Physics 9702 · Paper 1

Electrical Power

$P = VI$ is not new physics. Volts are joules per coulomb. Amps are coulombs per second. Multiply them and the coulombs cancel: joules per second. That is power.

P = VI and its two other forms

P = VI

Symbols

$P$ = power transferred (W)
$V$ = p.d. across the component (V)
$I$ = current in the component (A)

Substitute $V = IR$ into $P = VI$ and you get two more forms. Use whichever pair of quantities the question gives you.

P = I^2 R = \frac{V^2}{R}

Symbols

$R$ = resistance (\Omega)

Worked example

Power lost in a cable

A power station sends 100 A through cables of total resistance 3.0 Ω. How much power is wasted as heat in the cables?

We know I and R, so use $P = I^2R = 100^2 \times 3.0$ .
This gives $P = 3.0 \times 10^4$ W, which is 30 kW.

Answer

P = 30 \text{ kW}

Common mistake

For a fixed resistance, doubling the voltage does not double the power. It makes the power four times bigger, because

P = V^2/R

and V is squared. Squared terms are where these marks are lost.

Energy over a time: W = IVt

Power times time gives the energy transferred.

W = IVt

Symbols

$W$ = energy transferred (J)
$t$ = time (s)

Worked example

A light bulb's real numbers

A bulb is labelled “230 V, 15 W”. Find the current in it, and its resistance, when it runs normally.

Current: $I = \dfrac{P}{V} = \dfrac{15}{230} = 0.065$ A.
Resistance: $R = \dfrac{V^2}{P} = \dfrac{230^2}{15} \approx 3500$ Ω.

Answer

I \approx 0.065 \text{ A}, \quad R \approx 3500 \ \Omega

Your turn— tap to reveal the worked answer (9702-style)

A kettle is rated 2.5 kW at 230 V. Find the current it draws.

Answer: $I = \dfrac{P}{V} = \dfrac{2500}{230} \approx 11$ A. Remember to turn kW into W first. (This size of current is why a kettle gets a high-rated fuse.) (9702-style)

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Electrical Power

$P = VI$ is not new physics. Volts are joules per coulomb. Amps are coulombs per second. Multiply them and the coulombs cancel: joules per second. That is power.

P = VI and its two other forms

P = VI

Symbols

$P$ = power transferred (W)
$V$ = p.d. across the component (V)
$I$ = current in the component (A)

Substitute $V = IR$ into $P = VI$ and you get two more forms. Use whichever pair of quantities the question gives you.

P = I^2 R = \frac{V^2}{R}

Symbols

$R$ = resistance (\Omega)

Worked example

Power lost in a cable

A power station sends 100 A through cables of total resistance 3.0 Ω. How much power is wasted as heat in the cables?

We know I and R, so use $P = I^2R = 100^2 \times 3.0$ .
This gives $P = 3.0 \times 10^4$ W, which is 30 kW.

Answer

P = 30 \text{ kW}

Common mistake

For a fixed resistance, doubling the voltage does not double the power. It makes the power four times bigger, because

P = V^2/R

and V is squared. Squared terms are where these marks are lost.

Energy over a time: W = IVt

Power times time gives the energy transferred.

W = IVt

Symbols

$W$ = energy transferred (J)
$t$ = time (s)

Worked example

A light bulb's real numbers

A bulb is labelled “230 V, 15 W”. Find the current in it, and its resistance, when it runs normally.

Current: $I = \dfrac{P}{V} = \dfrac{15}{230} = 0.065$ A.
Resistance: $R = \dfrac{V^2}{P} = \dfrac{230^2}{15} \approx 3500$ Ω.

Answer

I \approx 0.065 \text{ A}, \quad R \approx 3500 \ \Omega

Your turn— tap to reveal the worked answer (9702-style)

A kettle is rated 2.5 kW at 230 V. Find the current it draws.

Answer: $I = \dfrac{P}{V} = \dfrac{2500}{230} \approx 11$ A. Remember to turn kW into W first. (This size of current is why a kettle gets a high-rated fuse.) (9702-style)