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Using Kirchhoff's Laws On A Single Loop Circuit

Key idea: Solve a single-loop circuit with multiple sources and internal resistances using Kirchhoff’s second law, then find V_ab and power supplied/absorbed by each source (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Recall circuit symbols and draw or interpret circuit diagrams.
  • Draw circuit diagrams containing sources, switches, resistors, meters, lamps, thermistors, light-dependent resistors and diodes.
  • Apply resistance and resistivity, interpret I–V characteristics and explain temperature effects.
  • Analyse e.m.f., terminal potential difference and internal resistance in real sources.
  • Analyse series, parallel and potential-divider resistor networks.
  • Combine capacitors in series and parallel.
  • Analyse charging and discharging in RC circuits using the time constant.
Syllabus note (9478)

This is an enrichment worked example using Kirchhoff’s laws (not explicitly listed in the 9478 syllabus), useful for analysing non-trivial d.c. circuits.

1. Definitions (Must Know)

A. Single-loop circuit

A single-loop circuit has only one closed loop, so there is only one current magnitude I throughout the loop.

B. Loop rule (Kirchhoff’s second law)

Around any closed loop:

∑ Δ V = 0

In practice: sum of e.m.f. rises = sum of IR drops (with a consistent sign convention).

C. Power supplied/absorbed by a source

Power associated with an e.m.f. ε is:

P_ε = Iε

If P_ε > 0, the source supplies power. If P_ε < 0, the source absorbs power (it is being charged).

2. Key Ideas (What Earns Marks)

  • In a single loop, the same current I flows through every component.
  • Choose a current direction. If you get I < 0, the real direction is opposite.
  • For any two nodes (A and B), V_ab = Vₐ-V_b is path-independent: using the “top path” or “bottom path” must give the same value.
Fast workflow
  1. Choose current direction. 2) Write one loop equation. 3) Solve for I. 4) Find V_ab by moving from a → b through any path. 5) Use signs to decide which source supplies power.

3. Detailed Explanations

A. Solving for the loop current

Write the loop equation in the chosen direction:

net e.m.f. = I(sum of resistances in loop)

B. Finding V_ab (two paths)

To find V_ab = Vₐ-V_b:

  • start from node a
  • add rises and drops along a path to node b
  • you should get the same V_ab from any valid path

C. Interpreting “power output of each battery”

Compute P_ε = Iε for each e.m.f.

  • positive: that battery is supplying power to the circuit
  • negative: that battery is absorbing power (charging)

4. Common Mistakes

  • Using different sign conventions in different parts of the same solution.
  • Forgetting that internal resistance contributes an Ir drop in the loop equation.
  • Treating V_ab as the e.m.f. of one battery (it is a node-to-node p.d.).
  • Saying both batteries “supply power” when one is opposing the current (it may be absorbing power).

5. Exam Tips

  • Always do a check: calculate V_ab using both paths. If they disagree, there is a sign error.
  • If two sources oppose, the smaller e.m.f. can be charged by the larger source.
  • Keep I to 3 s.f. during working; round at the end.

6. Worked Examples

Single loop with two opposing sourcesA single counterclockwise current I flows around a loop. From a to b along the upper path are a 3-ohm resistor, a 2-ohm resistor and a 12-volt source. Along the lower path are a 4-ohm resistor and an opposing 4-volt source. A 7-ohm resistor is on the right side.ab2 Ω12 V7 Ω4 Ω4 V3 ΩI

Modelled example 1

Find the loop current

Core

Problem

Using the circuit above, find the current in the loop.
Study the worked solution

  1. Choose a reference direction

    Method

    Take the direction driven by the 12 V source as positive.

    Reason

    The 4 V source then opposes the chosen traversal.

    Working

    εₙₑₜ = +12-4

  2. Combine the signed e.m.f.s

    Method

    The net e.m.f. is 8 V.

    Reason

    Opposing sources subtract.

    Working

    εₙₑₜ = 12-4 = 8 V

  3. Add every series resistance

    Method

    Rₜₒₜₐₗ = 16 Ω.

    Reason

    Internal and external resistances are all in the single loop.

    Working

    Rₜₒₜₐₗ = 4 + 7 + 2 + 3 = 16 Ω

  4. Calculate the current

    Method

    I = 0.50 A in the chosen direction.

    Reason

    The net source rise balances the total resistive drop.

    Working

    I = εₙₑₜ/Rₜₒₜₐₗ = 8/16 = 0.50 A

Guided practice 2

Find the potential difference V_ab

About 5 min

Problem

Using I = 0.50 A, find V_ab = Vₐ-V_b and verify it using both paths between the nodes.

Try this before viewing the solution

Unit: V

Hints

Hint 1: use the shorter top path
Along the top path, combine the 12 V rise with drops across the 2 Ω and 3 Ω resistances.
View solution step by step

  1. Evaluate the bottom path

    Method

    The bottom path gives V_ab = 9.5 V.

    Reason

    The 4 V source term and the drops associated with 4 + 7 Ω combine with the stated node orientation.

