GATE 2018 EE Q.6: solution and answer

Official answer: BSource: Official answer key, GATE 2018 EE (IIT Guwahati)

1. Let the line have series impedance $Z=R+jX=|Z|\angle \theta$ with $\theta ={\tan }^{-1}(X/R)$. Because the line is lossy, $R>0$, so $\theta <90^{\circ }$ and $\cos \theta >0$. Take the shunt admittance as negligible, so the ABCD constants are $A=D=1$, $B=Z$.

2. Power received at the receiving end, with $\delta$ the angle by which $V_1$ leads $V_2$: $$P_R=\frac{|V_1||V_2|}{|Z|}\cos (\theta -\delta )-\frac{|V_2{|}^2}{|Z|}\cos \theta$$

3. $P_R$ is largest when $\cos (\theta -\delta )=1$, that is $\delta =\theta$. The steady-state stability limit is this maximum: $$P_{max}=\frac{|V_1||V_2|}{|Z|}-\frac{|V_2{|}^2}{|Z|}\cos \theta$$

4. Compare with the lossless value $\left|\frac{V_1{V}_2}{X}\right|$. Since $|Z|=\sqrt{R^2+X^2}>X$, the first term satisfies $\frac{|V_1{V}_2|}{|Z|}<\frac{|V_1{V}_2|}{X}$.

5. The subtracted term $\frac{|V_2{|}^2}{|Z|}\cos \theta$ is positive because $\cos \theta >0$, so it lowers the limit further. Hence $P_{max}<\left|\frac{V_1{V}_2}{X}\right|$.

6. Check: as $R\to 0$, $|Z|\to X$, $\theta \to 90^{\circ }$ and $\cos \theta \to 0$, so $P_{max}\to \left|\frac{V_1{V}_2}{X}\right|$, the lossless value. Any resistance lowers the limit, so the answer is (B).