AQA · GCSE · Combined Science
Course version: 8464 · Version from Sept 2018
Power: power = potential difference × current
Electrical power is the rate at which a component transfers energy.
- 4 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
Electrical power is the rate at which a component transfers energy. When a current I flows through a component with potential difference V across it, each coulomb of charge carries energy V and I coulombs pass each second, so the energy transferred each second is V × I. This gives power = potential difference × current, with power in watts (W), potential difference in volts (V) and current in amperes (A). For example, a 12 V lamp drawing 0.50 A has power 12 × 0.50 = 6.0 W, so it transfers 6.0 J every second. Rearranged, V = P ÷ I and I = P ÷ V, which lets you find an unknown when the other two quantities are known.
What earns the marks
- State that power is the rate of energy transfer, measured in watts, where 1 W = 1 J/s.
- Substitute the correct values into power = potential difference × current, keeping V in volts and I in amperes.
- Calculate correctly, for example 12 V × 0.50 A = 6.0 W, and give the unit W.
- Rearrange the equation to find potential difference (V = P ÷ I) or current (I = P ÷ V) when required.
- Recognise that for a fixed potential difference, increasing the current increases the power transferred.
Common mistakes and how to put them right
Multiplying current by time instead of by potential difference.
Put it right: Power is V × I, while energy transferred is V × I × t, so check which quantity the question asks for.
Using milliamperes or kilovolts directly in the equation.
Put it right: Convert 250 mA to 0.25 A and 2 kV to 2000 V before substituting.
Writing the unit of power as joules.
Put it right: Power is measured in watts, and joules describe energy transferred, not the rate of transfer.
Exam tips
- Write the equation, then substitute values with units before calculating, so an examiner can follow your method even if the final number slips.
- Check the unit asked for; if the answer is required in kilowatts, divide the value in watts by 1000 at the end.
- For rearrangement questions, rearrange the equation symbolically first, then substitute, to reduce arithmetic errors.
Try the lesson
The start of the Power: power = potential difference × current lesson, made from checked teaching notes for this specification point.
Have a go first
A lamp is connected to a 12 V supply and a current of 2 A flows through it. Using power = potential difference × current, what is the power of the lamp?
Listen
Electrical power is the rate at which a component transfers energy. When a current I flows through a component with potential difference V across it, each coulomb of charge carries energy V and I coulombs pass each second, so the energy transferred each second is V × I. This gives power = potential difference × current, with power in watts (W), potential difference in volts (V) and current in amperes (A). For example, a 12 V lamp drawing 0.50 A has power 12 × 0.50 = 6.0 W, so it transfers 6.0 J every second. Rearranged, V = P ÷ I and I = P ÷ V, which lets you find an unknown when the other two quantities are known. A strong answer earns marks for these points. State that power is the rate of energy transfer, measured in watts, where 1 W = 1 J/s. Substitute the correct values into power = potential difference × current, keeping V in volts and I in amperes. Calculate correctly, for example 12 V × 0.50 A = 6.0 W, and give the unit W. Rearrange the equation to find potential difference (V = P ÷ I) or current (I = P ÷ V) when required. Exam tip: Write the equation, then substitute values with units before calculating, so an examiner can follow your method even if the final number slips.
Key words
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Power
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Relationship between potential difference, current and power
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Unit conversion and standard units
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Rearranging and applying equations
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See every definition
- Power
- The rate at which a component transfers energy, measured in watts, where 1 W = 1 J/s.
- Relationship between potential difference, current and power
- Power equals potential difference multiplied by current, with power in watts, potential difference in volts and current in amperes.
- Unit conversion and standard units
- Convert milliamperes to amperes and kilovolts to volts before substituting into the equation, keeping potential difference in volts and current in amperes.
- Rearranging and applying equations
- Rearranged, V = P ÷ I and I = P ÷ V, which lets you find an unknown when the other two quantities are known.
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