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Using Kirchhoff's Laws On A Charging Circuit

Key idea: Solve a charging circuit with internal resistances using Kirchhoff’s laws, then find power supplied by the source and power absorbed by the recharged battery (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. Kirchhoff’s laws in a multi-branch circuit

  • Junction rule: sum of currents into a junction equals sum out.
  • Loop rule: ∑ Δ V = 0 around a closed loop.

B. Battery being charged (sign convention)

For a battery (emf ε, internal resistance r) being charged (current enters its positive terminal), the terminal p.d. is:

V = ε + Ir

Power “delivered by the battery” is negative; it absorbs power into chemical energy.

2. Key Ideas (What Earns Marks)

  • Parallel branches have the same terminal p.d. across them.
  • Use the bulb branch to get the supply terminal p.d. quickly: V = IR.
  • For the supply with internal resistance: V = ε-Ir.
  • For a charging battery: V = ε + Ir.
Fast workflow
  1. Use junction rule to find the source current. 2) Use one branch to find the common p.d. across the parallel network. 3) Use V = ε-Ir to find the source internal resistance. 4) Use V = ε + Ir to find the battery emf.

3. Detailed Explanations

A. Use the “known branch” first

If one branch has a known current and resistance, it fixes the common potential difference across the parallel network:

V = IR

B. Treat internal resistances like resistors in series

Internal resistance is just a resistor in series with the ideal source inside the model.

Use:

  • source: V = ε-Ir
  • charging battery: V = ε + Ir

4. Common Mistakes

  • Getting the sign wrong for a charging battery (using V = ε-Ir instead of V = ε + Ir).
  • Confusing the emf of the source (12 V) with its terminal p.d. (which is lower when current flows).
  • Using P = IV with the wrong voltage (use emf for power supplied by the source’s emf).

5. Exam Tips

  • Do an energy check: total power supplied by sources should equal total power dissipated + power stored.
  • For a charging battery, split absorbed power:
    • chemical storage rate ≈ Iε
    • internal heating I²r

6. Worked Examples

Supply charging a battery through a parallel networkBetween nodes a and b are three parallel branches. The top branch has a 12-volt source and internal resistance r with current I to the right. The middle branch has battery emf epsilon with its positive terminal toward b and a 1-ohm internal resistance; 1 ampere flows left into that positive terminal. The bottom 3-ohm branch carries 2 amperes left.ab12 VrIε+1 Ω1 A3 Ω2 A

Modelled example 1

Find the current from the 12 V supply

Core

Problem

The bulb branch carries 2 A and the battery branch carries 1 A in the charging direction shown. Find the current supplied by the 12 V source.
Study the worked solution

  1. Read the junction directions

    Method

    The source current reaches the junction and splits into the bulb and battery branches.

    Reason

    Both stated branch currents leave the same junction.

    Working

    I_source = I_bulb + I_battery

  2. Apply charge conservation

    Method

    The supply current is 3 A.

    Reason

    Total current entering equals total current leaving.

    Working

    I = 2 + 1 = 3 A

Guided practice 2

Find the internal resistance of the 12 V supply

About 5 min

Problem

The bulb branch carries 2 A through 3 Ω, while the 12 V source supplies 3 A. Use the common branch p.d. and V = ε-Ir to find the source internal resistance.

Try this before viewing the solution

Unit: Ω

Hints

Hint 1: find the shared p.d. first
The 3 Ω bulb branch has known current, so V = IR fixes the voltage across every parallel branch.
View solution step by step

  1. Find the parallel-network p.d.

    Method

    The common terminal p.d. is 6 V.

    Reason

    Parallel branches share the same p.d., and the bulb branch is fully known.

    Working

    V = I_bulbR_bulb = (2)(3) = 6 V

  2. Apply the supplying-source relation

    Method

    6 = 12-3r.

    Reason

    The source terminal p.d. is reduced from its e.m.f. by the internal lost volts.

    Working

    V = ε-I_sourcer

  3. Calculate internal resistance

    Method

    r = 2 Ω.

    Reason

    The source loses 6 V internally at 3 A.

