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

Component X typically refers to a fuse or circuit breaker. Its purpose is to protect the electrical circuit by breaking the connection in case of excess current, thereby preventing potential damage or fire hazards.

To find the current (I), we will use the formula:

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

Where:

• E = 9000 J (energy transferred)
• V = 230 V (voltage)
• t = 1 minute = 60 seconds.

Substituting the values in:

$I = \frac{9000}{230 \times 60}$ $= \frac{9000}{13800}$ $\approx 0.65 A$

The useful energy transferred to the water is less than the total energy supplied because of energy losses in the system. These losses can occur due to factors like friction, heat dissipation, and inefficiencies in the pump’s mechanism, which prevent all the supplied energy from being converted to useful work.

The efficiency of the pump can be calculated using the formula:

$\text{efficiency} = \frac{\text{useful energy transferred by the pump}}{\text{total energy supplied to the pump}}$

Where:

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

Substituting the values:

$\text{efficiency} = \frac{8400}{9000} \approx 0.9333$

Converting this to a percentage:

$\text{efficiency} \approx 93.33\%$