- What electrical power means: the energy transferred each second by a circuit device.
- How electric current (flow of charge) and potential difference (energy transferred per unit charge) affect power.
- How to use P=EtP = \frac{E}{t}P=tE, P=I×VP = I \times VP=I×V, P=I2×RP = I^2 \times RP=I2×R, and E=I×V×tE = I \times V \times tE=I×V×t.
- How to choose the right equation and keep units under control.
A circuit device, such as a lamp, motor, heater or resistor, transfers energy. For example:
- a lamp transfers energy electrically to light and thermal stores
- a motor transfers energy electrically to kinetic and thermal stores
- a heater transfers energy electrically to thermal stores
The symbol for energy transferred is EEE. It is measured in joules, J.
To understand electrical power, you need three earlier circuit ideas.
Key circuit quantities
- Current, III, is the rate of flow of electric charge. It is measured in amperes, A.
- Potential difference, VVV, often called voltage, is the energy transferred per unit charge as charge passes through a component. It is measured in volts, V.
- Resistance, RRR, tells you how difficult it is for current to flow through a component. It is measured in ohms, Ω.
Power is not the same as energy. Energy is the total amount transferred. Power tells you how quickly that energy is transferred.
Power
Power is the energy transferred per second. A power of 1 watt means 1 joule of energy is transferred every second.
The symbol for power is PPP. It is measured in watts, W.
For Edexcel 1PH0, you need to recall and use:
P=EtP = \frac{E}{t}P=tE
where:
- PPP is power in watts, W
- EEE is energy transferred in joules, J
- ttt is time taken in seconds, s
Power is a rate
A high-power device transfers energy quickly. A low-power device transfers energy more slowly, but if it is left on for a long time it can still transfer a large total amount of energy.
Calculating power from energy and time
A phone charger transfers 18 000 J of energy in 30 minutes. Calculate its power.
- Convert the time into seconds because the equation uses seconds: t=30×60=1800 st = 30 \times 60 = 1800\ \text{s}t=30×60=1800 s.
- Use power as energy transferred per second: P=EtP = \frac{E}{t}P=tE.
- Substitute the values: P=18 000 J1800 s=10 WP = \frac{18\,000\ \text{J}}{1800\ \text{s}} = 10\ \text{W}P=1800 s18000 J=10 W.
- Interpret the result: the charger transfers 10 J of energy each second.
Forgetting to convert time
In power calculations, time must be in seconds. If the question gives minutes or hours, convert before substituting into P=EtP = \frac{E}{t}P=tE or E=I×V×tE = I \times V \times tE=I×V×t.
In a circuit device, power depends on two things:
- the current through the device
- the potential difference across the device
A bigger current means more charge passes through each second. A bigger potential difference means each unit of charge transfers more energy. So increasing either current or potential difference increases the power transferred.
For Edexcel 1PH0, you need to recall and use:
P=I×VP = I \times VP=I×V
where:
- PPP is electrical power in watts, W
- III is current in amperes, A
- VVV is potential difference in volts, V
To find the power of a real component, measure the current through it with an ammeter in series, and measure the potential difference across it with a voltmeter in parallel.

Finding current from power and voltage
A 24 W lamp is connected across a 12 V supply. Calculate the current through the lamp.
- The power and potential difference are for the same lamp, so use P=I×VP = I \times VP=I×V.
- Rearrange to make current the subject: I=PVI = \frac{P}{V}I=VP.
- Substitute the values: I=24 W12 V=2.0 AI = \frac{24\ \text{W}}{12\ \text{V}} = 2.0\ \text{A}I=12 V24 W=2.0 A.
- The lamp has a current of 2.0 A through it.
Using the wrong voltage
Use the potential difference across the device you are calculating the power for, not automatically the total supply voltage. This matters especially in circuits with components in series.
If a device has power PPP, then in time ttt it transfers energy EEE.
