A-Level — physicsMedium3 min read

Kirchhoff's Laws and Circuit Analysis

Kirchhoff's first and second laws; solving complex DC circuits

Tutor AI Team
Tutor AI Team

# Kirchhoff's Laws and Circuit Analysis — A-Level Physics

Kirchhoff's two laws provide the foundation for analysing any electrical circuit, no matter how complex.

1. Kirchhoff's First Law (Junction Rule)

The total current entering a junction equals the total current leaving it.

∑Iin=∑Iout\sum I_{\text{in}} = \sum I_{\text{out}}

This is based on conservation of charge — charge cannot accumulate at a junction.

2. Kirchhoff's Second Law (Loop Rule)

The sum of EMFs around any closed loop equals the sum of potential differences (voltage drops) around that loop.

∑ε=∑IR\sum \varepsilon = \sum IR

This is based on conservation of energy — the total energy gained per unit charge equals the total energy lost.

3. Series and Parallel Rules

Series: RT=R1+R2+...R_T = R_1 + R_2 + ...; current same everywhere; voltage shared.

Parallel: 1RT=1R1+1R2+...\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + ...; voltage same across all; current splits.

4. Solving Complex Circuits

Step-by-Step Method

  1. Label all currents with directions (if wrong, answer will be negative)
  2. Apply Kirchhoff's First Law at each junction
  3. Apply Kirchhoff's Second Law around each independent loop
  4. Follow sign conventions: EMF positive if traversed from − to +; IRIR drop is positive if traversed in current direction
  5. Solve the simultaneous equations

Worked Example: Two-Loop Circuit

Problem

A 12 V battery with internal resistance 1 Ω is connected to two resistors: 5 Ω and 20 Ω in parallel.

Rparallel=(5×20)/(5+20)=100/25=4R_{\text{parallel}} = (5 \times 20)/(5+20) = 100/25 = 4 Ω Rtotal=1+4=5R_{\text{total}} = 1 + 4 = 5 Ω Itotal=12/5=2.4I_{\text{total}} = 12/5 = 2.4 A Vacross parallel=I×Rp=2.4×4=9.6V_{\text{across parallel}} = I \times R_p = 2.4 \times 4 = 9.6 V I5Ω=9.6/5=1.92I_{5Ω} = 9.6/5 = 1.92 A; I20Ω=9.6/20=0.48I_{20Ω} = 9.6/20 = 0.48 A Check: 1.92+0.48=2.41.92 + 0.48 = 2.4 A ✓

Solution

Worked Example: Two EMF Sources

Problem

Two batteries (ε₁ = 6V, r₁ = 1Ω and ε₂ = 4V, r₂ = 2Ω) are connected in opposition with a 5Ω external resistor.

Loop: ε1−ε2=I(r1+r2+R)\varepsilon_1 - \varepsilon_2 = I(r_1 + r_2 + R) 6−4=I(1+2+5)=8I6 - 4 = I(1 + 2 + 5) = 8I I=0.25I = 0.25 A (in direction of the larger EMF)

Terminal PD of ε₁: V=ε1−Ir1=6−0.25=5.75V = \varepsilon_1 - Ir_1 = 6 - 0.25 = 5.75 V Terminal PD of ε₂: V=ε2+Ir2=4+0.5=4.5V = \varepsilon_2 + Ir_2 = 4 + 0.5 = 4.5 V (being charged)

Solution

6. Practice Questions

    1. At a junction, currents of 3 A and 5 A flow in. One current of 6 A flows out. Find the fourth current. (1 mark)
    1. A 9 V battery (internal resistance 0.5 Ω) is connected to a 4 Ω and 6 Ω resistor in series. Calculate the current and voltage across each resistor. (3 marks)
    1. Two 1.5 V cells (each with 0.3 Ω internal resistance) are connected in series with a 2.4 Ω lamp. Find the current and power dissipated in the lamp. (4 marks)

    Answers

    1. 3+5=6+I43 + 5 = 6 + I_4 → I4=2I_4 = 2 A (out).
    1. RT=0.5+4+6=10.5R_T = 0.5 + 4 + 6 = 10.5 Ω. I=9/10.5=0.857I = 9/10.5 = 0.857 A. V4Ω=3.43V_{4Ω} = 3.43 V, V6Ω=5.14V_{6Ω} = 5.14 V. Check: 3.43+5.14+0.43=9.03.43 + 5.14 + 0.43 = 9.0 V ✓

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Summary

  • K1 (junction): ∑Iin=∑Iout\sum I_{\text{in}} = \sum I_{\text{out}} (charge conservation)
  • K2 (loop): ∑ε=∑IR\sum \varepsilon = \sum IR (energy conservation)
  • Apply to each loop and junction to get simultaneous equations
  • Combine series/parallel rules for simpler circuits
Tags:
#a-level#physics#electricity#kirchhoffs-laws#circuit-analysis

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