⚡ To find the current
Divide both sides by V.
I = P ÷ V
current = power ÷ voltage
Mains Electricity » Electrical Power and Choosing Fuses
What you'll learn this session
Study time: 30 minutes
Edexcel spec: 2.4
- What the power rating on an appliance tells you
- How to use power = current × voltage, an equation you must learn
- How to rearrange the equation to find a current or a voltage
- How to work out the normal current of an appliance and pick the right fuse
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Unlock This CourseWhat is electrical power?
Every electrical appliance transfers energy. Some do it slowly and some do it very quickly. Power tells you how quickly: it is the energy transferred each second.
Power is measured in watts (W). An appliance with a power of 1 W transfers 1 joule of energy every second. So a 10 W lamp transfers 10 J each second, and a 2000 W appliance transfers 2000 J each second.
Look on the bottom or the back of any mains appliance and you will find a label. It gives a power rating and the voltage the appliance is made for, for example '230 V, 800 W'. Things that heat up usually have a high power rating. Things that only give out light or sound usually have a low one.
Key terms:
- Power: the energy transferred each second. It is measured in watts (W).
- Power rating: the power of an appliance when it is working normally, printed on its label.
- Normal current: the current in an appliance when it is working normally at the voltage it was made for.
Large powers are often given in kilowatts (kW). Remember, kilo means 1000, so always change kilowatts into watts before you calculate: multiply by 1000. A 2.3 kW tumble dryer has a power of 2300 W. A 0.46 kW blender has a power of 460 W.
The power equation
The power of an appliance depends on two things: the current in it and the voltage across it.
power = current × voltage
P = I × V
P is the power in watts (W), I is the current in amperes (A) and V is the voltage in volts (V). This equation is not given in the exam. You must learn it, in words and in symbols.
The equation makes sense. A larger current, or a larger voltage, means more energy is transferred each second. Double the current at the same voltage and the power doubles.
Worked example 1: finding the power
A torch lamp has a voltage of 3 V across it. The current in the lamp is 0.5 A. Calculate the power of the lamp.
P = I × V
P = 0.5 × 3
P = 1.5 W
Worked example 2: a mains appliance
A tumble dryer is connected to the 230 V mains supply. The current in it is 10 A. Calculate its power in kilowatts.
P = I × V
P = 10 × 230
P = 2300 W
2300 ÷ 1000 = 2.3 kW
Rearranging the equation
Exam questions often give you the power and ask for the current or the voltage. Rearrange P = I × V to get the one you need.
🔋 To find the voltage
Divide both sides by I.
V = P ÷ I
voltage = power ÷ current
In both cases the power goes on top. If you are not sure, check with easy numbers: 6 = 2 × 3, so 2 = 6 ÷ 3 and 3 = 6 ÷ 2.
Worked example 3: finding the current
A television has a power rating of 115 W. It is used on the 230 V mains supply. Calculate the current in the television.
P = I × V, so I = P ÷ V
I = 115 ÷ 230
I = 0.5 A
Worked example 4: finding the voltage
A car headlamp has a power of 60 W. The current in the headlamp is 5 A. Calculate the voltage across it.
P = I × V, so V = P ÷ I
V = 60 ÷ 5
V = 12 V
Choosing the right fuse
Remember, a fuse melts and breaks the circuit when the current is larger than its rating. The common ratings for plug fuses in the UK are 3 A, 5 A and 13 A. To choose between them you need the normal current of the appliance, and P = I × V gives you that.
- Read the power rating from the label. Change kW into W if you need to.
- Calculate the normal current using I = P ÷ V. For UK mains, V = 230 V.
- Choose the fuse with the rating just above the normal current.
