Correct Answer - Option 4 : 4.7%
Input = 2 kW ± 4%
Error in the input \(= \frac{4}{{100}} \times 2 \times {10^3} = 80\;W\)
Losses = 200 ± 5 W
Output = Input – Loses
= (2000 ± 80) – (200 ± 5)
= 1800 ± 85 W
Limiting error in the measurement of output
Losses \( = \frac{{85}}{{1800}} \times 100 = 4.72\% \)