Internal Resistance

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 Rightarrow V = ε.
• Power:
  • power supplied by source: Pₛₒᵤᵣcₑ = Iε
  • power delivered to external circuit: Pₗₒₐd = IV = I(ε-Ir)
  • power lost internally: Pᵢₙₜₑᵣₙₐₗ = I²r

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

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: eta = Pₗₒₐd/Pₛₒᵤᵣcₑ = 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

Example 1: Find current and terminal p.d.Core

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 (i) the current and (ii) the terminal p.d.

Show Answer

Total resistance: R + r = 2.0 + 0.50 = 2.50 Ω.

I = ε/R + r = 1.50/2.50 = 0.60 A

Terminal p.d. across R:

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

(Check: V = ε-Ir = 1.50-(0.60)(0.50) = 1.20 V.)

Example 2: Power lost internallyCore

Using Example 1, find the power dissipated inside the cell.

Show Answer

Pᵢₙₜₑᵣₙₐₗ = I²r = (0.60)²(0.50) = 0.18 W

Example 3: Find r from a V–I graphCore

A cell’s terminal p.d. changes from 1.50 V at I = 0 to 1.20 V at I = 0.60 A. Find r.

Show Answer

Use V = ε-Ir:

Δ V = -r Δ I

So:

r = 1.50-1.20/0.60 = 0.50 Ω

Example 4: Efficiency of power transfer to the loadCore

A source has internal resistance r = 1.0 Ω and is connected to an external resistor R = 4.0 Ω. Find the efficiency eta of power transfer to the load.

Show Answer

eta = Pₗₒₐd/Pₛₒᵤᵣcₑ = V/ε = R/R + r = 4.0/4.0 + 1.0 = 0.80 So eta = 80\%.

Example 5: Battery being charged (terminal p.d.)Core

A rechargeable battery has e.m.f. ε = 6.0 V and internal resistance r = 0.40 Ω. It is being charged with current I = 2.0 A (current enters the positive terminal). Find the terminal p.d.

Show Answer

For a charging battery: V = ε + Ir = 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:

Pₗₒₐd = I²R = (ε/R + r)²R

Differentiate with respect to R:

dPₗₒₐd/dR = ε²r-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.

8. Practice, Quiz and Next Step

Close your notes and use Internal Resistance in the supplied context below. This requires a constructed explanation or working, not recognition of an option.

Fresh context: An unfamiliar data set or physical system requires you to apply Internal Resistance while stating the model, regime and assumptions.

1. Retrieve: define internal resistance in your own words, including units, sign or conditions where relevant.
2. Represent: Choose and label an appropriate diagram, graph, table or symbolic model; derive or justify the relationship used.
3. Apply: Reach a conclusion, then evaluate it using units, uncertainty, a limiting case and one practical or modelling limitation.

Check the response before looking back

• The model, regime, coordinates and assumptions are explicit.
• The derivation or multi-step reasoning is visible rather than implied.
• The conclusion is tested against units, data quality and a limiting case.
• A practical control, uncertainty or model limitation is evaluated where applicable.

If one check fails, name that exact gap, revisit the matching explanation or worked example, and redo the task with different values or a different situation. Then use theA-Level Physics course hub orpractice browser for an independent re-test.