The primary coil of a transformer draws a current of 4 A from a 250 V ac supply. The secondary coil gives a current of 2 A at 450 V. The efficiency of the transformer is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 90%
The efficiency of a transformer is its output power divided by its input power. The power in each coil is the product of its current and voltage, \( p=iv \).
\( \begin{aligned} P_{\mathrm{in}} &= i_pv_p \\ &= 4\times250 \\ &= 1000\ \mathrm{W} \end{aligned} \)
\( \begin{aligned} P_{\mathrm{out}} &= i_sv_s \\ &= 2\times450 \\ &= 900\ \mathrm{W} \end{aligned} \)
\( \begin{aligned} \eta &= \frac{P_{\mathrm{out}}}{P_{\mathrm{in}}}\times100\% \\ &= \frac{900}{1000}\times100\% \\ &= 90\% \end{aligned} \)
From the above, 100 W of the input power is lost in the transformer. Hence, the efficiency is 90%.
NCERT (2026-27):
Transformers – “If the transformer is assumed to be 100% efficient (no energy losses), the power input is equal to the power output, and since p = i v,”
\( i_pv_p=i_sv_s \qquad (7.34) \)