    Working

    r = (12-6)/3 = 2 Ω

Common misconception 3

Find the emf of the rechargeable battery

Find and correct the mistake

Learner claim

The rechargeable-battery branch has terminal p.d. 6 V, current 1 A entering its positive terminal, and internal resistance 1 Ω. A learner uses the supplying relation V = ε-Ir and obtains ε = 7 V. Diagnose the sign and find the e.m.f.

Try this before viewing the solution

Unit: V

View solution step by step

  1. Identify the battery state

    Method

    The battery is being charged.

    Reason

    Current enters its positive terminal.

    Working

    Vₜₑᵣₘᵢₙₐₗ > ε while charging

  2. Use the charging relation

    Method

    6 = ε + (1)(1).

    Reason

    The applied terminal p.d. must overcome the e.m.f. and the internal resistive drop.

    Working

    V = ε + Ir

  3. Calculate

    Method

    ε = 5 V.

    Reason

    Subtract the 1 V internal drop from the 6 V terminal p.d.

    Working

    ε = 6-1 = 5 V

Examiner practice 4

Power delivered by the 12 V supply (its e.m.f.)

2 marks

Examination question

The 12 V source supplies 3 A. Calculate the power delivered by its e.m.f. [2 marks]

Try this before viewing the solution

Unit: W

View solution step by step

  1. Select the e.m.f. power relation

    1 mark

    Method

    P_ε = Iε.

    Reason

    The question asks for power delivered by the ideal e.m.f. part of the source.

    Working

    P₁₂ = Iε

  2. Calculate

    1 mark

    Method

    The source e.m.f. delivers 36 W.

    Reason

    The current leaves the source’s positive terminal in its supplying state.

    Working

    P₁₂ = (3)(12) = 36 W

Challenge 5

Power delivered by the rechargeable battery’s e.m.f. (expect negative)

Minimal support

Independent transfer

The 5 V rechargeable battery carries 1 A into its positive terminal. Using “power delivered by the battery e.m.f.” as the signed quantity, calculate its value and state the chemical-energy storage rate.

Try this before viewing the solution

Unit: W
Unit: W

Hints

Hint 1: separate signed output from absorbed rate
First find Iε as a magnitude, then apply the output sign convention.
View solution step by step

  1. Find the e.m.f. power magnitude

    Method

    Iε = 5 W.

    Reason

    The ideal e.m.f. converts electrical energy into chemical energy at this rate.

    Working

    (1)(5) = 5 W

  2. Apply the signed output convention

    Method

    Power delivered by the battery e.m.f. is -5 W.

    Reason

    The battery absorbs rather than supplies this e.m.f. power while charging.

    Working

    P_(ε,battery) = -Iε = -5 W

  3. State the storage rate

    Method

    Chemical energy increases at approximately 5 J s⁻¹.

    Reason

    The positive absorbed e.m.f. power is the chemical-storage rate in this model.

    Working

    P_chemical = +5 W

7. Mind Stretchers

Mind stretcher 1: Power absorbed by the battery (terminal p.d. × current)Extension

In Example 1, what is the power absorbed by the rechargeable battery as a whole device (terminal p.d. × current)? How is it split?

Show Answer

Battery terminal p.d. is 6 V and current is 1 A:

P_(battery, absorbed) = VI = (6)(1) = 6 W

It splits into:

  • chemical energy rate ≈ Iε = (1)(5) = 5 W
  • internal heating I²r = (1)²(1) = 1 W

Mind stretcher 2: Resistor power and energy conservationExtension

Find the power dissipated in each resistor and check energy conservation.

Show Answer

Source internal resistor: Pᵣ = I²r = 3²(2) = 18 W

Battery internal resistor: P_1Ω = 1²(1) = 1 W

Bulb: P_3Ω = 2²(3) = 12 W

Total dissipation: 18 + 1 + 12 = 31 W.

Net supplied by emfs: P₁₂ + P_(ε,battery) = 36 + (-5) = 31 W.

Mind stretcher 3: Optional (Enrichment)Extension

A. Solving using explicit loop equations

You can also solve by writing Kirchhoff loop equations for each loop, but using the “known branch fixes V” shortcut is usually faster and less error-prone for this kind of question.

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