GATE 2024 EE Q.50: solution and answer

Official answer: 692 to 693Source: Official answer key, GATE 2024 EE (IISc Bengaluru)

1. Read the figure. $Z_1$ is connected between lines R and B, and $Z_2$ between lines R and Y. The bottom side of the triangle is not a load. It is the pressure coil of $W_2$. An ideal pressure coil draws no current, so no load current flows between Y and B. $W_1$ has its current coil in line R and its pressure coil across R–B, with $+$ on R. $W_2$ has its current coil in line Y and its pressure coil across Y–B, with $+$ on Y. This is the two-wattmeter method with B as the common line.

2. Line voltages (rms), RYB sequence. The 400 V is the line-to-line rms value. Take $V_{RY}=400\angle 0^{\circ }\text{V}$. Then $V_{YB}=400\angle -120^{\circ }\text{V}$ and $V_{BR}=400\angle 120^{\circ }\text{V}$. Reversing these gives $V_{RB}=-V_{BR}=400\angle -60^{\circ }\text{V}$ and $V_{YR}=-V_{RY}=400\angle 180^{\circ }\text{V}$.

3. Line currents by KCL at the load nodes. The current in $Z_1$ from R to B is $V_{RB}/Z_1$, and the current in $Z_2$ from R to Y is $V_{RY}/Z_2$. So $$I_R=\frac{V_{RB}}{Z_1}+\frac{V_{RY}}{Z_2}=\frac{400\angle -60^{\circ }}{50\angle -90^{\circ }}+\frac{400\angle 0^{\circ }}{200\angle -30^{\circ }}=8\angle 30^{\circ }+2\angle 30^{\circ }=10\angle 30^{\circ }\text{A}$$ Only $Z_2$ takes current from line Y: $$I_Y=\frac{V_{YR}}{Z_2}=\frac{400\angle 180^{\circ }}{200\angle -30^{\circ }}=2\angle 210^{\circ }=2\angle -150^{\circ }\text{A}$$

4. Wattmeter reading. $W=\text{Re}\{V_{pc}{I}_{cc}^{\ast }\}$. Here $V_{pc}$ is the voltage from the $+$ end to the $-$ end of the pressure coil, and $I_{cc}$ is the current entering the $+$ end of the current coil. For a current $I=V/Z$, $V^{\prime }{I}^{\ast }=V^{\prime }{V}^{\ast }/Z^{\ast }$.

5. $W_1$. $$W_1=\text{Re}\{V_{RB}{I}_R^{\ast }\}=\text{Re}\left\{\frac{|V_{RB}{|}^2}{Z_1^{\ast }}\right\}+\text{Re}\left\{\frac{V_{RB}{V}_{RY}^{\ast }}{Z_2^{\ast }}\right\}=\text{Re}\left\{\frac{160000}{50\angle 90^{\circ }}\right\}+\text{Re}\left\{\frac{160000\angle -60^{\circ }}{200\angle 30^{\circ }}\right\}$$ $$W_1=\text{Re}\{3200\angle -90^{\circ }\}+\text{Re}\{800\angle -90^{\circ }\}=0\text{W}$$ Check: $I_R=10\angle 30^{\circ }\text{A}$ leads $V_{RB}=400\angle -60^{\circ }\text{V}$ by exactly $90^{\circ }$, so $W_1=400\times 10\times \cos 90^{\circ }=0$.

6. $W_2$. $$W_2=\text{Re}\{V_{YB}{I}_Y^{\ast }\}=\text{Re}\left\{\frac{V_{YB}{V}_{YR}^{\ast }}{Z_2^{\ast }}\right\}=\text{Re}\left\{\frac{(400\angle -120^{\circ })(400\angle -180^{\circ })}{200\angle 30^{\circ }}\right\}=\text{Re}\{800\angle 30^{\circ }\}=800\cos 30^{\circ }=692.82\text{W}$$ This agrees with $(400\angle -120^{\circ })(2\angle 150^{\circ })=800\angle 30^{\circ }$ from the current found above.

7. Check against total power. $Z_1$ is purely capacitive, so it takes no real power. $Z_2$ has the full 400 V line voltage across it, so $P=\frac{400^2}{200}\cos 30^{\circ }=692.82\text{W}$. In a three-wire system the two readings must add up to the total power, and they do: $W_1+W_2=0+692.82=692.82\text{W}$.

8. Answer. $|W_1-W_2|=|0-692.82|=400\sqrt{3}\approx 692.82\text{W}$. (Had the sequence been R-B-Y, the two readings would simply swap, to $W_1=692.82\text{W}$ and $W_2=0$. The magnitude of the difference would be the same.)