Is this formula given in the exam?

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

Why is transformer efficiency always less than 100%?

Real transformers lose some energy as heat, from eddy currents in the iron core and resistance in the copper windings. So the output power is always a little less than the input power.

Do I use voltage or power in the efficiency formula?

You use power, which is voltage multiplied by current. Work out VsIs for the output and VpIp for the input, then divide output by input and multiply by 100%.

Where does the wasted energy in a transformer go?

It becomes heat. The main causes are eddy currents induced in the core, resistive heating in the coils, and repeated magnetisation of the core, so transformers often warm up in use.

Transformer efficiency Formula, SPM Physics

Transformer efficiency

efficiency = (VsIs)/(VpIp) × 100%

Not given in the exam, recall it.

What it is for

Transformer efficiency is used in Electromagnetism (SPM Physics Form 5). Keep the SI units in every line of working. Electromagnetism

Symbols and SI units

Symbol	Meaning	SI unit
`Vs`	secondary voltage	V
`Is`	secondary current	A
`Vp`	primary voltage	V
`Ip`	primary current	A

Rearrangements

output power VsIs = efficiency × VpIp
input power VpIp = VsIs / efficiency

Worked example

Output power VsIs = 48 W, input power VpIp = 60 W. efficiency = (VsIs)/(VpIp) × 100% = (48 / 60) × 100% = 80%.

Common trap

It is output over input, so efficiency is always ≤ 100%. Use power (V × I) values, not voltage alone.

Understanding Transformer efficiency

Efficiency compares useful output power with input power: efficiency = (VsIs) / (VpIp) × 100%. Here Vs and Is are the secondary voltage (V) and current (A), giving the output power VsIs in watts; Vp and Ip are the primary voltage (V) and current (A), giving the input power VpIp in watts.

The two powers relate because a real transformer loses some energy as heat in the core and windings, so the output is always a little less than the input. Efficiency is a percentage, so it has no unit of its own.

To rearrange, treat efficiency as a fraction: output power VsIs = efficiency × VpIp finds the useful power, and input power VpIp = VsIs / efficiency finds the power drawn from the supply.

A worked example, step by step

A transformer draws Vp = 240 V and Ip = 2 A from the mains, and delivers Vs = 48 V at Is = 9 A to a device. Find its efficiency.

First the two powers, then the ratio:

Input power = VpIp = 240 V × 2 A = 480 W.

Output power = VsIs = 48 V × 9 A = 432 W.

efficiency = (VsIs) / (VpIp) × 100% = (432 W / 480 W) × 100% = 90%.

The output (432 W) is less than the input (480 W), so the answer is below 100%, as it must be. The 48 W difference is lost as heat.

The efficiency is 90%.

Common mistakes and how this is tested

The key trap is using voltages or currents on their own instead of the power products V × I. A second error is inverting the ratio (input over output), which wrongly gives a value above 100%. A third is forgetting the × 100% so the answer comes out as 0.9 instead of a percentage.

Under calculate, show both power lines with units before dividing, and give the percentage. Under define, state that efficiency is the ratio of output power to input power, expressed as a percentage.

Under explain, you may be asked where the lost energy goes, such as heat from eddy currents in the core or resistance in the coils.

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