B.5.5—Electrical power

Syllabus
First assessment 2025
Objective
Level
HL

Calculate Electrical Power

Electrical power

Power is the rate of electrical energy transfer. For a resistor,

P=IV=I2R=V2RP=IV=I^2R=\frac{V^2}{R}

Choose the convenient form

Use P=IVP=IV when current and voltage are given, P=I2RP=I^2R when current and resistance are given, and P=V2/RP=V^2/R when voltage and resistance are given.

Interpret the unit

A watt is a joule per second: 1W=1Js11\,\mathrm W=1\,\mathrm{J\,s^{-1}}. In a resistor, the transferred electrical energy becomes mainly internal energy and may produce heating.

Worked example from the mapped local textbook

A heater is rated 230V230\,\mathrm{V} and 1100W1100\,\mathrm{W}. Since voltage and power are known, use P=V2/RP=V^2/R:

R=V2P=23021100=48ΩR=\frac{V^2}{P}=\frac{230^2}{1100}=48\,\Omega

The rating means the heater transfers about 1100J1100\,\mathrm{J} each second when operated at 230V230\,\mathrm{V}.

Common trap

For alternating-current questions, distinguish peak values from mean or rms values. Use the convention and data supplied by the question.

B.5.5 Exam Analysis

Assessment in practice

1–2 marks
How it is assessed

The evidence includes a calculation from energy transferred each second and a mean-power question for an alternating supply; identify whether the stated voltage or current is peak or rms before using the equation.

Command terms

Calculate / Identify

What earns marks

Choose the power equation that matches the given quantities: P=IV, P=I²R or P=V²/R. Show the rearrangement and preserve the distinction between power, energy transferred per second, and peak or rms values in an AC question.

Watch for

Using a peak value directly in a mean-power calculation when the question requires rms quantities.

Representative question

Question 1

[Maximum number: 1]

The designers state that the energy transferred by the resistor every second is 15 J .

Calculate the current in the resistor.