If the electrician touches the live wire he will receive an electric shock - AQA - GCSE Physics - Question 13 - 2018 - Paper 1

To find the minimum diameter of wire, we first need to calculate the cross-sectional area, A, needed to safely carry the current. The power, P, of the shower is 13.8 kW, which is equal to 13,800 W. The current, I, can be calculated using the relationship:

$P = IV$

Rearranging gives: $I = \frac{P}{V} = \frac{13800}{230} = 60 A$

Using the formula for diameter:

$d = \sqrt{\frac{4A}{\pi}}$

Assuming a minimum cross-sectional area of 9.5 mm² (a standard), we can find d: $A = 9.5 \text{ mm}^2 = 9.5 \times 10^{-6} \text{ m}^2$

Substituting this value into the diameter formula, we get: $d = \sqrt{\frac{4 \times 9.5 \times 10^{-6}}{\pi}} \approx 3.50 \text{ mm} \text{ to } 3.55 \text{ mm}$

To calculate the resistance, R, we will use Ohm's law and the power formula. First, calculate the total charge, Q, that flows: $Q = 18000 \text{ C}$

The current, I, can also be calculated as: $I = \frac{Q}{t} = \frac{18000}{300} = 60\,A$

Now we can relate power, voltage, and resistance: $P = IV$ $V = IR$

Thus, $R = \frac{V}{I}$

Substituting our earlier calculations: $R = \frac{230}{60} \approx 3.83 \Omega$