GATE 2025 IN Q.44: solution and answer

Official answer: CSource: Official answer key, GATE 2025 IN (IIT Roorkee)

1. Total power. The load is balanced and the 30 kW is per phase, so the total three-phase power is $P=3\times 30=90\text{kW}$.

2. Power-factor angle. $\cos \varphi =\frac{\sqrt{3}}{2}$ lagging, so $\varphi =30^{\circ }$ and $\sin \varphi =\frac{1}{2}$.

3. Line volt-ampere product. For a balanced three-phase load, $P=\sqrt{3}{V}_L{I}_L\cos \varphi$, so $$V_L{I}_L=\frac{P}{\sqrt{3}\cos \varphi }=\frac{90}{\sqrt{3}\cdot \frac{\sqrt{3}}{2}}=\frac{90}{1.5}=60\text{kVA}$$

4. Two-wattmeter readings (balanced load, lagging pf): $W_1=V_L{I}_L\cos (30^{\circ }-\varphi )$ and $W_2=V_L{I}_L\cos (30^{\circ }+\varphi )$.

5. Substitute $\varphi =30^{\circ }$: $$\begin{array}{rl}{W}_1& =60\cos 0^{\circ }=60\text{kW}\\ W_2& =60\cos 60^{\circ }=30\text{kW}\end{array}$$

6. Check: $W_1+W_2=90\text{kW}=P$, and $\tan \varphi =\frac{\sqrt{3}(W_1-W_2)}{W_1+W_2}=\frac{\sqrt{3}\cdot 30}{90}=\frac{1}{\sqrt{3}}$, which gives $\varphi =30^{\circ }$ as required. The readings are 60 kW and 30 kW, option (C).