The battery had a power output of 230 mW when the resistance of the variable resistor was 36 Ω - AQA - GCSE Physics - Question 7 - 2021 - Paper 1

To find the current, we use the formula for power:

$P = IV$

Where:

• $P = 230 \text{ mW} = 0.230 \text{ W}$,
• $I$ is the current in amps,
• $V$ is the potential difference.

Rearranging for current gives: $I = \frac{P}{V}$

Using the resistance value of 36 Ω, we can express the potential difference as: $V = I \times R$ Substituting the current back into the equation gives: $V = 0.230 \text{ W} \times \frac{1}{I}$