GATE 2025 IN Q.65: solution and answer

Official answer: 590 to 610Source: Official answer key, GATE 2025 IN (IIT Roorkee)

1. Read the circuit as one series loop. The source drives $-j10\Omega$ and $10\Omega$ in series with a parallel pair between node A and the bottom wire. The pair is $(5-j5)\Omega$ in parallel with $j5\Omega$.

2. Combine the parallel pair with the product-over-sum rule: $$Z_p=\frac{(5-j5)(j5)}{(5-j5)+j5}=\frac{25+j25}{5}=5+j5\Omega$$

3. Add the series elements to get the total impedance seen by the source: $Z=10-j10+5+j5=15-j5\Omega$, so $|Z{|}^2=15^2+5^2=250$.

4. Apply Ohm's law in phasor form: $$I=\frac{100\angle 0^{\circ }}{15-j5}=\frac{100(15+j5)}{250}=6+j2\text{A}$$ so $|I{|}^2=36+4=40$.

5. Compute the complex power delivered by the source: $$S=V{I}^{\ast }=100(6-j2)=600-j200\text{VA}$$ This gives $P=\text{Re}(S)=600\text{W}$. Equivalently, $P=|I{|}^2\text{Re}(Z)=40\times 15=600\text{W}$. The source phasor is taken as rms, the usual GATE convention.

6. Check by adding up the power absorbed in each resistor. $V_A=I{Z}_p=(6+j2)(5+j5)=20+j40\text{V}$. The $5\Omega$ branch carries $I_5=\frac{20+j40}{5-j5}=\frac{(20+j40)(5+j5)}{50}=-2+j6\text{A}$, so $|I_5{|}^2=40$. The resistors absorb $40\times 10+40\times 5=400+200=600\text{W}$. This matches the source power, so the answer is $600.0\text{W}$.