Mini Physics
Part of Mini EducationSwitch subject
Search

Internal Resistance

Key idea: Relate e.m.f., terminal potential difference and internal resistance using V = ε − Ir, and solve power/efficiency problems for sources (A Level Physics).

  • GCE A-Level H2 Physics 2027
On this page

Learning objectives

  • Analyse e.m.f., terminal potential difference and internal resistance in real sources.

1. Definitions (Must Know)

A. Electromotive force (e.m.f.), ε

e.m.f., ε, of a source is the energy supplied by the source per unit charge:

ε = W/Q

Unit: volt (V) = J C⁻¹.

B. Terminal potential difference, V

Terminal potential difference, V, is the potential difference across the source’s terminals (what a voltmeter across the battery reads).

C. Internal resistance, r

Internal resistance, r, is the resistance inside a real source that causes energy to be dissipated inside the source when current flows.

2. Key Ideas (What Earns Marks)

  • A real source can be modelled as an ideal source of e.m.f. ε in series with an internal resistance r.
  • When current I flows: V = ε - Ir where Ir is the lost volts inside the source.
  • Open circuit: I = 0 ⇒ V = ε.
  • Power:
    • power supplied by source: P_source = Iε
    • power delivered to external circuit: P_load = IV = I(ε-Ir)
    • power lost internally: Pᵢₙₜₑᵣₙₐₗ = I²r
Quick diagnostic

If the current increases, Ir increases, so terminal p.d. V decreases.

Exam pitfall: terminal voltage at open circuit

The discharging relation V = ε-Ir also works at open circuit: I = 0, so V = ε. For a battery being charged, current enters its positive terminal and the relation becomes V = ε + Ir.

3. Detailed Explanations

A. Deriving V = ε-Ir (energy per unit charge)

Energy supplied per unit charge by the source is ε.

When current flows, some energy per unit charge is dissipated inside the source across r:

energy per unit charge lost = Ir

So the energy per unit charge available to the external circuit (the terminal p.d.) is:

V = ε-Ir

Equivalent circuit equation:

ε = V + Ir

B. Power balance

Multiply ε = V + Ir by I:

Iε = IV + I²r

Interpretation:

  • Iε is the rate energy is supplied by the source
  • IV is the rate energy is transferred to the external circuit
  • I²r is the rate energy is dissipated inside the source

C. I–V graph of a source

Rearrange:

V = ε-rI

This is a straight line:

  • intercept at I = 0 is ε
  • gradient is -r
Real source with internal resistance and external loadA circuit model with ideal electromotive force epsilon and internal resistance r inside the source boundary, connected in series to an external load R. Current leaves the positive terminal and voltage labels show epsilon equals V plus Ir during discharge.real source+−ideal e.m.f. εinternal rlost p.d. = IrRterminal p.d. VIdischarging: ε = V + Irsource energy = load transfer + internal heating
Discharging-source model: the ideal e.m.f. and internal resistance are in series. The terminal p.d. across the load is V = ε − Ir.
Check the current direction before choosing the sign

For a discharging source, current leaves the positive terminal and V < ε. For a source being charged, current enters the positive terminal and V > ε. At open circuit, I = 0 and V = ε.

D. Test the sign, graph and power balance

The explorer uses one graph convention: signed current is positive when it leaves the positive terminal. A discharging operating point therefore has I > 0; a charging point has I < 0. With that convention, the graph equation remains V = ε-rI in both cases.

Internal Resistance Explorer

Compare a source that is discharging with a cell being charged, then connect terminal p.d., the V–I graph, and the power balance.

BetaA LevelElectricityBest for: A Level current electricity
  • Source Modelling
  • Sign Conventions
  • Graph Interpretation
  • Power Balance

Open the full interactive simulation on its own page

Use the standalone simulation page for the live controls, SVG scene, run modes, and scoring flow.

Open Interactive Simulation

The lesson stays lightweight and links out to the dedicated simulation page.

4. Common Mistakes

  • Using the same sign without checking the current direction: use V = ε-Ir when discharging and V = ε + Ir when charging.
  • Forgetting that V is the terminal p.d. across the load, not across r.
  • Thinking ε changes when a load is connected (for this model, ε is a source property; V changes with I).
  • Mixing up internal resistance r with external resistance R.

5. Exam Tips

  • If given ε and r and asked for current with external resistance R: I = ε/(R + r) then V = IR.
  • Efficiency of power transfer to the load resistor: η = P_load/P_source = IV/Iε = V/ε = R/(R + r)
  • If you are given a V–I graph, read:
    • intercept → ε
    • magnitude of gradient → r

6. Worked Examples

Modelled example 1

Find current and terminal p.d.

Core

Problem

A cell has e.m.f. ε = 1.50 V and internal resistance r = 0.50 Ω. It is connected to an external resistor R = 2.0 Ω. Find the current and terminal p.d.
Study the worked solution

  1. Find the circuit current

    Method

    Add external and internal resistance.

    Reason

    The same current passes through R and r in the real-source model.

    Working

    I = ε/(R + r) = 1.50/(2.0 + 0.50) = 0.60 A

  2. Find terminal p.d.

    Method

    Use the p.d. across the external load.

    Reason

    The source terminals and load share the same two nodes.

    Working

    V = IR = (0.60)(2.0) = 1.20 V

  3. Check with lost volts

    Method

    Subtract the internal drop from the e.m.f.

