Electrical Power

This lesson covers electrical power and the equations used to calculate it, as required by the Edexcel GCSE Combined Science specification (1SC0). You will learn to use $P = IV$P=IV, $P = I^2R$P=I2R and $P = \frac{V^2}{R}$P=RV2​ and apply them to compare electrical devices.

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What Is Electrical Power?

Power is the rate at which energy is transferred (or the rate at which work is done).

$P = \frac{E}{t}$P=tE​

where:

• P = power in watts (W)
• E = energy transferred in joules (J)
• t = time in seconds (s)

Quantity	Unit	Equivalent
Power	Watt (W)	1 W = 1 J/s
	Kilowatt (kW)	1 kW = 1000 W

A 60 W lamp transfers 60 joules of energy every second.

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The Three Power Equations

For electrical circuits, power can be calculated using three related equations.

Equation 1: P = IV

$P = IV$P=IV

where:

• P = power in watts (W)
• I = current in amperes (A)
• V = potential difference in volts (V)

This is the primary power equation. It tells you that power equals the current flowing through a component multiplied by the potential difference across it.

Equation 2: P = I²R

Substituting $V = IR$V=IR into $P = IV$P=IV:

$P = I \times (IR) = I^2 R$P=I×(IR)=I2R

This is useful when you know the current and resistance but not the voltage.

Equation 3: P = V²/R

Substituting $I = \frac{V}{R}$I=RV​ into $P = IV$P=IV:

$P = \frac{V}{R} \times V = \frac{V^2}{R}$P=RV​×V=RV2​

This is useful when you know the voltage and resistance but not the current.

Equation	Use when you know...
$P = IV$P=IV	Current and voltage
$P = I^2R$P=I2R	Current and resistance
$P = V^2/R$P=V2/R	Voltage and resistance

Exam Tip: All three equations are derived from $P = IV$P=IV and $V = IR$V=IR. You need to be able to select the correct equation based on the quantities given in the question.

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Worked Examples

Worked Example 1: Using P = IV

A 230 V kettle draws a current of 10 A. What is the power?

$P = IV = 10 \times 230 = 2300 \text{ W} = 2.3 \text{ kW}$P=IV=10×230=2300 W=2.3 kW

Worked Example 2: Using P = I²R

A 3 A current flows through a 15 $\Omega$Ω heating element. What is the power?

$P = I^2 R = 3^2 \times 15 = 9 \times 15 = 135 \text{ W}$P=I2R=32×15=9×15=135 W

Worked Example 3: Using P = V²/R

A 12 V supply is connected to a 48 $\Omega$Ω resistor. What is the power dissipated?

$P = \frac{V^2}{R} = \frac{12^2}{48} = \frac{144}{48} = 3 \text{ W}$P=RV2​=48122​=48144​=3 W

Worked Example 4: Finding Current from Power

A 2000 W heater operates on a 230 V supply. What current does it draw?

Rearranging $P = IV$P=IV:

$I = \frac{P}{V} = \frac{2000}{230} = 8.7 \text{ A (1 d.p.)}$I=VP​=2302000​=8.7 A (1 d.p.)

Exam Tip: When a question asks "what fuse should be used?", first calculate the current using $I = P/V$I=P/V, then choose the fuse rating just above the normal operating current (e.g. 3 A, 5 A or 13 A).

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Comparing Electrical Devices

Device	Typical Power Rating	What This Means
LED lamp	5–10 W	Transfers 5–10 J per second to light and heat
Filament lamp	40–100 W	Transfers much more energy per second, mostly as heat
Kettle	2000–3000 W	Very high rate of energy transfer — heats water quickly
Phone charger	5–20 W	Relatively low power — slow energy transfer
Electric shower	7000–10 500 W	One of the highest-power household devices

A higher power rating means the device transfers energy faster, not necessarily that it uses more energy in total (that depends on how long it is on).

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Power and Resistance

From $P = I^2R$P=I2R, we can see that for a fixed current, a higher resistance means more power is dissipated. This is why:

• Long, thin wires have higher resistance and get hotter when carrying current.
• Heating elements are made from high-resistance wire.
• Thick copper cables are used for high-current connections to minimise unwanted heating.

From $P = V^2/R$P=V2/R, for a fixed voltage, a lower resistance means more power is dissipated and more current flows.

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Calculating Fuse Ratings

To choose the correct fuse for an appliance:

1. Find the current: $I = \frac{P}{V}$I=VP​
2. Choose the fuse rating that is just above the normal current.

Worked Example 5

A 920 W microwave operates on a 230 V supply. Should it use a 3 A or a 13 A fuse?

$I = \frac{P}{V} = \frac{920}{230} = 4.0 \text{ A}$I=VP​=230920​=4.0 A

A 3 A fuse would blow during normal use. A 13 A fuse is correct (or a 5 A fuse if available).

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Summary

• Power is the rate of energy transfer, measured in watts (W).
• Three power equations: $P = IV$P=IV, $P = I^2R$P=I2R, $P = \frac{V^2}{R}$P=RV2​.
• Choose the equation based on which quantities are given.
• A higher power rating means faster energy transfer.
• To select a fuse, calculate the current using $I = P/V$I=P/V and pick the rating just above.
• For a fixed current, more resistance means more power dissipated.
• For a fixed voltage, less resistance means more power dissipated.

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Extended Worked Examples

Worked Example 6 — Cross-checking with All Three Equations

A 5 A current flows through a 20 $\Omega$Ω resistor. Confirm the power using all three equations.

First find V: $V = IR = 5 \times 20 = 100 \text{ V}$V=IR=5×20=100 V.

$P = IV = 5 \times 100 = 500 \text{ W}$P=IV=5×100=500 W

$P = I^2R = 5^2 \times 20 = 500 \text{ W}$P=I2R=52×20=500 W

$P = V^2/R = 100^2/20 = 10000/20 = 500 \text{ W}$P=V2/R=1002/20=10000/20=500 W