Skip to content
Dabour HubLEARN · PRACTICE · EXPLORESign inStudent

4.3.4Electrons, Waves and Photons

Kirchhoff’s Laws in Circuits

5 min read 5 parts Lesson 30 of 62 Unit quiz

0%read

Section 1 of 5The Two Laws

4.3.4 Kirchhoff's Laws in Circuits

Together the laws allow any resistor circuit with known values to be analysed step by step.

First law
  • The total current entering any junction equals the total current leaving it, because charge is conserved.
Second law
  • Around any closed loop the sum of the e.m.f.s equals the sum of the p.d.s, because energy is conserved.
Junction
  • A junction is any point where three or more conductors meet and current can split or combine.
Partnership
  • The first law deals with currents at points and the second law with p.d.s around loops, so problems often need both.

Section 2 of 5Problem-Solving Method

1
Redraw the circuit and label every known current, resistance and e.m.f. directly on the diagram.
2
Combine resistors in series or parallel to find the total resistance seen by the source.
3
Find the main current from the e.m.f. divided by the total resistance, assuming negligible internal resistance.
4
Use the second law to find the p.d. across the parallel section, then find each branch current from I = V/R.
5
Check the first law at the junction: the branch currents must add up to the main current.

Section 3 of 5Typical Errors Avoided

Typical Errors Avoided
Branch currents
  • The current is shared between parallel branches and is not the same in each unless the resistances are equal.
Series p.d.s
  • The p.d.s across series components differ unless their resistances are equal, and they sum to the supply value.
Total resistance
  • Always use the series rule R = R₁ + R₂ and the parallel rule 1/R = 1/R₁ + 1/R₂ correctly.
Annotation
  • Writing values on the diagram as you go reveals missing information and prevents slips.

Section 4 of 5Worked Example

Question

  • A 12 V supply (negligible internal resistance) is connected to a 4.0 Ω resistor in series with a parallel pair of 6.0 Ω and 3.0 Ω resistors.
  • Find the current in each resistor.

  1. Parallel pair1/R = 1/6.0 + 1/3.0 = 1/2.0, so R = 2.0 Ω.
  2. Total resistance = 4.0 + 2.0 = 6.0 Ω, so the main current I = 12 / 6.0 = 2.0 A through the 4.0 Ω resistor.
  3. p.d. across 4.0 Ω = 2.0 × 4.0 = 8.0 V, so the parallel pair has 12 − 8.0 = 4.0 V (second law).
  4. Branch currents4.0 / 6.0 = 0.67 A and 4.0 / 3.0 = 1.33 A, which add to 2.0 A (first law).

Answer

  • 4.0 Ω: 2.0 A;
    • 6.0 Ω: 0.67 A;
    • 3.0 Ω: 1.3 A.

Section 5 of 5Exam Checklist & Tips

4.3.4 – Key Points for the Exam

  • Kirchhoff's first law The sum of currents into a junction equals the sum of currents out.
  • Junction A point in a circuit where conductors meet and current may divide.
  • Branch A path between two junctions that carries its own current.
  • Total resistance The single resistance that would draw the same current from the source.

Common Mistakes

  • Assuming equal current in each parallel branch whatever the resistances.
  • Adding parallel resistances directly instead of using reciprocals.
  • Forgetting to subtract the p.d. across the series resistor before finding the parallel p.d.
  • Not checking that branch currents sum to the main current.
\n