The mean power dissipated in a resistor is 47.5 µW when the root mean square (rms) voltage across the resistor is 150 mV - AQA - A-Level Physics - Question 23 - 2017 - Paper 2

Calculating Resistance:

1. Compute $V_{rms}^2$:

   $V_{rms}^2 = (150 \times 10^{-3})^2 = 2.25 \times 10^{-4}$

2. Calculate Resistance $R$:

   $R = \frac{2.25 \times 10^{-4}}{47.5 \times 10^{-6}} \approx 4.7368 \Omega$

Now substituting $R$ back to find $I_{rms}$:

$I_{rms} = \sqrt{\frac{P}{R}} = \sqrt{\frac{47.5 \times 10^{-6}}{4.7368}} \approx 0.0032 A = 3.2 \times 10^{-3} A$

The relationship between peak current ($I_{peak}$) and RMS current is:

$I_{peak} = I_{rms} \sqrt{2}$

Thus:

$I_{peak} = 3.2 \times 10^{-3} \sqrt{2} \approx 4.52 \times 10^{-3} A = 4.5 \times 10^{-3} A$

Therefore, the peak current in the resistor is approximately $4.5 \times 10^{-3} A$.