Is this formula given in the exam?

No. SPM Physics papers provide no formula sheet, so this must be recalled.

Why is the terminal p.d. less than the e.m.f.?

Because the current also flows through the source's own internal resistance r. The voltage Ir is used up inside the source (the 'lost volts'), so only V = ε − Ir is available at the terminals for the external circuit.

What happens to the terminal p.d. as more current is drawn?

It falls. A larger current means a larger Ir drop inside the source, so V = ε − Ir decreases. When the current is very small, the terminal p.d. is almost equal to the e.m.f.

How can I find the internal resistance from an experiment?

Plot terminal p.d. V against current I. The line has the form V = ε − Ir, so the y-intercept gives the e.m.f. ε and the magnitude of the gradient gives the internal resistance r.

Electromotive force Formula, SPM Physics

Electromotive force

ε = IR + Ir = V + Ir

Not given in the exam, recall it.

What it is for

Electromotive force is used in Electricity (SPM Physics Form 5). Keep the SI units in every line of working. Electricity

Symbols and SI units

Symbol	Meaning	SI unit
`ε`	electromotive force (e.m.f.)	V
`I`	current	A
`R`	external resistance	Ω
`r`	internal resistance	Ω
`V`	terminal potential difference	V

Rearrangements

V = ε − Ir
r = (ε − V) / I
I = ε / (R + r)

Worked example

A cell of ε = 1.5 V and r = 1.0 Ω drives a current I = 0.5 A. V = ε − Ir = 1.5 − (0.5 × 1.0) = 1.0 V (terminal p.d.).

Common trap

The terminal p.d. V is always less than the e.m.f. because of the 'lost volts' Ir across the internal resistance.

Understanding electromotive force

The equation ε = IR + Ir = V + Ir links a source's e.m.f. to the current it drives and the resistances in the circuit. Here ε is the electromotive force in volts (V), I is the current in amperes (A), R is the external resistance in ohms (Ω), r is the internal resistance in ohms (Ω), and V is the terminal potential difference in volts (V).

The e.m.f. is the total energy given per coulomb; part of it (Ir, the 'lost volts') is used inside the source, leaving the terminal p.d. V = ε − Ir for the external circuit.

Terminal p.d.: V = ε − Ir
Internal resistance: r = (ε − V) / I
Current: I = ε / (R + r)

A worked example, step by step

A cell of e.m.f. ε = 6.0 V and internal resistance r = 0.5 Ω is connected to an external resistor R = 2.5 Ω. Calculate the current and the terminal potential difference.

First find the current:

I = ε / (R + r)
I = 6.0 V / (2.5 Ω + 0.5 Ω) = 6.0 V / 3.0 Ω = 2.0 A

Then the terminal p.d.:

V = ε − Ir = 6.0 V − (2.0 A × 0.5 Ω)
V = 6.0 V − 1.0 V = 5.0 V

The current is 2.0 A and the terminal p.d. is 5.0 V, which is below the e.m.f. by the 1.0 V lost across the internal resistance.

Common mistakes and how this is tested

The key trap is treating the terminal p.d. as equal to the e.m.f. Once current flows, V is always less than ε because of the lost volts Ir.

Forgetting to include r when finding the current: use R + r, not R alone.
Mixing up R (external) and r (internal) in the formula.
Sign slips when rearranging for r = (ε − V) / I.

This is tested under define (state what e.m.f. means), calculate (find ε, I, r or V) and explain (why terminal p.d. falls when the current increases). A common graph plots V against I, where the y-intercept is ε and the gradient is −r.

Source: DSKP KSSM Physics Form 4 and 5 (Versi English), Sijil Pelajaran Malaysia: Format Pentaksiran mulai 2021, Fizik (4531) (Bahagian Pembangunan Kurikulum (BPK), KPM)

Written by the spmphysics.com.my editorial team.· Updated 5 Sept 2026