GATE 2020 EC Q.28: solution and answer

Official answer: CSource: Official answer key, GATE 2020 EC (IIT Delhi)

1. Reference and node voltages. The wire carrying $I$ joins $M$ and $J$ directly, so they are the same node. Take $V_M=V_J=0$. The upper source has + at $T$ and − at $M$, so $V_T=120\angle -90^{\circ }\text{V}$. The lower source has − at $M$ and + at $B$, so $V_B=+120\angle -30^{\circ }\text{V}$.

2. Currents in the two middle impedances (Ohm's law). Take downward as positive. Upper $Z$ (from $T$ to $J$): $I_{\text{up}}=\frac{V_T-V_J}{Z}=\frac{120\angle -90^{\circ }}{Z}$. Lower $Z$ (from $J$ to $B$): $I_{\text{low}}=\frac{V_J-V_B}{Z}=-\frac{120\angle -30^{\circ }}{Z}$.

3. KCL at node J. Two currents enter $J$: $I$ from the wire and $I_{\text{up}}$ from above. One current leaves: $I_{\text{low}}$ downward. So $I+I_{\text{up}}=I_{\text{low}}$, which gives $$I=-\frac{120\angle -90^{\circ }+120\angle -30^{\circ }}{Z}$$ The right-hand impedance is connected directly across $T$–$B$, and the two series sources fix that voltage. Its current flows around the source loop and never enters node $J$, so it does not appear in this KCL.

4. Phasor sum. $120\angle -30^{\circ }+120\angle -90^{\circ }=(103.92-j60)+(-j120)=103.92-j180=207.85\angle -60^{\circ }\text{V}$. With the minus sign, the numerator is $-207.85\angle -60^{\circ }=207.85\angle 120^{\circ }\text{V}$.

5. Impedance in polar form. $Z=80-j35=\sqrt{80^2+35^2}\angle -{\tan }^{-1}\frac{35}{80}=87.32\angle -23.63^{\circ }\Omega$.

6. Divide. $$I=\frac{207.85\angle 120^{\circ }}{87.32\angle -23.63^{\circ }}=2.38\angle \left(120^{\circ }+23.63^{\circ }\right)=2.38\angle 143.63^{\circ }\text{A}$$ In rectangular form this is $(-1.917+j1.412)\text{A}$.

7. Match to the options. $I=2.38\angle 143.63^{\circ }\text{A}$, which is option (C).