GATE 2025 IN – Question 60
The shown circuit is driven at 50 Hz by $V_s=25\sqrt2\angle30^\circ$ V. Both transformers are ideal. Average power in the 50 kΩ resistor is ______ W (rounded to two decimal places).
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Correct answer: 31.00 to 31.50
Explanation
Refer the 50 kΩ resistor through the second 1:10 transformer: $R=50000/10^2=500\ \Omega$.
Add the 400 Ω series resistor to get 900 Ω.
Refer 900 Ω through the first 1:10 transformer to get 9 Ω.
Including the 1 Ω source-side resistor gives 10 Ω, so source RMS current is $25\sqrt2/10$ A.
The load resistance referred through both transformers is $50000/10^4=5\ \Omega$.
Its power is $P=(25\sqrt2/10)^2(5)=\boxed{31.25\ \mathrm W}$.