AQA · GCSE · Combined Science

    Course version: 8464 · Version from Sept 2018

    Power: P = I² R

    This is the symbolic form of the power equation linking power P in watts, current I in amperes and resistance R in ohms.

      Study notes

      From the checked teaching notes for this specification point.

      The idea

      This is the symbolic form of the power equation linking power P in watts, current I in amperes and resistance R in ohms. It applies to any component obeying Ohm's law and describes the rate at which electrical energy is transferred to thermal energy in a resistor. Because current is squared, the equation shows that increasing current has a much larger effect on power than increasing resistance. For example, if I = 2 A and R = 5 Ω, then P = 2² × 5 = 4 × 5 = 20 W. The equation can be rearranged as I = √(P ÷ R) and R = P ÷ I². It is consistent with P = VI and V = IR, since substituting V = IR into P = VI gives P = I²R. Use SI units throughout and remember that the square applies only to the current.

      What earns the marks

      • Identify each symbol: P is power in watts, I is current in amperes and R is resistance in ohms.
      • Substitute numerical values into P = I² R, square the current and multiply by resistance.
      • Rearrange the equation to make current or resistance the subject, for example I = √(P ÷ R).
      • Show that P = I² R follows from combining P = VI with V = IR.
      • Use the equation to compare power dissipation in components with different currents or resistances.

      Common mistakes and how to put them right

      • Squaring the whole product IR instead of only the current.

        Put it right: Apply the square to I only, so P = I² × R.

      • Forgetting to take the square root when rearranging for current.

        Put it right: I = √(P ÷ R), not P ÷ R.

      • Mixing units, such as using mA with Ω and expecting W.

        Put it right: Convert current to A before calculating.

      Exam tips

      • Quote the equation in symbol form, then substitute values clearly before evaluating.
      • Check whether the question asks for power, current or resistance and rearrange before substituting.
      • Use the equation to justify why high-current circuits need thicker wires to limit heating.

      Try the lesson

      The start of the Power: P = I² R lesson, made from checked teaching notes for this specification point.

      Have a go first

      A resistor has a current of 2 A flowing through it and a resistance of 5 Ω. Use P = I² R to calculate the power dissipated by the resistor.

      Listen

      This is the symbolic form of the power equation linking power P in watts, current I in amperes and resistance R in ohms. It applies to any component obeying Ohm's law and describes the rate at which electrical energy is transferred to thermal energy in a resistor. Because current is squared, the equation shows that increasing current has a much larger effect on power than increasing resistance. For example, if I = 2 A and R = 5 Ω, then P = 2² × 5 = 4 × 5 = 20 W. The equation can be rearranged as I = √(P ÷ R) and R = P ÷ I². It is consistent with P = VI and V = IR, since substituting V = IR into P = VI gives P = I²R. Use SI units throughout and remember that the square applies only to the current. A strong answer earns marks for these points. Identify each symbol: P is power in watts, I is current in amperes and R is resistance in ohms. Substitute numerical values into P = I² R, square the current and multiply by resistance. Rearrange the equation to make current or resistance the subject, for example I = √(P ÷ R). Show that P = I² R follows from combining P = VI with V = IR. A common slip: Squaring the whole product IR instead of only the current. Instead, apply the square to I only, so P = I² × R. Exam tip: Quote the equation in symbol form, then substitute values clearly before evaluating.

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