    Working

    V_ab = 4 + I(4 + 7) = 4 + 0.50(11) = 9.5 V

  2. Evaluate the top path

    Method

    The top path also gives 9.5 V.

    Reason

    Node potential difference is path-independent in this steady d.c. circuit.

    Working

    V_ab = 12-I(2 + 3) = 12-0.50(5) = 9.5 V

  3. Use agreement as a check

    Method

    Vₐ-V_b = 9.5 V.

    Reason

    Matching paths show that the source and resistor signs are internally consistent.

    Working

    V_(ab,bottom) = V_(ab,top)

Common misconception 3

Power associated with each e.m.f.

Find and correct the mistake

Learner claim

Using I = 0.50 A, a learner says both sources supply power because both have positive e.m.f. values. Find each signed e.m.f. power and diagnose the claim.

Try this before viewing the solution

Unit: W
Unit: W

View solution step by step

  1. Evaluate the driving source

    Method

    The 12 V source supplies 6.0 W.

    Reason

    Its e.m.f. drives current in the loop direction.

    Working

    P₁₂ = Iε = (0.50)(12) = +6.0 W

  2. Evaluate the opposing source

    Method

    The 4 V source has signed power -2.0 W and absorbs 2.0 W.

    Reason

    Current enters its positive terminal, charging it rather than receiving energy from it.

    Working

    P₄ = -Iε = -(0.50)(4) = -2.0 W

  3. Diagnose the claim

    Method

    Only the 12 V source supplies e.m.f. power in this operating state.

    Reason

    The sign depends on current direction through a source, not merely on the positive magnitude printed on it.

    Working

    P₁₂ > 0; P₄ < 0

Examiner practice 4

Terminal p.d. of each source

4 marks

Examination question

Using I = 0.50 A, find the terminal p.d. of the 12 V source with r = 2 Ω and the 4 V source with r = 4 Ω. State which source is being charged. [4 marks]

Try this before viewing the solution

Unit: V
Unit: V

View solution step by step

  1. Calculate the supplying-source terminal p.d.

    1 mark

    Method

    V₁₂ = 11 V.

    Reason

    The 12 V source supplies current, so internal lost volts reduce its terminal p.d.

    Working

    V₁₂ = ε-Ir = 12-(0.50)(2) = 11 V

  2. Identify the charging source

    1 mark

    Method

    The 4 V source is being charged.

    Reason

    Current enters its positive terminal.

    Working

    P₄ = -2.0 W for the source-output convention.

  3. Calculate the charging-source terminal p.d.

    2 marks

    Method

    V₄ = 6.0 V.

    Reason

    A charging source has terminal p.d. ε + Ir.

    Working

    V₄ = 4 + (0.50)(4) = 6.0 V

Challenge 5

Power dissipated in the resistors (energy check)

Minimal support

Independent transfer

Using I = 0.50 A, find the total power dissipated in all 16 Ω of resistance and verify it against the signed e.m.f. powers + 6.0 W and -2.0 W.

Try this before viewing the solution

Unit: W
Unit: W

Hints

Hint 1: compare two energy accounts
Calculate I²Rₜₒₜₐₗ separately from P₁₂ + P₄.
View solution step by step

  1. Calculate total resistive dissipation

    Method

    The resistors dissipate 4.0 W.

    Reason

    Every internal and external resistance carries the same single-loop current.

    Working

    P_diss = I²Rₜₒₜₐₗ = (0.50)²(16) = 4.0 W

  2. Calculate net source output

    Method

    The e.m.f.s supply a net 4.0 W.

    Reason

    The charging source absorbs 2.0 W from the 6.0 W supplied by the driving source.

    Working

    Pₙₑₜ = 6.0 + (-2.0) = 4.0 W

  3. Verify conservation

    Method

    Net source power equals total resistive dissipation.

    Reason

    Energy supplied per second must equal energy absorbed and dissipated per second in steady state.

    Working

    Pₙₑₜ = P_diss = 4.0 W

7. Mind Stretchers

Mind stretcher 1: Split internal vs external dissipationExtension

In Example 1, find the power dissipated in the internal resistances and in the external resistors. Check energy conservation.

Show Answer

Current: I = 0.50 A.

Power in resistances (P = I²R):

  • internal resistances: I²(4 + 2) = (0.50)²(6) = 1.5 W
  • external resistors: I²(7 + 3) = (0.50)²(10) = 2.5 W

Total dissipation: 1.5 + 2.5 = 4.0 W.

Net power supplied by e.m.f.s: P₁₂ + P₄ = 6.0 + (-2.0) = 4.0 W.

They match.

Mind stretcher 2: What if the 4 V source is reversed?Extension

If the 4 V source is flipped so that it aids the 12 V source, what happens to the loop current magnitude? Explain briefly.

Show Answer

The net emf increases (it becomes 12 + 4 instead of 12-4), while the total resistance is unchanged.

So I = εₙₑₜ/Rₜₒₜₐₗ increases.

Mind stretcher 3: Optional (Enrichment)Extension

A. Terminal p.d. of each source

If you also want the terminal p.d. of each source, use V = ε-Ir for a supplying source and V = ε + Ir for a charging source (sign depends on current direction through the source).

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027