GATE 2018 EE – Question 57
A three-phase load is connected to a three-phase balanced supply as shown in the figure. If $V_{an}=100\angle0^\circ$ V, $V_{bn}=100\angle-120^\circ$ V and $V_{cn}=100\angle-240^\circ$ V (angles are considered positive in the anti-clockwise direction), the value of R for zero current in the neutral wire is ________ $\Omega$ (up to 2 decimal places).
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Correct answer: 5.74 to 5.80
Explanation
The phase currents are $I_a=\frac{100}{R}$, $I_b=\frac{100\angle-120^\circ}{j10}=10\angle-210^\circ$ and $I_c=\frac{100\angle-240^\circ}{-j10}=10\angle-150^\circ$. The sum of the last two is $10(\cos210^\circ+\cos150^\circ)=-17.32$ A (their imaginary parts cancel). The neutral current is zero when $\frac{100}{R}=17.32$, so $R=5.77\ \Omega$.