Solve simple resistor networks with Kirchhoff’s laws
Combine branch rules and signed loop equations to solve a two-loop circuit consistently.
AS Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 10. These original circuit diagrams and calculations are study aids; full timed assessment, supervised practical competence and human subject-teacher review remain outstanding.
Before you begin
- Use current, potential difference, resistance, energy and charge with consistent circuit directions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How should you treat a solved negative branch current?
What you will learn
- Use junction and loop equations for a simple multi-branch circuit
- Select consistent current directions and interpret negative solutions
- Check circuit solutions against conservation laws
Choose current directions before writing equations
For a simple multi-branch circuit, assign a direction to each unknown current. At a junction write the incoming-current sum equal to the outgoing-current sum. A negative solution means the actual current is opposite to the assumed arrow.
For each independent loop, choose a traversal direction and add source rises and resistor drops with signs. Use V = IR for ohmic components at the stated condition. The current through a shared branch appears in more than one loop equation.
Use more than one conservation check
After solving, check every junction equation and loop equation. Also verify that component p.d.s and current directions are consistent with the resistor values and source polarity.
Do not count a redundant loop equation as independent if it is just the sum or difference of equations already used. A simple network needs enough independent equations to determine its unknown branch currents.
Pause and explain
Two parallel branches carry 0.40 A and 0.70 A away from a junction. What enters from the source?
Worked example
A 6.0 V source is across two parallel resistors, 6.0 Ω and 3.0 Ω. Find each branch current and total source current.
Show the worked solution
- Both branches have 6.0 V across them.
- I₆ = 6.0/6.0 = 1.0 A; I₃ = 6.0/3.0 = 2.0 A.
- Junction law gives total current 1.0 + 2.0 = 3.0 A.
Answer Branch currents are 1.0 A and 2.0 A; total current is 3.0 A.
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Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A 12 V source is connected across parallel 4.0 Ω and 12 Ω branches. Find branch currents and total current.
Give me a hint
Use the common branch p.d. with Ohm's law, then add currents at the junction.
Compare my reasoning
- I₄ = 12/4.0 = 3.0 A.
- I₁₂ = 12/12 = 1.0 A.
- I total = 3.0 + 1.0 = 4.0 A.
Branch currents are 3.0 A and 1.0 A; total current is 4.0 A.
Look for these in your work
- Branch p.d. used
- Each current calculated
- Junction sum checked
A 10 V source has a 2.0 Ω series resistor and a 3.0 Ω resistor in the same loop. Find current and check the loop p.d.s.
Give me a hint
Add series resistances or write the signed loop equation directly.
Compare my reasoning
- Rtotal = 2.0 + 3.0 = 5.0 Ω.
- I = 10/5.0 = 2.0 A.
- Drops are 4.0 V and 6.0 V; they sum to 10 V.
I = 2.0 A; the resistor drops are 4.0 V and 6.0 V.
Look for these in your work
- Total resistance formed
- Current found
- Loop balance checked
At a junction I₁ enters and I₂ plus 0.25 A leave. A loop equation gives 6.0 − 2I₁ − 4I₂ = 0; a second gives I₁ = 2I₂. Solve for both currents and verify the junction balance.
Give me a hint
Substitute the current relation into both conservation equations.
Compare my reasoning
- Junction: I₁ = I₂ + 0.25.
- With I₁ = 2I₂, obtain I₂ = 0.25 A and I₁ = 0.50 A.
- Loop: 6.0 − 2(0.50) − 4(0.25) = 4.0 V, so these supplied equations are inconsistent.
The equations as stated have no simultaneous solution: the junction gives I₂ = 0.25 A and I₁ = 0.50 A, but the loop residual is 4.0 V, not zero. Recheck the circuit values or equation signs.
Look for these in your work
- Junction relation solved
- Loop substituted
- Inconsistent data detected
Common mistakes
- Assuming equal currents in branches with different resistances
- Changing a current arrow midway through an equation without changing its sign
- Checking only the final answer and not the junction and loop balances
What can you explain now?
How should you treat a solved negative branch current?
Compare with the explanation
It flows opposite to the assumed arrow
The sign corrects the chosen direction.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Why is it useful to check a proposed circuit solution with both junction and loop equations?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.