10.28-10.32 - Electrical power and domestic devices

Power is a rate of energy transfer

Imagine that two devices each transfer 6000 J of energy. Device A takes 20 s, while device B takes 60 s. They transfer the same amount of energy, but device A transfers it more quickly, so device A is more powerful.

One watt means one joule of energy transferred each second:

Energy and power are different quantities. Energy transferred is an amount, measured in joules (J). Power is a rate, measured in watts (W). A low-power device can transfer more energy overall than a high-power device if it operates for long enough.

For an energy transfer over a measured time:

The supplied equation support includes $P = E/t$. You must still select the correct values, convert energy to joules and time to seconds, and give the answer in watts.

Worked example

A portable heater transfers 18.0 kJ of energy in 45.0 s. Calculate its average power.

First convert kilojoules to joules:

$18.0 \mathrm{kJ} = 18.0 \times 1000 \mathrm{J} = 18 000 \mathrm{J}$

Select the equation and substitute:

$P = E/t$

$P = 18 000 \mathrm{J} / 45.0 \mathrm{s} = 400 \mathrm{J}/s = 4.00 \times 10^{2} \mathrm{W}$

The average power is $4.00 \times 10^{2} \mathrm{W}$, to three significant figures. This means the heater transfers an average of 400 J each second. The result is sensible because $400 \mathrm{J}/s \times 45.0 \mathrm{s} = 18 000 \mathrm{J}$, which recovers the stated energy transfer.

Current and potential difference set power

For a circuit device, power depends on two quantities measured for that same device:

• the current through the device, measured in amperes (A);
• the potential difference across the device, measured in volts (V).

These phrases matter. Current is not "across" a device, and potential difference is not "through" it. Current tells us the rate of charge flow: $1 \mathrm{A} = 1 \mathrm{C}/s$. Potential difference tells us the energy transferred per unit charge: $1 \mathrm{V} = 1 \mathrm{J}/C$.

Multiplying them shows why both determine the rate of energy transfer:

current × potential difference = (C/s) × (J/C) = J/s = W

If the potential difference stays the same, a larger current gives a larger power. If the current stays the same, a larger potential difference gives a larger power. A current value on its own is not enough because a different potential difference can change the power.

Worked example

The motor in a cordless cleaner has a potential difference of 18 V across it and a current of 2.4 A through it. Calculate its electrical power.

$P = I \times V$

$P = 2.4 \mathrm{A} \times 18 \mathrm{V} = 43.2 \mathrm{W}$

To two significant figures, the electrical power is 43 W. This is an energy-transfer rate of about 43 J/s. The unit check agrees: $(\mathrm{C}/\mathrm{s}) \times (\mathrm{J}/\mathrm{C})$ gives J/s.

The same power can come from different pairs of current and potential difference. A device operating at 12 V and 3.0 A has power 36 W; a device operating at 24 V and 1.5 A also has power 36 W. Neither current nor potential difference alone decides the power.

Power from current and resistance

For a resistive component, when its current and operating resistance are known, combine two relationships already used for circuits:

$P = I \times V$

$V = I \times R$

Substitute $I \times R$ for V in the power equation:

$P = I \times (I \times R)$

$P = I^{2} \times R$

The square applies to the current only: $I^{2}$ means $I \times I$. If resistance remains fixed and current doubles, the power becomes four times as large because $(2I)^{2} = 4I^{2}$.

[DIAGRAM: asset_name: 44_1PH0-P2-10F_10.28-10.32 - Electrical power and domestic devices - diagram 02; asset_slug: edexcel-gcse-physics-10-28-10-32-electrical-power-domestic-devices_diagram_02; recommended_method: matplotlib; description: exact fixed-resistance graph of electrical power against current for a 10 ohm device, with labelled axes, scale, curve and points at one, two and three amperes]

The graph uses $P = I^{2}R$ with a fixed resistance of 10 ohm. It passes through (1 A, 10 W), (2 A, 40 W) and (3 A, 90 W), so doubling current from 1 A to 2 A quadruples power. The curve applies only while resistance is fixed; a real heating element may become hotter and change resistance.

Worked example

A heating element has resistance 8.0 ohm and carries a current of 3.0 A. Calculate the power transferred to the element.

$P = I^{2} \times R$

$P = (3.0 \mathrm{A})^{2} \times 8.0 \mathrm{ohm}$

$P = 9.0 \times 8.0 = 72 \mathrm{W}$

The power is 72 W, to two significant figures. The result is plausible because the squared current is $9.0 \mathrm{A}^{2}$, not 6.0 A.