During one year, 1.25 × 10^18 J of energy was transferred from the National Grid - AQA - GCSE Physics - Question 5 - 2021 - Paper 1

To calculate the mean current, we use the formula:

$I = \frac{Q}{t}$

Where ( Q = 27,000 ) C and ( t = 24 \text{ hours} = 24 \times 60 \times 60 = 86,400 \text{ s} ).

Substituting these values gives:

$I = \frac{27,000 \text{ C}}{86,400 \text{ s}} \approx 0.3125 \text{ A}$

Therefore, the mean current in the cable is approximately 0.3125 A.

To calculate the useful power output, use the formula:

$Useful \ Power \ Output = Efficiency \times Power \ Input$

Given the efficiency of the solar cells is 0.15 and the total power input is 7.8 kW,

Convert 7.8 kW to watts:

$7.8 \text{ kW} = 7800 \text{ W}$

Now, substitute into the formula:

$Useful \ Power \ Output = 0.15 \times 7800 \text{ W} = 1170 \text{ W}$

Thus, the useful power output of the solar cells is 1170 W.

It is unlikely that all electricity in the UK can be generated by solar systems because:

1. The UK experiences long periods of low sunlight, especially during the winter months.
2. Solar panels rely on adequate sunlight exposure, and the available solar radiation may not meet demand during cloudy days.
3. A large area of land would be needed to cover with solar cells to produce sufficient energy, which may not be feasible.

Additionally, the efficiency of solar panels is limited, meaning a significant amount of surface area would be required to produce the needed electricity.