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    Power and energy in circuits: P = VI, P = I²R, E = Pt and E = QV

    By Engr. Mehmood Kakar · Updated

    Quick answer

    Use P = V × I when you know pd and current, P = I² × R when you know current and resistance, then E = P × t for the energy transferred. E = Q × V gives the same energy from charge and pd. Power is in watts (joules per second), so time must be in seconds to get energy in joules.

    The four equations

    EquationIn wordsUnits
    P = V Ipower = potential difference × currentW = V × A
    P = I² Rpower = current² × resistanceW = A² × Ω
    E = P tenergy transferred = power × timeJ = W × s
    E = Q Venergy transferred = charge flow × potential differenceJ = C × V

    All four are on the AQA equation sheet. You still have to choose the right one and rearrange it.

    Which equation to use

    • Given pd and current, finding power → P = VI
    • Given current and resistance, finding power → P = I²R
    • Given power and time, finding energy → E = Pt
    • Given charge and pd, finding energy → E = QV

    If a question gives pd and resistance but no current, find the current first with I = V ÷ R, then use either power equation.

    P = VI worked

    A kettle is rated 2.3 kW and runs on the 230 V mains. Calculate the current.

    1. Convert: P = 2.3 kW = 2300 W
    2. Rearrange: I = P ÷ V
    3. I = 2300 ÷ 230
    4. I = 10 A

    P = I²R worked

    A current of 3.0 A flows through a 20 Ω resistor. Calculate the power.

    1. P = I² × R = 3.0² × 20
    2. Square the current first: 3.0² = 9.0
    3. P = 9.0 × 20
    4. P = 180 W

    A heater has a resistance of 25 Ω and a power of 100 W. Calculate the current. (Higher)

    1. Rearrange: I² = P ÷ R = 100 ÷ 25 = 4.0
    2. Square root: I = √4.0
    3. I = 2.0 A

    Energy: E = Pt and E = QV

    How much energy does the 2.3 kW kettle transfer in 3 minutes?

    1. Convert: P = 2300 W, t = 3 × 60 = 180 s
    2. E = P × t = 2300 × 180
    3. E = 414 000 J (414 kJ)

    The two energy equations agree. A 12 V lamp taking 2.0 A for 5 minutes:

    Find the energy transferred by a 12 V, 2.0 A lamp in 5 minutes, two ways.

    1. t = 5 × 60 = 300 s
    2. Way 1: P = VI = 12 × 2.0 = 24 W, so E = Pt = 24 × 300 = 7200 J
    3. Way 2: Q = It = 2.0 × 300 = 600 C, so E = QV = 600 × 12 = 7200 J
    4. E = 7200 J either way

    Why the National Grid uses high pd

    Power stations send electricity across the country at very high pd. For the same power, P = VI means a higher pd needs a smaller current. The power wasted heating the cables is P = I²R, so cutting the current cuts that waste sharply: halve the current and the heating loss falls to a quarter.

    Step-up transformers raise the pd before the cables, and step-down transformers lower it to 230 V for homes. The transformer guide has the turns-ratio calculations.

    Common mistakes

    • Minutes or hours in E = Pt. Convert to seconds for an answer in joules.
    • Kilowatts left as kW. 2.3 kW is 2300 W.
    • Squaring the wrong thing. In P = I²R only the current is squared.
    • Missing the square root. When finding current from P = I²R, the last step is √.

    Practice

    1. A hairdryer takes 5.0 A from the 230 V mains. Calculate its power.

    P = V × I = 230 × 5.0 → P = 1150 W

    2. A 60 W lamp is on for 2 hours. How much energy does it transfer?

    t = 2 × 3600 = 7200 s → E = P × t = 60 × 7200 → E = 432 000 J

    3. 150 C of charge flows through a component with a pd of 9.0 V across it. How much energy is transferred?

    E = Q × V = 150 × 9.0 → E = 1350 J

    4. A 0.50 A current flows through a 40 Ω resistor. Calculate the power.

    P = I² × R = 0.50² × 40 = 0.25 × 40 → P = 10 W

    Frequently asked questions

    What is the difference between P = VI and P = I²R?

    They give the same power. P = VI uses the potential difference and current. P = I²R uses the current and resistance; it comes from putting V = IR into P = VI. Pick the one that matches the quantities the question gives you.

    What unit is electrical power measured in?

    Watts (W). One watt is one joule of energy transferred per second. 1 kW = 1000 W.

    How do you calculate the energy used by an appliance?

    Multiply its power in watts by the time in seconds: E = P × t. The answer is in joules. If you know the charge instead, use E = Q × V.

    Why is electricity transmitted at high voltage?

    For the same power, a higher pd means a smaller current (P = VI). The energy wasted heating the cables depends on I²R, so a smaller current wastes much less energy. Step-up transformers raise the pd for the cables and step-down transformers lower it for homes.

    Engr. Mehmood Kakar

    Written by Engr. Mehmood Kakar

    Physics educator and engineer (MSc Physics, B.Ed, M.Ed) with 8 years of hands-on teaching, from college students to teacher training. About the author · LinkedIn (opens in a new tab)

    Checked against the AQA GCSE Physics (8463) specification on 15 November 2026.