5 (a) An electric water pump is powered by the 230 V mains supply - Edexcel - GCSE Physics - Question 5 - 2022 - Paper 1

Component X serves as a fuse. Its purpose is to protect the circuit by breaking the flow of electricity in case of a fault, such as excessive current, which can cause overheating and potential fires.

To calculate the current, we use the formula: $I = \frac{E}{V \times t}$ Where:

• E = 9000 J
• V = 230 V
• t = 60 s
  Substituting the values gives: $I = \frac{9000}{230 \times 60} = \frac{9000}{13800} \approx 0.65 A$

The useful energy transferred to the water is less than the total energy supplied to the pump due to energy losses. This loss can occur in the form of heat dissipated to the surroundings, friction within the pump, or energy used in sound production. Only part of the input energy is converted to useful work.

The efficiency can be calculated using the formula: $efficiency = \frac{useful \ energy \ transferred \ by \ the \ pump}{total \ energy \ supplied \ to \ the \ pump}$ Substituting the values gives: $efficiency = \frac{8400}{9000} \approx 0.93$
Thus, the efficiency of the pump is approximately 93%.