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Physics · Electricity

Electrical power

Use \(P = VI\) and \(P = I^2R\) to calculate the power of electrical devices.

  • 6 key terms
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Warm-up

Answer each one, then check.

  1. 1

    What is the unit of power?

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    Watt (W)

  2. 2

    Write the equation linking energy and time for power.

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    \(P = E \div t\)

  3. 3

    What is \(V = IR\) used for?

    Show answerHide answer

    Linking pd, current and resistance

  4. 4

    Convert 2.3 kW to watts.

    Show answerHide answer

    2300 W

  5. 5

    Work out \(3^2\).

    Show answerHide answer

    9

Learning Objectives

  1. 1Explain how the power transfer in a device is related to the pd across it and the current through it.
  2. 2Recall and apply \(P = VI\).
  3. 3Recall and apply \(P = I^2R\).
  4. 4Rearrange to find current, pd or resistance.

ELECTRICAL POWER

power \(=\) potential difference \(\times\) current \(P = VI\)

Also \(P = I^2R\) (power = current squared × resistance). Power is in watts (W), pd in volts (V), current in amperes (A) and resistance in ohms (Ω).

Using P = VI

A kettle has a current of 10 A when connected to the 230 V mains. Calculate its power.

Show the solutionHide the solution
  1. 1 Write the equation \(P = VI\)
  2. 2 Substitute \(P = 230 \times 10\)
  3. 3 Answer \(P = 2300\) W

Answer2300 W (2.3 kW)

Using P = I²R

A current of 6.0 A flows through a 5.0 Ω resistor. Calculate the power.

Show the solutionHide the solution
  1. 1 Write the equation \(P = I^2R\)
  2. 2 Square the current first \(6.0^2 = 36\)
  3. 3 Substitute \(P = 36 \times 5.0\)
  4. 4 Answer \(P = 180\) W

Answer180 W

Finding the Current

A 60 W lamp is connected to the 230 V mains. Calculate the current.

Show the solutionHide the solution
  1. 1 Rearrange \(I = P \div V\)
  2. 2 Substitute \(I = 60 \div 230\)
  3. 3 Answer \(I = 0.26\) A

Answer0.26 A

Which Equation?

Look at what you are given.

  • V and I

    Use: \(P = VI\)

  • I and R

    Use: \(P = I^2R\)

  • V and R

    Use: Find I first with \(I = V \div R\), then \(P = VI\)

  • E and t

    Use: \(P = E \div t\)

Common Mistakes

Take care with squares.

  • Squaring

    Square only the current, then multiply by the resistance.

  • Units

    Convert mA to A and kW to W first.

  • Rating labels

    The power rating of an appliance is at its normal pd.

Rating Plate

A hair dryer is rated 1500 W at 230 V. Calculate the current. A toaster takes 4.0 A at 230 V; find its power.

1. Rearrange P = VI for current.

2. Then calculate the toaster power.

A good answer shows: Hair dryer: \(1500 \div 230 = 6.5\) A. Toaster: \(230 \times 4.0 = 920\) W.

Can I...?

  1. 1Recall P = VI.
  2. 2Recall P = I²R.
  3. 3Choose the right equation.
  4. 4Rearrange for current.
  5. 5Square the current correctly.
  6. 6Convert units to W and A.
  7. 7Explain what power means.
  8. 8Give the unit.

Summary & Exam Focus

  • \(P = VI\).
  • \(P = I^2R\).
  • Power is in watts.
  • Higher power means faster energy transfer.

Exam focus

A kettle draws 10 A from the 230 V mains. Calculate its power. (2 marks) (2 marks)

Choose P = VI, substitute and give W.

Key terms

The vocabulary this lesson expects you to use. Each one is linked from the first place it appears above.

Power
The rate at which energy is transferred.
Watt
One joule per second.
Power rating
The power an appliance transfers in normal use.
Kilowatt
1000 watts.
Potential difference
Energy transferred per coulomb of charge.
Current
The rate of flow of charge.

Practice questions

Have a go at each one before you open its answer.

  1. Question 1 Calculate 2 marks

    A kettle is connected to the 230 V mains and the current is 10 A. Calculate the power of the kettle. Use the equation: power = potential difference × current

    Show answerHide answer

    Model answer

    \(P = 230 \times 10 = 2300\) W

    Mark scheme

    • Correct substitution — 1 mark
    • 2300 W — 1 mark
  2. Question 2 Calculate 3 marks

    A current of 6.0 A flows through a resistor of resistance 5.0 Ω. Calculate the power. Use the equation: power = (current)² × resistance

    Show answerHide answer

    Model answer

    \(P = 6.0^2 \times 5.0 = 180\) W

    Mark scheme

    • Squares the current — 1 mark
    • Correct substitution — 1 mark
    • 180 W — 1 mark
  3. Question 3 Calculate 3 marks

    A lamp has a power of 60 W and works at 230 V. Calculate the current in the lamp.

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    Model answer

    \(I = P \div V = 60 \div 230 = 0.26\) A

    Mark scheme

    • Rearranges to I = P ÷ V — 1 mark
    • Correct substitution — 1 mark
    • 0.26 A — 1 mark
  4. Question 4 Explain 2 marks

    A student says that the power of a device only depends on the current. Explain why the student is not correct.

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    Model answer

    Power depends on both the current and the potential difference (P = VI), or on the current and the resistance (P = I²R).

    Mark scheme

    • Power = V × I — 1 mark
    • Depends on pd (or resistance) as well as current — 1 mark
  5. Question 5 Calculate 4 marks

    An electric heater has a resistance of 46 Ω and is connected to a 230 V supply. Calculate the power of the heater.

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    Model answer

    \(I = 230 \div 46 = 5.0\) A; \(P = 230 \times 5.0 = 1150\) W

    Mark scheme

    • Rearranges V = IR — 1 mark
    • 5.0 A — 1 mark
    • Correct substitution into P = VI — 1 mark
    • 1150 W — 1 mark

Quick check

  1. The equation for electrical power is...

    1. A\(P = V \div I\)
    2. B\(P = V + I\)
    3. C\(P = VI\)
    4. D\(P = I \div V\)
    Show answerHide answer

    C: \(P = VI\)

    Power is pd times current.

  2. A 12 V, 3 A device has a power of...

    1. A4 W
    2. B15 W
    3. C9 W
    4. D36 W
    Show answerHide answer

    D: 36 W

    12 × 3 = 36 W.

  3. A 2 A current through a 5 Ω resistor gives a power of...

    1. A20 W
    2. B10 W
    3. C50 W
    4. D100 W
    Show answerHide answer

    A: 20 W

    \(2^2 \times 5 = 20\) W.

  4. A higher current at the same pd gives a...

    1. Asmaller power
    2. Bgreater power
    3. Cpower of zero
    4. Dlower resistance
    Show answerHide answer

    B: greater power

    P = VI.

  5. The unit of power is the...

    1. Ajoule
    2. Bampere
    3. Cwatt
    4. Dohm
    Show answerHide answer

    C: watt

    Watts, W.

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