Calculate terminal potential difference with internal resistance
Use the source model ε = V + Ir to relate e.m.f., terminal p.d. and lost volts.
Section 10 · D.C. circuits
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 is the internal voltage drop when current I flows through internal resistance r?
What you will learn
- Explain how internal resistance affects terminal p.d.
- Use ε = V + Ir and V = ε − Ir
- Interpret Ir as potential difference across source internal resistance
A real source has resistance inside it
A real source can be modelled as an ideal e.m.f. ε in series with an internal resistance r. When current I flows, the internal p.d. is Ir, sometimes called lost volts. The terminal p.d. V available to the external circuit is ε − Ir.
The energy balance per unit charge is ε = V + Ir. With no current, the internal drop is zero and terminal p.d. equals e.m.f. Under load, a positive current through positive internal resistance makes the terminal p.d. smaller than ε.
Use the model with consistent current direction
For a source delivering current, substitute ε, I and r into V = ε − Ir. The product Ir has units volts. If the source is being charged, the direction of energy transfer changes; do not apply the discharging sign without checking the stated current direction.
The external circuit receives energy at the terminal p.d. The remainder per coulomb is transferred inside the source, commonly as heating. The model describes the source behavior at the given current and does not imply r is identical for every real cell or operating condition.
Pause and explain
A source delivers current through internal resistance. How does terminal p.d. compare with e.m.f.?
Worked example
A cell has e.m.f. 1.5 V and internal resistance 0.20 Ω. Find terminal p.d. when it supplies 0.50 A.
Show the worked solution
- Lost volts = Ir = 0.50 × 0.20 = 0.10 V.
- V = ε − Ir = 1.5 − 0.10.
- Check that terminal p.d. is below the e.m.f. while the cell delivers current.
Answer V = 1.4 V.
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Test this topic →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.
A source has ε = 9.0 V, r = 0.50 Ω and supplies 2.0 A. Calculate terminal p.d.
Give me a hint
Find the internal drop Ir, then subtract from e.m.f.
Compare my reasoning
- Ir = 2.0 × 0.50 = 1.0 V.
- V = 9.0 − 1.0.
- State volts.
V = 8.0 V.
Look for these in your work
- Internal drop calculated
- Correct sign applied
- Terminal voltage stated
A 6.0 V source has terminal p.d. 5.4 V while supplying 0.30 A. Find internal resistance.
Give me a hint
The difference ε − V is the internal p.d. Ir.
Compare my reasoning
- Ir = 6.0 − 5.4 = 0.60 V.
- r = 0.60/0.30.
- Give Ω.
r = 2.0 Ω.
Look for these in your work
- Lost volts found
- Equation rearranged
- Resistance unit stated
A cell's current doubles while its e.m.f. and internal resistance remain constant. Explain the change in lost volts and terminal p.d. while it supplies a circuit.
Give me a hint
Use Ir and hold ε and r fixed.
Compare my reasoning
- Doubling I doubles Ir.
- V = ε − Ir, so the terminal p.d. falls by the additional internal drop.
- The remaining energy per coulomb available externally is smaller.
Lost volts double and terminal p.d. decreases by the increase in Ir, for this constant-ε, constant-r source model.
Look for these in your work
- Current scaling applied
- Terminal-voltage consequence derived
- Model conditions retained
Common mistakes
- Adding Ir to e.m.f. for a discharging cell
- Using current in milliamperes with resistance in ohms and forgetting the conversion
- Claiming terminal p.d. exceeds e.m.f. while a passive real source delivers current
What can you explain now?
What is the internal voltage drop when current I flows through internal resistance r?
Compare with the explanation
Ir
Ohm's law gives the p.d. across internal resistance.
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Make it stick
Why does the terminal p.d. fall when a source with internal resistance supplies more current?
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.