A-Level Physics October/November 2024 Q8(c): Use data from (b), including your answer in (b)(iii), to show by calculation that the m…

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  The power dissipated in the resistor is given by the formula $P = I^2R$.

  From part (b), the peak current is $I_0 = 3.5$ A and the r.m.s. current is $I_{rms} = 2.47$ A. The resistance is $R = 680 \, \Omega$.

  First, calculate the peak power ($P_0$) using the peak current ($I_0$): $P_0 = I_0^2 R$ $P_0 = (3.5)^2 \times 680$ $P_0 = 12.25 \times 680 = 8330$ W

  Next, calculate the mean power ($\langle P \rangle$) using the r.m.s. current ($I_{rms}$): $\langle P \rangle = I_{rms}^2 R$ $\langle P \rangle = (2.47)^2 \times 680$ $\langle P \rangle = 6.1009 \times 680 = 4148.6$ W $\approx 4150$ W (3 s.f.)

  Finally, compare the mean power to half of the peak power: Half of peak power = $\frac{P_0}{2} = \frac{8330}{2} = 4165$ W.

  The calculated mean power (4150 W) is equal to half the peak power (4165 W) within the precision of the value for $I_{rms}$ used. This shows that the mean power is half of the peak power.

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  How the marks are awarded

  • C1 — Stating or using the correct formula for power, P = I²R.
  • M1 — Correctly substituting values to calculate either the peak power (3.5² × 680) or the mean power (2.47² × 680).
  • A1 — Calculating both peak power (8330 W) and mean power (≈4150 W) correctly, and then making a clear comparison (e.g., by dividing the peak power by 2) to conclude that the mean power is half the peak power.

  Common mistakes

  • Confusing peak and r.m.s. values, for example, calculating mean power using the peak current ($P = I_0^2 R$).
  • Simply stating that mean power is half of peak power without performing any calculations, which fails to meet the 'show by calculation' instruction.
  • Making a calculation error when squaring the current or multiplying by the resistance.
  • Using a different power formula like P = V²/R or P = VI without first correctly calculating the required voltage values, which is less direct and more prone to error.

  Examiner tip: Always be clear on the distinction between peak and r.m.s. values, and ensure you use the correct one ($I_0$ for peak power, $I_{rms}$ for mean power) in power calculations.

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