Because P=I×VP = I \times VP=I×V and P=EtP = \frac{E}{t}P=tE, you can combine them to get:
E=I×V×tE = I \times V \times tE=I×V×t
For this spec point, Edexcel says you must use this equation. That means you must be able to select it, substitute into it and rearrange it; it is not labelled as a recall-and-use equation in the extract. It is still worth learning because it comes directly from the two power equations.
In the equation:
- EEE is energy transferred in joules, J
- III is current in amperes, A
- VVV is potential difference in volts, V
- ttt is time in seconds, s
Calculating energy transferred by a motor
A motor has a current of 2.0 A through it and a potential difference of 12 V across it. It runs for 5 minutes. Calculate the energy transferred.
- Convert the time into seconds: t=5×60=300 st = 5 \times 60 = 300\ \text{s}t=5×60=300 s.
- Use the equation connecting current, potential difference and time: E=I×V×tE = I \times V \times tE=I×V×t.
- Substitute the values: E=2.0 A×12 V×300 s=7200 JE = 2.0\ \text{A} \times 12\ \text{V} \times 300\ \text{s} = 7200\ \text{J}E=2.0 A×12 V×300 s=7200 J.
- The motor transfers 7200 J, which is 7.2 kJ.
A quick units check
Watts are joules per second. So if your answer is in W, it should describe a rate of energy transfer. If your answer is in J, it should describe a total amount of energy transferred.
Sometimes you are given the current and resistance, but not the potential difference. In that case, use the resistance version of the power equation.
For Edexcel 1PH0, you need to recall and use:
P=I2×RP = I^2 \times RP=I2×R
where:
- PPP is electrical power in watts, W
- III is current in amperes, A
- RRR is resistance in ohms, Ω
This equation comes from combining P=I×VP = I \times VP=I×V with V=I×RV = I \times RV=I×R. You do not need to prove it in detail, but you should understand the idea: resistance affects the potential difference needed for a particular current, so it affects the power transferred.
Using resistance to find power
A resistor has a resistance of 20 Ω. The current through it is 0.40 A. Calculate the power transferred by the resistor.
- The question gives current and resistance, but not potential difference, so use P=I2×RP = I^2 \times RP=I2×R.
- Square the current: I2=0.402=0.16 A2I^2 = 0.40^2 = 0.16\ \text{A}^2I2=0.402=0.16 A2.
- Multiply by the resistance: P=0.16×20 Ω=3.2 WP = 0.16 \times 20\ \Omega = 3.2\ \text{W}P=0.16×20 Ω=3.2 W.
- The resistor transfers energy at a rate of 3.2 W.
Forgetting the square
In P=I2×RP = I^2 \times RP=I2×R, the current is squared. If the current doubles, the power transferred in the same resistance becomes four times bigger, not just twice as big.
A good way to choose is to look at which quantities you know and which one you need.
- If you have energy and time, use P=EtP = \frac{E}{t}P=tE.
- If you have current and potential difference, use P=I×VP = I \times VP=I×V.
- If you have current and resistance, use P=I2×RP = I^2 \times RP=I2×R.
- If you need energy transferred and have current, potential difference and time, use E=I×V×tE = I \times V \times tE=I×V×t.
Rearranging safely
Do the same operation to both sides. For example, from P=I×VP = I \times VP=I×V, divide both sides by VVV to get I=PVI = \frac{P}{V}I=VP.
A power rating tells you how quickly a device transfers energy under its normal operating conditions. For example, a 100 W heater transfers 100 J of energy every second when it is working as intended.
A more powerful device is not necessarily “better”; it just transfers energy faster. Whether that is useful depends on the purpose of the device.
Power ratings depend on conditions
The power written on a device usually assumes it is connected to its correct potential difference. If the potential difference changes, the current and power may also change.
In the exam
- Check that time is in seconds before using any equation involving power or energy.
- Match the values to the same component: current through it, potential difference across it, resistance of it.
- Write the equation, substitute with units, then check whether the answer should be in W or J.
Check yourself
- A device transfers 600 J in 20 s. What is its power?
- Why does increasing either current or potential difference increase electrical power?
- A 5 Ω resistor has a current of 3 A through it. Which power equation would you use?