Question 1 |

A balanced delta connected load consisting of the series connection of one resistor (R=15 \Omega) and a capacitor (C=212.21 \mu \mathrm{F}) in each phase is connected to three-phase, 50 \mathrm{~Hz}, 415 \mathrm{~V} supply terminals through a line having an inductance of \mathrm{L}=31.83 \mathrm{mH} per phase, as shown in the figure. Considering the change in the supply terminal voltage with loading to be negligible, the magnitude of the voltage across the terminals V_{AB} in Volts is ____ (Round off to the nearest integer).

585 | |

632 | |

225 | |

415 |

Question 1 Explanation:

Given circuit :

where,

\begin{aligned} X_{L} & =2 \pi \times 50 \times 31.83 \times 10^{-3}=10 \Omega \\ R & =15 \Omega \\ X_{C} & =\frac{1}{2 \pi \times 50 \times 212.21 \times 10^{-6}}=15 \Omega \end{aligned}

Convert \Delta to star :

\begin{aligned} Z_{Y} & =\frac{Z_{\Delta}}{3} \\ & =\frac{15-j 15}{3}=5-j 5 \end{aligned}

Now, per phase equivalent circuit :

Using voltage division,

\begin{aligned} \mathrm{V}_{\mathrm{AN}} & =\frac{5-\mathrm{j} 5}{5-\mathrm{j} 5+10 \mathrm{j}} \times \frac{415 \angle 0^{\circ}}{\sqrt{3}} \\ & =\frac{415}{\sqrt{3}} \angle-90^{\circ} \mathrm{V} \\ \therefore \quad \quad\left|\mathrm{V}_{\mathrm{AB}}\right| & =\sqrt{3} \times \frac{415}{\sqrt{3}}=415 \mathrm{~V} \end{aligned}

where,

\begin{aligned} X_{L} & =2 \pi \times 50 \times 31.83 \times 10^{-3}=10 \Omega \\ R & =15 \Omega \\ X_{C} & =\frac{1}{2 \pi \times 50 \times 212.21 \times 10^{-6}}=15 \Omega \end{aligned}

Convert \Delta to star :

\begin{aligned} Z_{Y} & =\frac{Z_{\Delta}}{3} \\ & =\frac{15-j 15}{3}=5-j 5 \end{aligned}

Now, per phase equivalent circuit :

Using voltage division,

\begin{aligned} \mathrm{V}_{\mathrm{AN}} & =\frac{5-\mathrm{j} 5}{5-\mathrm{j} 5+10 \mathrm{j}} \times \frac{415 \angle 0^{\circ}}{\sqrt{3}} \\ & =\frac{415}{\sqrt{3}} \angle-90^{\circ} \mathrm{V} \\ \therefore \quad \quad\left|\mathrm{V}_{\mathrm{AB}}\right| & =\sqrt{3} \times \frac{415}{\sqrt{3}}=415 \mathrm{~V} \end{aligned}

Question 2 |

For the three-bus power system shown in the
figure, the trip signals to the circuit breakers B_1 to B_9 are provided by overcurrent relays R_1 to R_9, respectively, some of which have directional
properties also. The necessary condition for the
system to be protected for short circuit fault at
any part of the system between bus 1 and the
R-L loads with isolation of minimum portion of
the network using minimum number of
directional relays is

\mathrm{R}_{3} and \mathrm{R}_{4} are directional overcurrent relays blocking faults towards bus 2 | |

R_{3} and R_{4} are directional overcurrent relays blocking faults towards bus 2 and R_{7} is directional overcurrent relay blocking faults towards bus 3 | |

R_{3} and \mathrm{R}_{4} are directional overcurrent relays blocking faults towards Line 1 and Line 2, respectively, R_{7} is directional overcurrent relay blocking faults towards Line 3 and R5 is directional overcurrent relay blocking faults towards bus 2 | |

\mathrm{R}_{3} and \mathrm{R}_{4} are directional overcurrent relays blocking faults towards Line 1 and Line 2, respectively. |

Question 2 Explanation:

B_3,B_4 and B_5,B_7 are directional over current relays.

All remaining are ( B_1,B_2,B_6,B_8,B_9)

Now directional over current relays.

Question 3 |

The most commonly used relay, for the protection of an alternator against loss of
excitation, is

offset Mho relay. | |

over current relay. | |

differential relay | |

Buchholz relay. |

Question 4 |

In a 132 kV system, the series inductance up to the point of circuit breaker locationis 50 mH. The shunt capacitanceat the circuit breaker terminal is 0.05 \mu F. The critical value of resistance in ohms required to be connected across the circuit breaker contacts which will give no transient oscillation is_____

100 | |

250 | |

500 | |

1000 |

Question 4 Explanation:

\begin{aligned} L&=50mH\\ C&=0.05\mu F\\ R_{cr}&=\frac{1}{2}\sqrt{\frac{L}{C}}\\ &=\frac{1}{2}\sqrt{\frac{50 \times 10^{-3}}{0.05 \times 10^{-6}}}\\ &=500\Omega \end{aligned}

Question 5 |

The total impedance of the secondary winding, leads, and burden of a 5 A CT is 0.01 \Omega. If the fault current is 20 times the rated primary current of the CT, the VA output of the CT is ________

50 | |

100 | |

150 | |

200 |

Question 5 Explanation:

\begin{aligned} I_{sec}&=5 \times 20=100A \\ V &=I_{sec}R =100 \times 0.01\\ &=1V \\ \text{VA } &\text{output of the CT} \\ &= VI_{sec}=100 \times 1\\ &= 100VA \end{aligned}

There are 5 questions to complete.