Kirchhoff's Laws

Overview

Kirchhoff's laws are fundamental tools for analyzing complex circuits. They are based on conservation of charge and conservation of energy.

Kirchhoff's Current Law (KCL)

Statement

The sum of currents entering a node equals the sum of currents leaving that node.

$$\sum I_{\text{in}} = \sum I_{\text{out}}$$

Or equivalently:

$$\sum I = 0 \quad \text{(at any node)}$$

Physical Basis

Conservation of charge: charge cannot accumulate at a node.

Application

At any junction in a circuit, assign current directions and write:

• Currents entering: positive
• Currents leaving: negative

Kirchhoff's Voltage Law (KVL)

Statement

The sum of all voltages around any closed loop equals zero.

$$\sum V = 0 \quad \text{(around any closed loop)}$$

Physical Basis

Conservation of energy: a charge gains as much energy as it loses in traversing a closed path.

Sign Conventions

Traversing the loop:

• Through resistor in current direction: $-IR$ (voltage drop)
• Through resistor against current: $+IR$ (voltage rise)
• Through battery from − to +: $+\varepsilon$ (voltage rise)
• Through battery from + to −: $-\varepsilon$ (voltage drop)

Applying Kirchhoff's Laws

Step-by-Step Method

1. Label all currents with directions (if wrong, answer will be negative)
2. Label all nodes
3. Write KCL equations at nodes ($n-1$ equations for $n$ nodes)
4. Write KVL equations for independent loops
5. Solve the system of equations

Number of Equations Needed

• If $b$ = number of branches, $n$ = number of nodes
• Need $b$ equations total
• KCL gives $(n-1)$ independent equations
• KVL gives $(b - n + 1)$ independent equations

Multi-Loop Circuit Analysis

Independent Loops

A loop is independent if it contains at least one branch not in other loops.

Mesh Currents

Assign a current to each mesh (inner loop), circulating in same direction (usually clockwise).

Examples

Example 1: Simple KCL

At a node: $I_1 = 3$ A enters, $I_2 = 1$ A enters, $I_3$ leaves. Find $I_3$.

$$I_1 + I_2 = I_3$$ $$I_3 = 3 + 1 = 4 \text{ A}$$

Example 2: Two-Loop Circuit

$\varepsilon_1 = 12$ V, $\varepsilon_2 = 6$ V, $R_1 = 4$ Ω, $R_2 = 2$ Ω, $R_3 = 3$ Ω

Loop 1 (left): $\varepsilon_1 - I_1R_1 - I_3R_3 = 0$

Loop 2 (right): $\varepsilon_2 - I_2R_2 - I_3R_3 = 0$

Node: $I_1 + I_2 = I_3$

Substituting:

$$12 - 4I_1 - 3I_3 = 0$$ $$6 - 2I_2 - 3I_3 = 0$$ $$I_1 + I_2 = I_3$$

From equations 1 and 2:

$$4I_1 + 3I_3 = 12 \to 4I_1 + 3(I_1 + I_2) = 12 \to 7I_1 + 3I_2 = 12$$ $$2I_2 + 3I_3 = 6 \to 2I_2 + 3(I_1 + I_2) = 6 \to 3I_1 + 5I_2 = 6$$

Solving:

$$I_1 = 1.5 \text{ A}, \quad I_2 = 0.3 \text{ A}, \quad I_3 = 1.8 \text{ A}$$

Example 3: Three Resistors with Two Batteries

$\varepsilon_1 = 10$ V, $\varepsilon_2 = 5$ V, $R_1 = 2$ Ω, $R_2 = 4$ Ω, $R_3 = 6$ Ω

Using mesh currents $I_1$ (left loop), $I_2$ (right loop):

Loop 1: $\varepsilon_1 - I_1R_1 - (I_1 - I_2)R_2 = 0$

$$10 - 2I_1 - 4(I_1 - I_2) = 0$$ $$10 - 6I_1 + 4I_2 = 0$$ $$6I_1 - 4I_2 = 10 \quad \text{...(1)}$$

Loop 2: $-\varepsilon_2 - I_2R_3 - (I_2 - I_1)R_2 = 0$

$$-5 - 6I_2 - 4(I_2 - I_1) = 0$$ $$-5 + 4I_1 - 10I_2 = 0$$ $$4I_1 - 10I_2 = 5 \quad \text{...(2)}$$

From (1): $I_1 = (10 + 4I_2)/6$

Substituting into (2):

$$4\frac{10 + 4I_2}{6} - 10I_2 = 5$$ $$40 + 16I_2 - 60I_2 = 30$$ $$-44I_2 = -10$$ $$I_2 = 0.227 \text{ A}, \quad I_1 = 1.82 \text{ A}$$

Example 4: Wheatstone Bridge

Find condition for balanced bridge (no current through galvanometer).

When balanced:

$$\frac{R_1}{R_2} = \frac{R_3}{R_4}$$

Or:

$$R_1 R_4 = R_2 R_3$$

Example 5: Current Divider Verification

Total current $I = 6$ A splits between $R_1 = 4$ Ω and $R_2 = 12$ Ω in parallel.

Using KCL and KVL:

$$I_1 + I_2 = 6$$ $$I_1 R_1 = I_2 R_2 \quad \text{(same voltage)}$$ $$4I_1 = 12I_2$$ $$I_1 = 3I_2$$

Solving:

$$3I_2 + I_2 = 6$$ $$I_2 = 1.5 \text{ A}, \quad I_1 = 4.5 \text{ A}$$

Verify with current divider formula:

$$I_1 = I \times \frac{R_2}{R_1 + R_2} = 6 \times \frac{12}{16} = 4.5 \text{ A} \checkmark$$

Problem-Solving Tips

1. Redraw circuit if needed for clarity
2. Assume current directions (can be arbitrary)
3. Be consistent with sign conventions
4. Check answers by verifying KVL around loops
5. Power delivered should equal power consumed

Common Mistakes to Avoid

• Wrong sign when traversing voltage source
• Forgetting internal resistance
• Using wrong current in shared branches
• Not enough independent equations