Potential difference and power

AS Level Physics · Electricity. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

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The idea

Charge carries energy around a circuit. The potential difference (p.d.) across a component is the energy transferred per unit charge as charge passes through it. A p.d. of 12 V means each coulomb transfers 12 J in that component.

Power is the energy transferred each second. Each coulomb transfers V joules and I coulombs pass each second, so the power is V × I.

Putting V = IR into P = VI gives two more forms. Choose the form that uses the quantities you know and the one that stays constant. For a fixed resistance, doubling the current or the p.d. makes the power four times as large.

Rules to remember

potential difference = energy transferred ÷ charge (V = W/Q); 1 V = 1 J C⁻¹
Energy gained by a charge q moved through a p.d. V: W = qV
power = p.d. × current (P = VI); 1 W = 1 J s⁻¹
power = current² × resistance (P = I²R)
power = p.d.² ÷ resistance (P = V²/R)
energy transferred = power × time (W = Pt = VIt), with t in seconds
Fixed R: P is proportional to I² and to V²

Common mistake

Students forget the square. If the current in a fixed resistor rises by 50%, the power is multiplied by 1.5² = 2.25, not by 1.5.

Worked example

A 50 Ω resistor carries a current of 0.20 A for 2.0 minutes. How much energy is transferred in the resistor?

P = I²R = 0.20² × 50 = 0.040 × 50 = 2.0 W.
t = 2.0 × 60 = 120 s.
W = Pt = 2.0 × 120 = 240 J.

Answer: The energy transferred is 240 J.

Practice questions

Try each one, then open the answer.

1. The current in a resistor of fixed resistance is increased by 50%. By what percentage does the power dissipated increase?

A
50%
B
125%
C
150%
D
225%

Show answer

Answer: B. P = I²R, so P becomes 1.5² = 2.25 times as large. The increase is 1.25 times the original, i.e. 125%.

2. Which expression gives the power dissipated in a resistor R carrying current I?

A
IR
B
I²R
C
I/R
D
IR²

Show answer

Answer: B. P = VI and V = IR, so P = I²R (or V²/R).

3. A lamp is rated 60 W when connected to a 230 V supply. What is its resistance when working normally?

A
0.26 Ω
B
3.8 Ω
C
230 Ω
D
880 Ω

Show answer

Answer: D. P = V²/R, so R = V²/P = 230² ÷ 60 = 880 Ω. 3.8 Ω comes from V/P, which is not a resistance.

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