_{\max} = \frac{V_{\text{th}}^2}{4 R_{\text{th}}} = \frac{(30)^2}{4 \times 15} = \frac{900}{60} = 15\ \text{W}$

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Example 2: Wheatstone Bridge Resistance via Delta-Wye Conversion

Problem: A Wheatstone bridge circuit has four arm resistors $R_1 = 10\ \Omega$, $R_2 = 20\ \Omega$, $R_3 = 30\ \Omega$, $R_4 = 60\ \Omega$, and a detector resistor $R_5 = 30\ \Omega$ connected across the center nodes. Calculate the total equivalent input resistance $R_{\text{eq}}$ of the bridge network.

Solution:

1. Identify upper Delta formed by $R_1 = 10\ \Omega$, $R_2 = 20\ \Omega$, and $R_5 = 30\ \Omega$ ($R_a = 10$, $R_b = 20$, $R_c = 30$).
2. Compute sum of Delta resistors: $R_{\text{sum}} = 10 + 20 + 30 = 60\ \Omega$.
3. Convert Delta to equivalent Wye branches $R_A, R_B, R_C$: $R_A = \frac{10 \times 20}{60} = 3.333\ \Omega, \quad R_B = \frac{10 \times 30}{60} = 5.0\ \Omega, \quad R_C = \frac{20 \times 30}{60} = 10.0\ \Omega$
4. Combine series branches with lower resistors $R_3 = 30\ \Omega$ and $R_4 = 60\ \Omega$: $\text{Left Branch} = R_B + R_3 = 5.0 + 30 = 35.0\ \Omega$ $\text{Right Branch} = R_C + R_4 = 10.0 + 60 = 70.0\ \Omega$
5. Combine parallel branches and add stem resistor $R_A$: $R_{\text{parallel}} = 35.0 \parallel 70.0 = \frac{35 \times 70}{35 + 70} = \frac{2450}{105} = 23.333\ \Omega$ $R_{\text{eq}} = R_A + R_{\text{parallel}} = 3.333 + 23.333 = 26.67\ \Omega$