Resistive Networks

A resistive network is a circuit composed solely of resistors, voltage sources, and current sources. The primary goal of analysis is to determine the voltage across and current through every element.

Ohm’s Law

The relationship between voltage ( $V$ ), current ( $I$ ), and resistance ( $R$ ) is linear:

$V = I \cdot R$

Unit: Ohms ( $\Omega$ )
Power Dissipation: $P = V \cdot I = I^2 \cdot R = \frac{V^2}{R}$ (measured in Watts)

Kirchhoff’s Laws

Kirchhoff’s Current Law (KCL): The algebraic sum of currents entering a node is zero. $\sum I_{in} = \sum I_{out}$
Kirchhoff’s Voltage Law (KVL): The algebraic sum of all voltages around any closed loop in a circuit is zero. $\sum V = 0$

Resistor Combinations

To simplify complex circuits, we combine resistors into equivalent resistances ( $R_{eq}$ ).

\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}

Configuration	Formula	Characteristics
Series	$R_{eq} = R_1 + R_2 + ... + R_n$	Same current flows through each resistor.
Parallel	$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}$	Same voltage across each resistor.

Voltage and Current Division

These rules allow for quick calculations without solving full systems of equations.

Voltage Divider (Series)

Used to find the voltage across a specific resistor in a series string:

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

Current Divider (Parallel)

Assume there are two resistors $R_1$ and $R_2$ connected in parallel. The current through $R_1$ is:

$I_1 = I_{total} \cdot \left( \frac{R_2}{R_1 + R_2} \right)$

Network Analysis Techniques

Nodal Analysis

Based on KCL. You select a reference node (ground) and solve for the voltages at the remaining nodes.

Best for: Circuits with many parallel branches and current sources.

Mesh Analysis

Based on KVL. You assign “mesh currents” to the smallest loops in a circuit and solve for them.

Best for: Planar circuits with many series components and voltage sources.

Problem: Find the total current ( $I_t$ ) supplied by a $12\text{V}$ battery connected to a series-parallel combination where $R_1 = 4\Omega$ is in series with a parallel pair of $R_2 = 6\Omega$ and $R_3 = 3\Omega$ .

Step 1: Simplify the parallel branch ( $R_2$ and $R_3$ )

$R_p = \frac{R_2 \cdot R_3}{R_2 + R_3} = \frac{6 \cdot 3}{6 + 3} = \frac{18}{9} = 2\Omega$

Step 2: Find the equivalent resistance ( $R_{eq}$ )

$R_{eq} = R_1 + R_p = 4\Omega + 2\Omega = 6\Omega$

Step 3: Apply Ohm’s Law

$I_t = \frac{V}{R_{eq}} = \frac{12\text{V}}{6\Omega} = 2\text{A}$

Key Theorems for Simplification

Thevenin’s Theorem: Any linear circuit can be replaced by an equivalent voltage source ( $V_{th}$ ) in series with a resistor ( $R_{th}$ ).
Norton’s Theorem: Any linear circuit can be replaced by an equivalent current source ( $I_n$ ) in parallel with a resistor ( $R_n$ ).
Maximum Power Transfer: Maximum power is delivered to a load when the load resistance ( $R_L$ ) equals the Thevenin resistance ( $R_{th}$ ) of the source.