AS · Practice questions · E.M.F. and internal resistance

The voltage that sags.

Six original Cambridge-style questions on V = E − Ir and finding E and r.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

Lost volts inside.

E.M.F.: energy per charge supplied.
Terminal: V = E − Ir.
Lost: Ir inside the cell.

Analysis

[2 marks]

Define the electromotive force of a source and give its unit.

The e.m.f. is the energy supplied per unit charge by the source ✓
The unit is the volt (V) ✓

Analysis

[3 marks]

A cell of e.m.f. 6.0 V and internal resistance 0.40 Ω supplies a current of 3.0 A. Find the terminal potential difference.

V = E − Ir = 6.0 − (3.0 × 0.40) ✓
V = 6.0 − 1.2 = 4.8 V ✓

Analysis

[3 marks]

A battery of e.m.f. 9.0 V and internal resistance 1.0 Ω is connected to a 5.0 Ω resistor. Find the current and the terminal potential difference.

I = E/(R + r) = 9.0/(5.0 + 1.0) = 1.5 A ✓
V = IR = 1.5 × 5.0 = 7.5 V ✓

Analysis

[2 marks]

State the condition under which the terminal potential difference of a cell equals its e.m.f.

When no current is drawn from the cell ✓
because then the lost volts Ir = 0, so V = E ✓

Analysis

[3 marks]

A cell of e.m.f. 1.5 V and internal resistance 0.30 Ω is short-circuited (external resistance about zero). Find the current.

With R ≈ 0, I = E/r ✓
I = 1.5/0.30 = 5.0 A ✓

Analysis

[3 marks]

A graph of terminal p.d. against current for a cell is a straight line with intercept 6.0 V and gradient −0.50 V A⁻¹. State the e.m.f. and the internal resistance.

The intercept gives the e.m.f.: E = 6.0 V ✓
The magnitude of the gradient gives the internal resistance: r = 0.50 Ω ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.