Learn7408Paper 1 · Paper 1ElectricityPotential difference and power

Electricity · Subtopic 3 of 6

Potential Difference and Power

The potential difference across a component is the energy transferred from electrical to other forms per unit charge: $V=\dfrac WQ$ , so $1$ V $=1$ J C⁻¹. Electrical power is $P=VI$ ; combining with $V=IR$ gives $P=I^2R=\dfrac{V^2}R$ . Choose whichever form uses the quantities you know.

Cambridge 9702 Paper 1Cambridge 9702 Paper 2Edexcel YPH11 WPH12AQA 7408 Paper 1≈ 30 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Define potential difference and use $V=\dfrac WQ$ .
Calculate the energy gained by a charge accelerated through a p.d.
Use $P=VI$ , $P=I^2R$ and $P=\dfrac{V^2}R$ .
Calculate energy transferred, $W=VIt$ .

Pause and think

A $2.3$ kW kettle designed for $230$ V is used on a $115$ V supply. Does it take twice as long to boil?

Section 2 of 14

Core idea

Start hereSame heater, half the voltage

Take a 230 V heater abroad to a 115 V country. Half the voltage, so half the power? No: a quarter. The heater's resistance is fixed, and power depends on voltage squared.

The big idea

$V=\frac WQ$ ; $P=VI=I^2R=\frac{V^2}R$ .

1. Definitions

Three forms of power

$P=\frac Wt=\frac{VQ}t=VI$
$W=VQ$ ; $I=Q/t$ .
$=I(IR)=I^2R$
$V=IR$ .
$=\frac VR\cdot V=\frac{V^2}R$
$I=V/R$ .

Power dissipated in a 10 Ω resistor

Doubling the current quadruples the heating. That is why power lines use small currents at high voltage.

Worked example10 Ω with 2.0 A

p.d., power, energy in 5 min.

Goal: Three values.
Strategy: Formulas.

$V=20\text{ V}$
$IR$ .
$P=40\text{ W}$
$I^2R$ .
$E=40\times300=12000\text{ J}$
$Pt$ .

Answer

20

40

12

kJ.

✓ Check

VI=40

✓.

Worked exampleThe heater

230 V, 1.2 kW; then on 115 V.

Goal: $I$ , $R$ , new power.
Strategy: $R$ fixed.

$I=\frac{1200}{230}=5.2\text{ A}$
$P/V$ .
$R=\frac{230^2}{1200}=44\ \Omega$
$V^2/P$ .
$P=\frac{115^2}{44}=300\text{ W}$
Half $V$ : quarter $P$ .

Answer

5.2

44\ \Omega

;

300

✓ Check

1200/4=300

✓.

Edge caseThe electronvolt An electron through 1 V gains

1.6\times10^{-19}

=1

eV.

Pause and think

Energy to an electron through 2 kV?

Summary

$V=W/Q$ .
Three forms of $P$ .

Volt: Joule per coulomb.

HistoryVolta's pile. In

1800

Alessandro Volta stacked discs of zinc and copper separated by brine-soaked cloth and produced a steady current: the first battery. Napoleon made him a count. The unit of potential difference is named after him.

Fun factElectric eels. An electric eel can produce pulses of up to about

860

V from thousands of stacked cells called electrocytes, each giving only about

0.15

V. It's a living version of Volta's pile, wired in series.

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