B5-4. Kirchhoff's Laws and Circuit Topologies

1. Circuit Topologies and Equivalent Resistance

Real-world circuits consist of webs of components. By applying fundamental conservation laws, we can collapse massive networks into equivalent resistances.

Series Circuits:

There is only one topological path for charge. Voltage divides across the components.

$$R_{\text{eq}} = R_1 + R_2 + R_3 + \dots$$

Parallel Circuits:

Multiple paths exist between two shared nodes. Voltage is identical, but the current divides.

$$\dfrac{1}{R_{\text{eq}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + \dots$$

2. Kirchhoff's Circuit Laws

For complex multi-loop circuits (like bridge circuits or multi-battery setups), simple series/parallel rules completely fail. We must apply Kirchhoff's laws as rigorous systems of equations.

Kirchhoff's First Law (Current / Junction Rule):

A direct consequence of the Conservation of Charge. The sum of all currents entering any node must exactly equal the sum of currents leaving it.

$$\sum I = 0$$

Kirchhoff's Second Law (Voltage / Loop Rule):

A direct consequence of the Conservation of Energy. In any closed loop in a circuit, the algebraic sum of the EMFs provided must equal the sum of the potential drops ($IR$) across components.

$$\sum \Delta V = 0 \quad \text{or} \quad \sum \varepsilon = \sum I R$$

Example 1: Rigorous Multi-Loop Analysis

Problem: A circuit contains two adjacent loops. The left loop is powered by a $12$ V battery, and the right loop by an opposing $6$ V battery. They share a central branch containing a $4$ Ω resistor. The outer branches each contain a $2$ Ω resistor. Find the total current flowing down the shared central branch.

Solution:

Step 1: Assign loop currents. Let $I_1$ flow clockwise in the left loop, and $I_2$ flow anti-clockwise in the right loop. Both merge and flow down the central $4$ Ω resistor ($I_{\text{center}} = I_1 + I_2$).
Step 2: Apply KVL to the Left Loop.
$12 = 2(I_1) + 4(I_1 + I_2) \quad \Rightarrow \quad 12 = 6I_1 + 4I_2 \quad \text{--- (Eq 1)}$
Step 3: Apply KVL to the Right Loop.
$6 = 2(I_2) + 4(I_1 + I_2) \quad \Rightarrow \quad 6 = 4I_1 + 6I_2 \quad \text{--- (Eq 2)}$
Step 4: Solve the simultaneous equations.
From Eq 2: $I_1 = \dfrac{6 - 6I_2}{4} = 1.5 - 1.5I_2$.
Substitute into Eq 1: $12 = 6(1.5 - 1.5I_2) + 4I_2 \quad \Rightarrow \quad 12 = 9 - 9I_2 + 4I_2$
$3 = -5I_2 \quad \Rightarrow \quad I_2 = -0.6$ A. (The negative sign simply means it flows clockwise).
Calculate $I_1$: $I_1 = 1.5 - 1.5(-0.6) = 1.5 + 0.9 = 2.4$ A.
Step 5: Calculate the central current.
$I_{\text{center}} = I_1 + I_2 = 2.4 - 0.6 = \mathbf{1.8 \text{ A}}$.

3. The Potential Divider

A series circuit forces a total voltage to divide proportionally across its components based on their resistance. This is exceptionally useful for sensor logic circuits.

The Potential Divider Formula:

$$V_{\text{out}} = V_{\text{in}} \left( \dfrac{R_2}{R_1 + R_2} \right)$$

Example 2: Logic Sensor Design

Problem: A $9.0$ V supply is connected in series with a fixed $5.0$ kΩ resistor ($R_1$) and a thermistor ($R_2$). At room temperature, the thermistor has a resistance of $15.0$ kΩ. In boiling water, its resistance drops to $1.0$ kΩ. Calculate the voltage output across the thermistor ($V_{\text{out}}$) in both environments.