E.m.f. and internal resistance

Real batteries are not perfect: some energy is wasted inside them because of their internal resistance r. The electromotive force (e.m.f.) ε is the total energy given to each coulomb; the terminal p.d. is what is left for the circuit. That is why a battery’s voltage drops when a big current is drawn – like car headlights dimming when the engine starts.

Key facts

  • ε = V + Ir, so terminal p.d. V = ε − Ir. The “lost volts” are Ir.
  • ε = I(R + r), so I = ε ÷ (R + r).
  • With no current flowing, a voltmeter across the terminals reads ε.
  • A graph of V against I is a straight line: y-intercept = ε, gradient = −r.
  • Cells in series: e.m.f.s add and internal resistances add.
  • Maximum power is delivered to the external resistor when R = r.

Worked examples

Example 1

A 6.0 V battery with internal resistance 0.5 Ω is connected to a 2.5 Ω lamp. Find the current and the terminal p.d.

  1. I = ε ÷ (R + r) I = 6.0 ÷ (2.5 + 0.5) = 2.0 A
  2. Terminal p.d. V = IR = 2.0 × 2.5 = 5.0 V

Answer: 2.0 A; 5.0 V

Example 2

A voltmeter reads 9.0 V across a battery on open circuit, and 8.1 V when it supplies 1.5 A. Find r.

  1. Open circuit: ε = 9.0 V.
  2. Lost volts 9.0 − 8.1 = 0.9 V
  3. r = lost volts ÷ I r = 0.9 ÷ 1.5 = 0.6 Ω

Answer: 0.6 Ω

Example 3

A cell gives 1.40 V at 0.20 A and 1.10 V at 0.80 A. Find its e.m.f. and internal resistance.

  1. Write V = ε − Ir twice 1.40 = ε − 0.20r, 1.10 = ε − 0.80r
  2. Subtract 0.30 = 0.60r, so r = 0.50 Ω
  3. Back-substitute ε = 1.40 + 0.20 × 0.50 = 1.50 V

Answer: ε = 1.50 V, r = 0.50 Ω