Kirchhoff's Laws and Circuit Analysis

Kirchhoff's two laws are the foundation of all circuit analysis. They are based on two fundamental conservation laws — conservation of charge and conservation of energy. Together, they allow us to solve any circuit, no matter how complex.

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Kirchhoff's First Law (KCL — Current Law)

Statement: The sum of the currents entering any junction equals the sum of the currents leaving that junction.

Equivalently: the algebraic sum of currents at any junction is zero.

$\sum I_{in} = \sum I_{out}$∑Iin​=∑Iout​

or

$\sum I = 0 \text{ (using sign convention: positive for currents entering, negative for leaving)}$∑I=0 (using sign convention: positive for currents entering, negative for leaving)

Physical basis: Conservation of charge. Charge cannot accumulate at a junction — what flows in must flow out.

Worked Example 1 — KCL at a Junction

At a junction, three wires carry currents of 3.0 A, 5.0 A, and 2.0 A. The first two are entering the junction. What is the current in the third wire, and in which direction?

Solution:

Sum entering = 3.0 + 5.0 = 8.0 A

Current leaving through wire 3 = 8.0 − 2.0 = 6.0 A (leaving)

Wait — let me re-read. If the third wire carries 2.0 A, and the first two carry 3.0 A and 5.0 A entering, then:

If the third wire is leaving: we need I₃ such that 3.0 + 5.0 = I₃

Actually, there must be more wires. Let me restate:

At a junction, four wires meet. I₁ = 3.0 A (entering), I₂ = 5.0 A (entering), I₃ = 2.0 A (leaving). Find I₄.

Currents in = Currents out: 3.0 + 5.0 = 2.0 + I₄

I₄ = 8.0 − 2.0 = 6.0 A (leaving the junction)

Worked Example 2 — Multiple Junctions

In a circuit, 2.0 A flows from A to B through a 10 Ω resistor. At junction B, the current splits: 1.2 A goes through a 15 Ω resistor. What current flows through the other branch?

Solution:

By KCL at junction B: 2.0 = 1.2 + I₂

I₂ = 2.0 − 1.2 = 0.8 A

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Kirchhoff's Second Law (KVL — Voltage Law)

Statement: The sum of the EMFs around any closed loop in a circuit equals the sum of the potential differences (IR drops) around that loop.

$\sum \varepsilon = \sum IR$∑ε=∑IR

Or equivalently: the algebraic sum of all voltages (EMFs and p.d.s) around any closed loop is zero.

Physical basis: Conservation of energy. The total energy supplied per coulomb (EMFs) equals the total energy dissipated per coulomb (p.d.s) around any closed path.

Sign Convention for KVL

When traversing a loop in a chosen direction:

• EMF is positive if you pass through the source from − to + (in the direction of the EMF)
• EMF is negative if you pass from + to − (against the EMF)
• IR drop is positive if you traverse in the direction of current flow
• IR drop is negative if you traverse against the current flow

Worked Example 3 — Single Loop

A circuit contains a 12 V battery with internal resistance 0.5 Ω, connected to two resistors in series: R₁ = 4.5 Ω and R₂ = 7.0 Ω. Find the current.

Solution:

Applying KVL around the loop:

ε = IR₁ + IR₂ + Ir

12 = I(4.5 + 7.0 + 0.5) = I × 12.0

I = 12 / 12.0 = 1.0 A

p.d. across R₁ = 1.0 × 4.5 = 4.5 V p.d. across R₂ = 1.0 × 7.0 = 7.0 V p.d. across r = 1.0 × 0.5 = 0.5 V

Check: 4.5 + 7.0 + 0.5 = 12.0 V = ε ✓

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Solving Multi-Loop Circuits

For circuits with multiple loops and/or multiple sources, use the following systematic approach:

Step-by-Step Method

1. Label all currents with assumed directions (if you guess wrong, the answer will be negative — that's fine)
2. Apply KCL at each junction to relate the currents
3. Apply KVL around enough independent loops to generate enough equations
4. Solve the simultaneous equations

Worked Example 4 — Two-Loop Circuit

Consider a circuit with two batteries and three resistors:

• Left loop: Battery ε₁ = 6.0 V with R₁ = 2.0 Ω
• Right loop: Battery ε₂ = 4.0 V with R₂ = 3.0 Ω
• Shared branch (middle): R₃ = 5.0 Ω
• Both batteries have the positive terminal at the top

Let I₁ flow clockwise in the left loop through R₁, I₂ flow clockwise in the right loop through R₂, and I₃ flow downward through R₃.

KCL at the top junction:

I₁ = I₂ + I₃ ... (assuming I₁ enters, I₂ and I₃ leave)

Wait — let me set this up more carefully.

Define: I₁ flows through ε₁ and R₁ (left branch), I₂ flows through ε₂ and R₂ (right branch), I₃ flows through R₃ (middle branch).

At the top junction: I₁ = I₃ + I₂ → I₃ = I₁ − I₂ ... (1)