Q2GATE 2018MCQ

Consider the two bus power system network with given loads as shown in the figure. All the values shown in the figure are in per unit. The reactive power supplied by generator $G_1$ and $G_2$ are $Q_{G1}$ and $Q_{G2}$ respectively. The per unit values of $Q_{G1}, Q_{G2}$ , and line reactive power loss ( $Q_{loss}$ ) respectively are

🚀 Solve in Practice Mode

📖 Explanation

At $G_2$ load demand is 20 pu, $G_2$ supplies only 15 pu. Remaining supplied by $G_1$ through transmission line.

\begin{aligned} P_S&=\left | \frac{V_SV_R}{X_L} \right | \sin \delta \\ 5&=\frac{1 \times 1}{0.1} \sin \delta \\ \delta &=30^{\circ}\\ Q_S&=\frac{V_S^2}{X_L}-\frac{V_SV_R}{X_L} \cos \delta \\ &=\frac{1^2}{0.1}-\frac{1 \times 1 }{0.1} \cos 30^{\circ}\\ &=1.34 p.u.\\ Q_R&=\left | \frac{V_SV_R}{X_L} \right | \cos \delta -\left | \frac{V_R^2}{X_L} \right | \\ 5&=\frac{1 \times 1}{0.1} \cos 30^{\circ}-\frac{1^2}{0.1} \\ &=-1.34p.u.\\ Q_{loss}&+Q_S-Q_R\\ &=1.34-(-1.34)=2.68p.u.\\ \text{At }G_1: \; Q_{G1}&=Q_{load}+Q_S\\ &=5+1.34=6.34p.u.\\ \text{At }G_2: \; Q_{G2}&=Q_{load}-(-Q_R)\\ &=10-(-1.34)=11.34p.u. \end{aligned}

Q12GATE 2007MCQ

A 230 V (Phase), 50 Hz, three-phase, 4-wire, system has a phase sequence ABC. A unity power-factor load of 4 kW is connected between phase A and neutral N. It is desired to achieve zero neutral current through the use of a pure inductor and a pure capacitor in the other two phases. The value of inductor and capacitor is

🚀 Solve in Practice Mode

📖 Explanation

Taking $V_{AN}$ as the reference

\begin{aligned} V_{AN}&=230\angle 0^{\circ} V\\ V_{BN}&=230\angle -120^{\circ}V \\ V_{CN}&=230\angle 120^{\circ}V \\ I_A&= \frac{P}{V_{AN} \cos \phi }\\ &= \frac{4 \times 10^3}{230 \times 1}\\ &=17.4\angle 0^{\circ} \end{aligned}

If an pure inductor is present in phase B, then $I_B$ lags $V_{BN}$ by $90^{\circ}$ . If a pure capacitor is present in phase C, then $I_C$ leads $V_{CN}$ by $90^{\circ}$ . If current through neutral is to be zero.

\begin{aligned} I_B \sin 30^{\circ} &=I_C \sin 30^{\circ} \\ \Rightarrow \; I_B &=I_C=I \\ I_A&=I_B \cos 30^{\circ} +I_C \cos 30^{\circ} \\ &=2I \cos 30^{\circ}=\sqrt{3}I \\ |I_B|&=|I_C|=|I|=\frac{|I_A|}{\sqrt{3}} \\ &= \frac{17.04}{\sqrt{3}}=10.05\\ |I_B|&=\frac{|V_{BN}|}{\omega L} \\ 10.05 &=\frac{230}{2 \pi \times 50 \times L} \\ L&=72.847 mH \\ |I_C| &=|V_{CN}| \cdot \omega C\\ 10.05&=230 \times 2 \times \pi \times 50 \times C \\ C&=139.087 \mu F \end{aligned}

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