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4.4.1Electrons, Waves and Photons

E.m.f and Internal Resistance

5 min read 5 parts Lesson 34 of 62 Unit quiz Virtual lab

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Part 1 of 5E.m.f. and Terminal P.d.

4.4.1 E.m.f and Internal Resistance

A real cell is best modelled as a source of e.m.f. in series with its own internal resistance.

E.m.f. (ε)
  • The energy transferred to each coulomb of charge by the source, measured in volts despite not being a force.
Open circuit
  • With no current flowing, the p.d. across the cell terminals equals the e.m.f., which a high-resistance voltmeter can measure.
Terminal p.d.
  • The p.d. across the cell terminals while current flows; it equals the p.d. across the external load resistor.
Comparison
  • Whenever a cell has internal resistance and supplies current, the terminal p.d. is lower than the e.m.f.

Part 2 of 5Internal Resistance

Internal Resistance
Origin
  • Charge must pass through the chemicals and materials inside the cell, which oppose current just like any resistor.
Symbol
  • Internal resistance is written r and is measured in ohms, with the load resistor written R.
Energy loss
  • Some energy per coulomb is transferred as heat inside the cell, which is why a cell warms up in use.
Dependence
  • The greater the current drawn, the larger the p.d. across r and the further the terminal p.d. falls.

Part 3 of 5Lost Volts

1
Charge passing through the cell gains ε joules per coulomb from the chemical reaction.
2
Part of that energy is transferred inside the cell, so the charge loses energy per coulomb equal to Ir.
3
The quantity Ir is known as the lost volts, v, and is the p.d. across the internal resistance.
4
The remaining energy per coulomb is delivered to the external circuit, giving terminal p.d. V = ε − Ir.
5
If the internal resistance is negligible, lost volts are zero and the terminal p.d. is effectively the e.m.f.

Part 4 of 5Worked Example

Question

  • A cell of e.m.f.
  • 9.0 V and internal resistance 0.50 Ω is connected to a 4.0 Ω resistor.
  • Calculate the current, the lost volts and the terminal p.d.

  1. Total resistance = R + r = 4.0 + 0.50 = 4.5 Ω.
  2. Current I = ε / (R + r) = 9.0 / 4.5 = 2.0 A.
  3. Lost volts v = Ir = 2.0 × 0.50 = 1.0 V.
  4. Terminal p.d.
    • V = ε − Ir = 9.0 − 1.0 = 8.0 V, which equals IR = 2.0 × 4.0 = 8.0 V.
      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)

Answer I = 2.0 A, lost volts = 1.0 V, terminal p.d. = 8.0 V.

Part 5 of 5Exam Checklist & Tips

4.4.1 – Key Points for the Exam

  • E.m.f. Energy transferred to each coulomb of charge by a source, in volts.
  • Internal resistance The resistance of the materials inside a cell or power supply.
  • Terminal p.d. The p.d. across a cell's terminals when it delivers current.
  • Lost volts
    • The p.d.
    • Ir across the internal resistance, representing energy wasted inside the cell.
  • Load resistor The external resistance connected across the cell.

Common Mistakes

  • Believing e.m.f. is a force; it is energy per unit charge and is measured in volts.
  • Saying the terminal p.d. equals the e.m.f. even when current flows through a cell with internal resistance.
  • Leaving the internal resistance out when calculating the total resistance of the circuit.
  • Mixing up r (internal) and R (external) in the equations.
Try it in the virtual labE.m.f. and internal resistance of a cell
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