Power $P$ is dissipated in a resistor of resistance $R$ carrying a direct current $I$ - AQA - A-Level Physics - Question 24 - 2022 - Paper 2

To determine the power dissipated in the second resistor, we first recall the formula for power in a resistor:

$P = I^2 R$

1. Find the current across the first resistor: For the first resistor, the power $P$ is given by the equation: $P = I^2 R$ Rearranging gives us: I = rac{P}{R}

2. Determine the effective current for the second resistor: The second resistor has double the resistance, so: $R_{2R} = 2R$ The peak current in the second resistor is still $I$, but for alternating current, we use the RMS value, which is:

   $$$$

oot{2}{}} $$

3. Calculate the power in the second resistor: Using the RMS current: $P_{2R} = I_{ ext{RMS}}^2 (2R)$ Substitute the RMS current: P_{2R} = rac{I^2}{2} (2R) = I^2 R

4. Relate this back to $P$: Since we know from earlier that: $P = I^2 R$ Thus, the power dissipated in the second resistor is: $P_{2R} = P$

Therefore, the power dissipated in the second resistor is $P$.