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 the fuse. Its primary purpose is to protect the circuit by melting and breaking the circuit in case of an overload, thereby preventing damage or fire hazards.

To find the current, we can 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 the values:

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

Calculating:

$I = \frac{9000}{13800} = 0.65217 A$

Thus, rounding, the current is approximately 0.65 A.

The useful energy transferred to the water is 8400 J, while the total energy supplied to the pump is 9000 J. The difference arises due to energy losses in the system, such as heat loss due to friction and inefficiency in energy conversion within the pump. Not all supplied energy is converted into useful work.

The 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 = 8400 J
• Total energy = 9000 J

Substituting the values:

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

Expressing this as a percentage:

$\text{Efficiency} = 0.9333 \times 100 = 93.33\%$

Thus the efficiency of the pump is approximately 93.33%.