An electric water pump is powered by the 230 V mains supply - Edexcel - GCSE Physics Combined Science - Question 3 - 2022 - Paper 1

Component X is a fuse, which is designed to protect the appliance by breaking the circuit if the current exceeds safe levels, thereby preventing damage or fire hazards.

To calculate the current, use the given formula:

$I = \frac{E}{V \cdot t}$

Where:

• $E = 9000$ J
• $V = 230$ V
• $t = 1 \text{ minute} = 60$ seconds

Substituting the values gives:

$I = \frac{9000}{230 \times 60} = \frac{9000}{13800} \approx 0.6522 \text{ A}$

Rounding this gives approximately $0.65$ A.

The useful energy transferred to the water is less than the total energy supplied due to energy losses in the system. These losses may occur from friction, heat dissipation, or other inefficiencies, meaning that not all supplied energy is converted into useful kinetic energy.

The efficiency can be calculated using:
$\text{efficiency} = \frac{\text{useful energy transferred by the pump}}{\text{total energy supplied to the pump}}$

Here:

• Useful energy transferred = 8400 J
• Total energy supplied = 9000 J

So, the efficiency is: $\text{efficiency} = \frac{8400}{9000} \approx 0.9333$

This indicates an efficiency of approximately 93.33%.