    Reason

    For a discharging cell, ε = V + Ir.

    Working

    V = 1.50-(0.60)(0.50) = 1.20 V

Guided practice 2

Power lost internally

About 4 min

Problem

A cell with internal resistance 0.50 Ω supplies current 0.60 A. Find the power dissipated inside the cell.

Try this before viewing the solution

Unit: W

Hints

Hint 1: locate the dissipation
The relevant resistance is r, not the external load R.
Hint 2: choose the power form
Use Pᵢₙₜₑᵣₙₐₗ = I²r.
View solution step by step

  1. Select the internal component

    Method

    Apply resistor power to r.

    Reason

    The entire source current passes through its internal resistance.

    Working

    Pᵢₙₜₑᵣₙₐₗ = I²r

  2. Evaluate

    Method

    Obtain 0.18 W.

    Reason

    Current must be squared.

    Working

    Pᵢₙₜₑᵣₙₐₗ = (0.60)²(0.50) = 0.18 W

Common misconception 3

Identify what determines lost volts

Find and correct the mistake

Learner claim

A battery’s terminal p.d. decreases from its open-circuit value when current flows. A learner says current I and external resistance R are sufficient to calculate this decrease because the load p.d. is IR. Diagnose the claim and identify the quantities that are sufficient.

Try this before viewing the solution

Sufficient quantities

View solution step by step

  1. Separate terminal p.d. from its decrease

    Method

    Write the discharging-source relation.

    Reason

    IR is the terminal p.d. across the load, while the requested decrease is the internal p.d.

    Working

    V = ε-Ir

  2. Calculate the decrease

    Method

    Subtract terminal p.d. from the open-circuit value.

    Reason

    At open circuit V = ε, so the magnitude of the decrease is the lost volts.

    Working

    Δ V = ε-V = Ir

Examiner practice 4

Infer the load and account for source power

6 marks

Examination question

A source has e.m.f. 12.0 V and internal resistance 0.50 Ω. Its terminal p.d. is 10.0 V while supplying a resistor. Calculate the current, load resistance and transfer efficiency. Show that the source-power account closes. [6 marks]

Try this before viewing the solution

View solution step by step

  1. Use lost volts to find current

    1 mark

    Method

    Divide the internal p.d. by r.

    Reason

    The difference ε-V appears across the internal resistance.

    Working

    I = (ε-V)/r = (12.0-10.0)/0.50 = 4.0 A

  2. Infer the load resistance

    1 mark

    Method

    Use the terminal p.d. across the external resistor.

    Reason

    The load carries the same current in the series real-source model.

    Working

    R = V/I = 10.0/4.0 = 2.5 Ω

  3. Find efficiency

    2 marks

    Method

    Form useful load power divided by source power.

    Reason

    The current cancels from VI/(ε I).

    Working

    η = V/ε = 10.0/12.0 = 0.833 = 83.3%

  4. Close the power account

    2 marks

    Method

    Compare source, load and internal power.

    Reason

    Energy conservation requires Iε = IV + I²r.

    Working

    P_source = 48.0 W, P_load = 40.0 W, Pᵢₙₜₑᵣₙₐₗ = 8.0 W; 48.0 W = 40.0 W + 8.0 W

Challenge 5

Battery being charged (terminal p.d.)

Minimal support

Independent transfer

A rechargeable battery has e.m.f. ε = 6.0 V and internal resistance r = 0.40 Ω. It is charged by a 2.0 A current entering its positive terminal. Find the terminal p.d. and explain why it exceeds the e.m.f.

Try this before viewing the solution

Hints

Hint 1: reverse the operating direction
During charging, the external supply must overcome both the battery e.m.f. and the internal resistive drop.
View solution step by step

  1. Choose the charging relation

    Method

    Add the internal drop to the e.m.f.

    Reason

    Current entering the positive terminal reverses the discharging energy-flow condition.

    Working

    V = ε + Ir

  2. Evaluate

    Method

    Obtain 6.8 V.

    Reason

    The applied terminal p.d. supplies energy both to chemical storage and internal heating.

    Working

    V = 6.0 + (2.0)(0.40) = 6.8 V

7. Mind Stretchers

Mind stretcher 1: Why terminal p.d. drops with smaller RExtension

Explain why connecting a smaller external resistance makes a battery’s terminal p.d. drop more.

Show Answer

Smaller R increases current I = ε/(R + r).

As I increases, the lost volts Ir increases, so V = ε-Ir decreases more.

Mind stretcher 2: Maximum power transferExtension

Show that maximum power is delivered to the load when R = r. (You may use calculus or another method.)

Show Answer

Power in the load is:

P_load = I²R = (ε/(R + r))²R

Differentiate with respect to R:

dP_load/dR = ε²(r-R)/((R + r)³).

The derivative is zero when R = r and changes from positive to negative there, so the load power is maximum at R = r.

At this point, the transfer efficiency is only R/(R + r) = 1/2. Maximum power transfer is therefore not the same as maximum efficiency.

Mind stretcher 3: Optional (Enrichment)Extension

A. Maximum power transfer (beyond typical A Level workload)

The result “maximum power to the load when R = r” is useful, but many A Level questions only require you to compute power values using P = IV and P = I²R rather than prove the condition.

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