IB — physicsMedium2 min read

Series and Parallel Circuits and Kirchhoff's Laws

Series and parallel resistor rules; Kirchhoff's junction and loop laws; solving DC circuits

Tutor AI Team
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# Series and Parallel Circuits and Kirchhoff's Laws — IB Physics

1. Series Circuits

  • Same current through all components: IT=I1=I2I_T = I_1 = I_2
  • Voltages add: VT=V1+V2+...V_T = V_1 + V_2 + ...
  • Resistances add: RT=R1+R2+...R_T = R_1 + R_2 + ...

2. Parallel Circuits

  • Same voltage across all branches: VT=V1=V2V_T = V_1 = V_2
  • Currents add: IT=I1+I2+...I_T = I_1 + I_2 + ...
  • 1RT=1R1+1R2+...\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + ...

3. Kirchhoff's Laws

First Law (Junction): ∑Iin=∑Iout\sum I_{\text{in}} = \sum I_{\text{out}} (conservation of charge)

Second Law (Loop): ∑ε=∑IR\sum \varepsilon = \sum IR (conservation of energy)

4. Potential Divider

Vout=Vin×R2R1+R2V_{\text{out}} = V_{\text{in}} \times \frac{R_2}{R_1 + R_2}

With sensors (thermistor, LDR): variable output voltage for monitoring conditions.

Worked Example: Example 1

Problem

3 Ω and 6 Ω in parallel: 1/R=1/3+1/6=3/6=1/21/R = 1/3 + 1/6 = 3/6 = 1/2 → R=2R = 2 Ω. In series with 4 Ω: total = 6 Ω. With 12 V battery: I=12/6=2I = 12/6 = 2 A. Vparallel=2×2=4V_{\text{parallel}} = 2 \times 2 = 4 V. I3Ω=4/3=1.33I_{3Ω} = 4/3 = 1.33 A. I6Ω=4/6=0.67I_{6Ω} = 4/6 = 0.67 A.

Solution

Worked Example: Potential Divider

Problem

9 V supply, 3 kΩ and 6 kΩ. VoutV_{\text{out}} across 6 kΩ: Vout=9×6/9=6V_{\text{out}} = 9 \times 6/9 = 6 V.

Solution

6. Practice Questions

    1. Three resistors (2Ω, 3Ω, 6Ω) are in parallel. Find total resistance. (2 marks)
    1. A 6V battery drives 4Ω and 8Ω in series. Find current and PD across each. (3 marks)
    1. Explain how a potential divider with an LDR can be used as a light sensor. (3 marks)

    Answers

    1. 1/R=1/2+1/3+1/6=3/6+2/6+1/6=11/R = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 1 → R=1R = 1 Ω.
    1. I=6/12=0.5I = 6/12 = 0.5 A. V4Ω=2V_{4Ω} = 2 V. V8Ω=4V_{8Ω} = 4 V.

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Summary

  • Series: same II, VV adds, RR adds
  • Parallel: same VV, II adds, 1/R1/R adds
  • K1: charge conservation at junctions; K2: energy conservation around loops
  • Potential divider: Vout=VinR2/(R1+R2)V_{\text{out}} = V_{\text{in}} R_2/(R_1+R_2)
Tags:
#ib#physics#electricity#circuits#kirchhoffs-laws#series#parallel

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