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

The component labelled X is a fuse. Its primary purpose is to act as a safety device that protects the circuit from excessive current. If there is a fault in the system causing too much current to flow, the fuse will blow and break the circuit, preventing potential damage or fire.

To find the current, use the formula:

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

Given:

• Energy (E) = 9000 J
• Voltage (V) = 230 V
• Time (t) = 1 minute = 60 seconds

Substituting these values into the equation:

$I = \frac{9000}{230 \times 60}$

Calculating this gives: $I = \frac{9000}{13800} = 0.65217 A$

Rounding this value gives approximately 0.65 A.

The useful energy transferred to the water is less than the total energy supplied to the pump due to energy losses in the system. These losses can occur because of factors such as friction, heat, and sound, which dissipate some of the energy supplied to the pump. Consequently, not all the energy input is converted into useful work for the water.

Efficiency can be calculated using the formula:

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

Given:

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

Substituting these values gives:

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

To express this as a percentage, multiply by 100: $0.9333 \times 100 = 93.33\%$

Thus, the efficiency of the pump is approximately 93.33%.