GATE 2020 EE Q.18: solution and answer

Official answer: 1.700 to 1.750Source: Official answer key, GATE 2020 EE (IIT Delhi)

1. KCL at the junction: the current through A1 is the phasor sum of the two branch currents, $I_1=I_2+I_3=1\angle 10^{\circ }+1\angle 70^{\circ }\text{A}$.

2. Magnitude of the sum of two phasors: $|I_2+I_3{|}^2=|I_2{|}^2+|I_3{|}^2+2|I_2||I_3|\cos (70^{\circ }-10^{\circ })=1+1+2\cos 60^{\circ }=3$, so $|I_1|=\sqrt{3}$. Equivalently, for two equal magnitudes, $|I_2+I_3|=2\cdot 1\cdot \cos \left(\frac{70^{\circ }-10^{\circ }}{2}\right)=2\cos 30^{\circ }=\sqrt{3}$, and the resultant lies on the bisector, at $\frac{10^{\circ }+70^{\circ }}{2}=40^{\circ }$.

3. Check in rectangular form: $1\angle 10^{\circ }=0.9848+j0.1736$, $1\angle 70^{\circ }=0.3420+j0.9397$, so $I_1=1.3268+j1.1133\text{A}$.

4. $$|I_1|=\sqrt{1.3268^2+1.1133^2}=\sqrt{3.000}=\sqrt{3}=1.7321\text{A},I_1=1.732\angle 40^{\circ }\text{A}$$

5. An ammeter reads only the magnitude of the current, not its phase, so A1 reads $\sqrt{3}\approx 1.732\text{A}$.