GATE 2025 IN – Question 65
For the circuit shown, the active power supplied by the source is ______ W (rounded to one decimal place).
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Correct answer: 590 to 610
Explanation
The parallel load is $Z_p=(5-j5)\parallel(j5)=(5-j5)j5/5=5+j5\ \Omega$.
Adding series impedance $10-j10$ gives $Z=15-j5\ \Omega$.
With a 100 V RMS source, $P=V^2\operatorname{Re}Z/|Z|^2=10000(15)/(225+25)=\boxed{600.0\ \mathrm W}$.