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A Level · Cambridge (CIE) · 9702

Apply Kirchhoff’s laws to junctions and loops

Use charge conservation at junctions and energy conservation around complete circuit loops.

D.C. circuits · Cambridge International AS & A Level Physics 9702 · 10.2

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

What physical conservation principle underlies Kirchhoff’s first law?

What you will learn

  • Recall Kirchhoff's first law and relate it to conservation of charge
  • Recall Kirchhoff's second law and relate it to conservation of energy
  • Use signed current and potential changes to solve simple circuits

The junction law is conservation of charge

Kirchhoff's first law states that the total current entering a junction equals the total current leaving it. This is a consequence of conservation of charge: charge cannot build up indefinitely at an ideal circuit junction, so the rate of charge flow in balances the rate flowing out.

Choose a direction for each branch current. If a solved current is negative, the actual direction is opposite to the chosen arrow. This sign convention lets one equation handle unknown branch directions consistently.

The loop law is conservation of energy

Kirchhoff's second law states that the algebraic sum of potential changes around a complete closed loop is zero. This is a consequence of conservation of energy: energy supplied per charge by sources balances energy transferred per charge by components around that loop.

Choose a loop direction and record source rises and component drops with consistent signs. A component drop is V = IR for an ohmic resistor. The sum of rises equals the sum of drops for the selected loop.

junction: ΣI entering = ΣI leaving · closed loop: ΣΔV = 0

Pause and explain

What should the signed potential changes around a closed loop sum to?

Put the idea to work

Worked example

At a junction, 2.0 A enters and 0.70 A and I leave. Find I. In a loop with a 12 V source and 4.0 V and V drops, find V.

Show the worked solution
  1. Junction: 2.0 = 0.70 + I, so I = 1.3 A.
  2. Loop: 12 − 4.0 − V = 0.
  3. Thus V = 8.0 V.

Answer I = 1.3 A; V = 8.0 V.

Try a topic test

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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.

1 · Guided

At a junction, 1.8 A enters. Two outgoing branch currents are 0.45 A and I. Find I.

Give me a hint

Set total entering current equal to total leaving current.

Compare my reasoning
  1. 1.8 = 0.45 + I.
  2. I = 1.8 − 0.45.
  3. State the branch direction as leaving, as drawn.

I = 1.35 A leaving the junction.

Look for these in your work

  • Junction equation written
  • Unknown isolated
  • Direction interpreted
2 · Independent

A 9.0 V source drives a series loop with drops of 2.5 V and V. Find V using Kirchhoff’s second law.

Give me a hint

The signed rise and drops around the loop sum to zero.

Compare my reasoning
  1. 9.0 − 2.5 − V = 0.
  2. V = 9.0 − 2.5.
  3. Check the drops total the source rise.

V = 6.5 V.

Look for these in your work

  • Loop signs consistent
  • Unknown p.d. found
  • Energy balance checked
3 · Transfer

At a junction 0.60 A enters; two currents labelled leaving are 0.90 A and I. Find I and explain its sign.

Give me a hint

Use the directions shown in the signed junction equation.

Compare my reasoning
  1. Write 0.60 = 0.90 + I.
  2. I = −0.30 A.
  3. The negative sign means actual current is 0.30 A entering, opposite the drawn leaving arrow.

I = −0.30 A relative to the chosen arrow; actual current is 0.30 A into the junction.

Look for these in your work

  • Signed equation correct
  • Negative value retained
  • Physical direction explained

Common mistakes

  • Adding currents without distinguishing entering from leaving
  • Mixing potential rises and drops with inconsistent signs
  • Interpreting a negative solved current as impossible instead of reversed direction
Recall without your notes

What can you explain now?

What physical conservation principle underlies Kirchhoff’s first law?

Compare with the explanation

Conservation of charge

Current entering balances current leaving at a junction.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

How do the two Kirchhoff laws express separate conservation ideas?

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.

Keep learning Physics

Course scope and assessment

One hundred and sixty-two original lessons cover all 12 published subject-content outcomes in Topic 1 (sections 1.1–1.4), all nine outcomes in Topic 2 (section 2.1), all 13 outcomes in Topic 3 (sections 3.1–3.3), all 13 outcomes in Topic 4 (sections 4.1–4.3), all 11 outcomes in Topic 5 (sections 5.1–5.2), all 10 outcomes in Topic 6 (sections 6.1–6.2), all 16 outcomes in Topic 7 (sections 7.1–7.5), all 12 outcomes in Topic 8 (sections 8.1–8.4), all 15 outcomes in Topic 9 (sections 9.1–9.3), all 16 outcomes in Topic 10 (sections 10.1–10.3), all 18 outcomes in Topic 11 (sections 11.1–11.2), all seven outcomes in Topic 12 (sections 12.1–12.2), all 12 outcomes in Topic 13 (sections 13.1–13.4), all eight outcomes in Topic 14 (sections 14.1–14.3), and all nine outcomes in Topic 15 (sections 15.1–15.3), plus all four outcomes in Topic 16 (sections 16.1–16.2), and all ten outcomes in Topic 17 (sections 17.1–17.3), plus all 12 outcomes in Topic 18 (sections 18.1–18.5), all nine outcomes in Topic 19 (sections 19.1–19.3), all 19 outcomes in Topic 20 (sections 20.1–20.5), and all eight outcomes in Topic 21 (sections 21.1–21.2), and all 17 outcomes in Topic 22 (sections 22.1–22.4), and all 13 outcomes in Topic 23 (sections 23.1–23.2), and all 16 outcomes in Topic 24 (sections 24.1–24.3), and all 11 outcomes in Topic 25 (sections 25.1–25.3), of the 2025–2027 syllabus. Each lesson has guided, independent and application practice with explained answers. All 25 syllabus topics now have mapped teaching and practice; the 2028–2030 changes outside Topics 23–25 still need reconciliation. Online study cannot demonstrate supervised Paper 3 laboratory competence. Paper 5 planning and analysis, full timed cumulative assessment and human subject-teacher review remain outstanding.

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