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

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5 (a) An electric water pump is powered by the 230 V mains supply. Figure 12 shows the inside of the plug on the cable to the pump. (i) One wire in the plug is the ... show full transcript

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

Step 1

5 (a) (i) One wire in the plug is the earth wire.

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Answer

The correct answer is B: live and neutral. The live wire carries the current to the appliance, while the neutral wire returns it to the power source.

Step 2

5 (a) (ii) Describe the purpose of the component labelled X.

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Answer

Component X is the fuse. Its purpose is to protect the appliance from excessive current by melting or breaking the circuit if the current exceeds a safe level, thereby preventing potential damage or fire.

Step 3

5 (b) Calculate the current in the pump motor.

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Answer

To calculate the current, we use the formula:

I=EV×tI = \frac{E}{V \times t}

Where:

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

Now substituting the values:

I=9000230×60=9000138000.65AI = \frac{9000}{230 \times 60} = \frac{9000}{13800} \approx 0.65 A

Step 4

5 (c) (i) Explain why the useful energy transferred to the water is different from the total energy supplied to the pump.

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Answer

The useful energy transferred to the water is less than the total energy supplied because some energy is lost through various forms such as heat to the surroundings or mechanical inefficiencies. Specifically, in this case, 600 J of energy is dissipated as thermal energy.

Step 5

5 (c) (ii) Calculate the efficiency.

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Answer

The efficiency can be calculated using the formula:

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

Substituting the values:

efficiency=84009000=0.93393%\text{efficiency} = \frac{8400}{9000} = 0.933 \approx 93\%

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