GATE 2025 EC Q.20: solution and answer

Official answer: ASource: Official answer key, GATE 2025 EC (IIT Roorkee)

1. All three elements are connected directly across the source. Each current is arrowed from the + rail down to the − rail, so each element sees the full source voltage in the same direction as its current: $V=1\angle 90^{\circ }\text{V}=j\text{V}$.

2. The values given are admittances, so Ohm's law in phasor form is $I=YV$: multiply by the admittance, do not divide. In polar form the magnitudes multiply and the angles add. Here $|V|=1$ and $\angle V=90^{\circ }$.

3. Capacitor, $Y_C=j0.25=0.25\angle 90^{\circ }\text{S}$: $$i_C=(j0.25)(j)=-0.25=0.25\angle 180^{\circ }\text{A}$$ It leads $V$ by $90^{\circ }$, as a capacitor current should.

4. Inductor, $Y_L=-j0.1=0.1\angle -90^{\circ }\text{S}$: $$i_L=(-j0.1)(j)=0.1=0.1\angle 0^{\circ }\text{A}$$ It lags $V$ by $90^{\circ }$, as an inductor current should.

5. Resistor, $Y_R=0.2\angle 0^{\circ }\text{S}$: $$i_R=(0.2)(j)=0.2\angle 90^{\circ }\text{A}$$ It is in phase with $V$.

6. Only option (A) matches all three currents. Options (B) and (D) have magnitudes 4, 10 and 5, which you get by dividing by the admittance ($1/0.25$, $1/0.1$, $1/0.2$). Option (C) has the right magnitudes but the wrong phases for $i_C$ and $i_L$.