Figure 1 shows how the National Grid connects a power station to consumers - AQA - GCSE Physics - Question 1 - 2023 - Paper 1

We start with the power loss equation:

$$P = I^2R$$

Given:

• Power loss, $P = 1.60 \times 10^9 \, W$
• Current, $I = 2000 \, A$

Now, substituting the given values:

$$1.60 \times 10^9 = (2000)^2 \times R$$

This simplifies to:

$$R = \frac{1.60 \times 10^9}{2000^2} = 400 \, \Omega$$

The equation linking efficiency, total energy input (E_in) and useful energy output (E_out) is:

$$\text{Efficiency} = \frac{\text{Useful Energy Output}}{\text{Total Energy Input}}$$

Given:

• Total energy input, $E_{in} = 34.2 \, GJ$
• Efficiency, $\text{Efficiency} = 0.992$

Using the efficiency equation:

$$E_{out} = \text{Efficiency} \times E_{in} = 0.992 \times 34.2$$

Calculating:

$$E_{out} \approx 33.9 \, GJ$$