Figure 3 shows an LED torch - AQA - GCSE Physics - Question 2 - 2020 - Paper 1

To calculate the total charge flow through the cells, we need to substitute the given values into the equation:

• Current ($I$) = 50 mA = 0.050 A
• Time ($t$) = 14 400 s.

Now, using the formula:

$Q = 0.050 A \times 14 400 s$

Calculating this gives:

$Q = 720 C$

The equation that links efficiency ($\eta$), total power input ($P_{input}$), and useful power output ($P_{output}$) is:

$\eta = \frac{P_{output}}{P_{input}}$

where:

• $\eta$ is the efficiency (as a decimal or percentage),
• $P_{output}$ is the useful power output in watts (W),
• $P_{input}$ is the total power input in watts (W).

To find the useful power output of the LED, we can rearrange the efficiency equation:

$P_{output} = \eta \times P_{input}$

Given:

• Efficiency ($\eta$) = 0.75
• Total power input ($P_{input}$) = 0.24 W.

Substituting the values:

$P_{output} = 0.75 \times 0.24 W$

Calculating this yields:

$P_{output} = 0.18 W$