Electrical power and energy

Power and energy

P = VI and W = VIt, where P is in watts, V in volts, I in amperes, t in seconds and W (energy) in joules. These come from V = W/Q and I = Q/t.

Derived forms

Substituting V = IR into P = VI gives

• P = I² R (useful when you know the current, for example in series)
• P = V² / R (useful when you know the p.d., for example in parallel)

Choose the form that uses the quantities you already have.

Worked example

A heater runs on 230 V and draws 8.0 A for 10 minutes. Find its power and the energy transferred.

1. P = VI = 230 x 8.0 = 1840 W
2. W = Pt = 1840 x 600

Answer: P = 1.84 kW; energy = 1.10 x 10⁶ J

Check yourself

1. A component has 12 V across it and 2.5 A through it. Find its power in W.
   Show answer

   30 W. P = VI = 12 x 2.5 = 30 W.

2. A current of 0.50 A flows through a 40 Ω resistor. Find the power in W.
   Show answer

   10 W. P = I² R = 0.25 x 40 = 10 W.

3. A 529 Ω lamp is connected to 230 V. Find its power in W.
   Show answer

   100 W. P = V² / R = 52900 / 529 = 100 W.

4. A 6.0 V supply drives 0.30 A through a motor for 5.0 minutes. Find the energy transferred in J.
   Show answer

   540 J. W = VIt = 6.0 x 0.30 x 300 = 540 J.

5. Find the current in a 60 W lamp running on 230 V, in A, to 3 s.f.
   Show answer

   0.261 A. I = P / V = 60 / 230 = 0.261 A.

6. A 2.5 kW heater runs on 230 V. Find its resistance in Ω, to 3 s.f.
   Show answer

   21.2 Ω. R = V² / P = 52900 / 2500 = 21.2 Ω.

7. The current through a fixed resistor is doubled. What happens to the power?
   • It doubles
   • It becomes 4 times as large
   • It halves
   • It stays the same
   Show answer

   It becomes 4 times as large. P = I² R, so P is proportional to I².

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Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics specification (XPH11/YPH11), Issue 3
• Pearson qualifications: International Advanced Levels Physics (2018)

Last updated 7 November 2026. Spotted a mistake? Tell us.