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Learn7408Paper 1 · Paper 1ElectricityE.m.f. and internal resistance
Electricity · Subtopic 2 of 6

E.m.f. and Internal Resistance

A real source has an e.m.f. EE and an internal resistance rr. When it drives a current II through an external resistance RR, some energy is dissipated inside the source: E=I(R+r)E=I(R+r), so the terminal p.d. is V=E−IrV=E-Ir. The "lost volts" IrIr grow with current. A graph of VV against II is a straight line with intercept EE and gradient −r-r.

Cambridge 9702 Paper 1Cambridge 9702 Paper 2Edexcel YPH11 WPH12AQA 7408 Paper 1≈ 35 minRevision 7 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Distinguish e.m.f. from terminal p.d. in terms of energy.
  • Use E=I(R+r)E=I(R+r) and V=E−IrV=E-Ir.
  • Find EE and rr from a graph of VV against II.
  • Explain why car batteries need very low internal resistance.
Pause and think

A car's headlights dim slightly when the starter motor turns. The battery hasn't run flat, so why?

Section 2 of 14

Core idea

Start hereDimming headlights

Start a car with the headlights on and they dim. The battery hasn't weakened; it's delivering a huge current to the starter motor, and some of its voltage is used up inside the battery itself.

The big idea

E=I(R+r)=V+IrE=I(R+r)=V+Ir. Terminal p.d. falls as current rises. VV–II graph: intercept EE, gradient −r-r. Max power when R=rR=r.

1. Internal resistance

Terminal p.d. against current: intercept E, gradient −r
0.511.522.5312345670current I (A)terminal p.d. V (V)V = E − IrE = 6 V (no current)short circuit: 3 A
Gradient −2-2: r=2r=2 Ω. With no current, the voltmeter reads the full e.m.f.
E=V+IrE=V+Ir
where
EE
e.m.f.: energy per coulomb supplied.
VV
terminal p.d.
IrIr
"lost volts" inside the source.
Worked exampleE = 6.0 V, r = 2.0 Ω, R = 4.0 Ω

Current, terminal p.d., efficiency.

Goal
Values.
Strategy
Total resistance R+rR+r.
  1. I=66=1.0 AI=\frac6{6}=1.0\text{ A}
    E/(R+r)E/(R+r).
  2. V=4.0 V, Ir=2.0 VV=4.0\text{ V},\ Ir=2.0\text{ V}
    Split.
  3. 46=67%\frac46=67\%
    Useful/total.
Answer1.01.0 A; 4.04.0 V; 67%67\%.
✓ Check4+2=64+2=6 ✓.

2. Measuring E and r

A cell with internal resistance driving a load
ErRthe "cell"
E=I(R+r)E=I(R+r). With E=6E=6 V, r=2r=2 Ω and R=4R=4 Ω: I=1I=1 A, terminal p.d. =4=4 V, lost volts =2=2 V.
Straight-line form
  1. V=E−rIV=E-rI
    Rearrange.
  2. y=c+mx with y=V, x=Iy=c+mx\text{ with }y=V,\ x=I
    Intercept EE, gradient −r-r.
Worked exampleMaximum power

Same cell; choose RR for maximum load power.

Goal
RR, PP.
Strategy
Result: R=rR=r.
  1. R=2 Ω, I=64=1.5 AR=2\ \Omega,\ I=\frac6{4}=1.5\text{ A}
    Match.
  2. P=1.52×2=4.5 WP=1.5^2\times2=4.5\text{ W}
    I2RI^2R.
Answer4.54.5 W (efficiency only 50%).
✓ CheckR=4R=4: 12×4=41^2\times4=4 W, less ✓.
Common misconceptione.m.f. as a force It's energy per charge, measured in volts.

Summary

  • E=V+IrE=V+Ir.
  • Intercept EE, gradient −r-r.
Key words
Electromotive force
Energy converted to electrical per unit charge.
HistoryDaniell's steady cell. Early voltaic piles faded within minutes as hydrogen bubbles built up on the copper plate, raising internal resistance. In 18361836 John Daniell separated the two metals into different solutions, creating a cell with a steady e.m.f. of about 1.11.1 V that powered the first telegraph networks.
Fun factWhy you can't start a car with AA batteries. Eight AA cells give 1212 V, the same e.m.f. as a car battery. But each has an internal resistance around 0.20.2 Ω, totalling 1.61.6 Ω, so they can deliver at most about 7.57.5 A. A car battery's internal resistance is about 0.010.01 Ω, enough for hundreds of amps.
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