The table shows this done for some appliances on the 230 V mains.
| Appliance | Power rating | Normal current (I = P ÷ V) | Fuse to use |
|---|---|---|---|
| Television | 115 W | 115 ÷ 230 = 0.5 A | 3 A |
| Blender | 460 W | 460 ÷ 230 = 2 A | 3 A |
| Lawnmower | 920 W | 920 ÷ 230 = 4 A | 5 A |
| Dishwasher | 1840 W | 1840 ÷ 230 = 8 A | 13 A |
| Tumble dryer | 2.3 kW = 2300 W | 2300 ÷ 230 = 10 A | 13 A |
Why 'just above'? There are two ways to get it wrong.
📉 Rating too low
If the rating is below the normal current, the fuse melts as soon as the appliance is switched on, even though nothing is wrong. A 5 A fuse is no use in the dishwasher, which takes 8 A.
📈 Rating too high
If the rating is far above the normal current, a fault could make the current much too large without melting the fuse. With a 13 A fuse in the television, a fault current of 10 A would keep flowing, and the wires could overheat and start a fire.
Worked example 5: choosing a fuse
A sandwich toaster has a power rating of 0.805 kW. It is used on the 230 V mains. Which fuse should be fitted in its plug: 3 A, 5 A or 13 A?
P = 0.805 × 1000 = 805 W
I = P ÷ V
I = 805 ÷ 230
I = 3.5 A
The normal current is 3.5 A. A 3 A fuse is below this, so it would melt in normal use. A 13 A fuse is far above it. The rating just above 3.5 A is 5 A, so fit a 5 A fuse.
Common mistakes
- Leaving the power in kilowatts. 2.3 ÷ 230 gives 0.01 A, which is wrong. Change 2.3 kW to 2300 W first.
- Dividing the wrong way round. For current, it is power ÷ voltage, not voltage ÷ power.
- Choosing the fuse just below the normal current. It must be just above.
- Choosing 13 A every time 'to be safe'. A fuse that is much too large gives poor protection.
- Stopping after the current. If the question asks for a fuse, finish by naming the rating.
An investigation you could try
The specification suggests an optional investigation: measuring the power of low-voltage electrical items, such as a 12 V lamp or a small electric motor. It is not a required practical, but the method is a good way to see P = I × V at work.
- Connect the item to a low-voltage power supply.
- Put an ammeter in series with the item to measure the current in it.
- Put a voltmeter in parallel with the item to measure the voltage across it.
- Switch on and record both readings.
- Calculate the power using P = I × V.
- Repeat with other low-voltage items and compare their powers.
For example, if a lamp has readings of 12 V and 2.0 A, its power is 2.0 × 12 = 24 W. If a small motor has readings of 6 V and 0.4 A, its power is 0.4 × 6 = 2.4 W. The lamp transfers ten times as much energy each second as the motor.
Exam-style question
A steam cleaner has a label that reads '230 V, 1.61 kW'.
(a) State the equation linking power, current and voltage. [1 mark]
(b) Calculate the current in the steam cleaner when it is working normally. [3 marks]
(c) Fuses rated at 3 A, 5 A and 13 A are available. State which fuse should be fitted in the plug of the steam cleaner and explain your choice. [3 marks]
(d) The steam cleaner has a small indicator lamp. The voltage across the lamp is 2 V and the current in it is 0.02 A. Calculate the power of the lamp. [2 marks]
Model answer
(a) power = current × voltage, or P = I × V (1)
(b) Power = 1.61 × 1000 = 1610 W (1)
I = P ÷ V = 1610 ÷ 230 (1)
I = 7 A (1)
(c) The 13 A fuse (1). The 3 A and 5 A fuses have ratings below the normal current of 7 A, so they would melt when the cleaner is working normally (1). 13 A is the rating just above the normal current, so the fuse will not melt in normal use but will melt if a fault makes the current too large (1).
(d) P = I × V = 0.02 × 2 (1)
P = 0.04 W (1)
Exam tip
In a fuse question like part (c), always use your calculated current in the explanation. Compare each fuse rating with it: say which ratings are below the normal current and which one is just above it. Naming the fuse with no reason gets only the first mark.