A transformer has an efficiency of 80%. It works at 3 kW and 120 V. If the secondary voltage is 240 V, what will be the secondary current?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 10 A
The efficiency of a transformer is the ratio of the output power (secondary) to the input power (primary). The 3 kW at 120 V is the input; only 80% of it reaches the secondary.
\( \displaystyle \eta = \frac{P_{\mathrm{out}}}{P_{\mathrm{in}}} \)
\( \displaystyle P_{\mathrm{out}} = 0.8 \times 3000\,\mathrm{W} = 2400\,\mathrm{W} \)
The secondary current follows from the output power and the secondary voltage.
\( \displaystyle I_{s} = \frac{P_{\mathrm{out}}}{V_{s}} = \frac{2400\,\mathrm{W}}{240\,\mathrm{V}} = 10\,\mathrm{A} \)