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4.4.2Electrons, Waves and Photons

Calculating E.m.f

5 min read 5 parts Lesson 35 of 62 Unit quiz

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Section 1 of 5Definitions in Equation Form

4.4.2 Calculating E.m.f

The e.m.f. of a cell can be found from energy and charge or from circuit measurements.

Energy form
  • ε = E/Q, where E is the energy transferred to the charge in joules and Q is the charge in coulombs.
Circuit form
  • ε = I(R + r), because the e.m.f. drives current through both the load R and the internal resistance r.
Terminal form
  • ε = V + Ir, where V = IR is the terminal p.d. and Ir is the lost volts.
Maximum p.d.
  • The e.m.f. is the greatest p.d. a source can provide, reached only when no current is drawn.

Section 2 of 5Relationships to Remember

Relationships to Remember
Current
  • Rearranging gives I = ε / (R + r), showing that internal resistance reduces the current for a given load.
Lost volts
  • v = Ir = ε − V, so lost volts increase when the load resistance is reduced and more current flows.
Terminal p.d.
    • V = ε − Ir = IR, so the terminal p.d. is the product of the current and the load resistance.
      What the symbols mean: E or ε = e.m.f. (V) · I = current (A) · R = external resistance (Ω) · r = internal resistance (Ω) · V = terminal p.d. (V)
Special case
  • With r = 0 the equations simplify to ε = IR, which is the situation in questions saying the internal resistance is negligible.

Section 3 of 5Calculation Strategy

1
List every quantity given, labelling which resistance is the load R and which is the internal resistance r.
2
Add R and r to obtain the total resistance of the complete circuit including the cell.
3
Find the current from I = ε / (R + r), or find ε from I(R + r) if the current is known.
4
Calculate the lost volts as Ir and the terminal p.d. as ε − Ir or IR, checking that both routes agree.
5
State the answer with units and sensible significant figures, using volts for ε, V and v.

Section 4 of 5Worked Example

Question

  • A battery of e.m.f.
  • 7.3 V and internal resistance 0.30 Ω is connected in series with a 9.5 Ω resistor.
  • Find the current and the lost volts.

  1. Total resistance = 9.5 + 0.30 = 9.8 Ω.
  2. Current I = 7.3 / 9.8 = 0.745 A (0.74 A to 2 significant figures).
  3. Lost volts = Ir = 0.745 × 0.30 = 0.22 V.
  4. Check terminal p.d. = IR = 0.745 × 9.5 = 7.08 V and 7.3 − 0.22 = 7.08 V.

Answer I = 0.74 A and lost volts = 0.22 V.

Section 5 of 5Exam Checklist & Tips

4.4.2 – Key Points for the Exam

  • E.m.f. equation ε = E/Q, energy transferred per unit charge.
  • Total resistance R + r, the sum of load and internal resistance.
  • Terminal p.d. V = ε − Ir, the p.d. available to the external circuit.
  • Lost volts v = Ir, the p.d. across the internal resistance.

Common Mistakes

  • Using ε = IR and forgetting to include r in the circuit's total resistance.
  • Treating lost volts as a fixed property of the cell; they depend on the current drawn.
  • Quoting the terminal p.d. as the e.m.f. in a question where current is flowing.
  • Mixing joules and voltse.m.f. is joules per coulomb, not joules.
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