Question 1 |

The two-bus power system shown in figure (i) has one alternator supplying a synchronous motor load through a Y-\Delta transformer. The positive, negative and zero-sequence diagrams of the system are shown in figures (ii), (iii) and (iv), respectively. All reactances in the sequence diagrams are in p.u. For a bolted line-to-line fault (fault impedance = zero) between phases 'b' and 'c' at bus 1, neglecting all pre-fault currents, the magnitude of the fault current (from phase 'b' to 'c') in p.u. is _____ (Round off to 2 decimal places).

7.22 | |

5.47 | |

6.98 | |

9.82 |

Question 1 Explanation:

From positive sequence network :

\begin{aligned} X_{1} & =\mathrm{j}(0.3 \| 0.2) \\ & =\mathrm{j} 0.12 \mathrm{pu} \end{aligned}

From negative sequence network :

\begin{aligned} \mathrm{X}_{2} & =\mathrm{j}(0.3 \| 0.2) \\ & =\mathrm{j} 0.12 \mathrm{pu} \end{aligned}

For \mathrm{L}-\mathrm{L} fault,

Fault current, I_{f}=\frac{\sqrt{3} E}{\left(X_{1}+X_{2}\right)} =\frac{\sqrt{3} \times 1}{0.12+0.12} =7.217 \mathrm{pu}

\begin{aligned} X_{1} & =\mathrm{j}(0.3 \| 0.2) \\ & =\mathrm{j} 0.12 \mathrm{pu} \end{aligned}

From negative sequence network :

\begin{aligned} \mathrm{X}_{2} & =\mathrm{j}(0.3 \| 0.2) \\ & =\mathrm{j} 0.12 \mathrm{pu} \end{aligned}

For \mathrm{L}-\mathrm{L} fault,

Fault current, I_{f}=\frac{\sqrt{3} E}{\left(X_{1}+X_{2}\right)} =\frac{\sqrt{3} \times 1}{0.12+0.12} =7.217 \mathrm{pu}

Question 2 |

The valid positive, negative and zero sequence impedances (in p.u.), respectively, for
a 220 kV, fully transposed three-phase transmission line, from the given choices are

1.1, 0.15 and 0.08 | |

0.15, 0.15 and 0.35 | |

0.2, 0.2 and 0.2 | |

0.1, 0.3 and 0.1 |

Question 2 Explanation:

We have,

X_0 \gt X_1=X_2

(for 3-\phi transposed transmission line)

X_0 \gt X_1=X_2

(for 3-\phi transposed transmission line)

Question 3 |

A 30 kV, 50 Hz, 50 MVA generator has the positive, negative, and zero sequence reactancesof 0.25 pu, 0.15 pu, and 0.05 pu, respectively. The neutral of the generator is grounded with a reactance so that the fault current for a bolted LG fault and that of a bolted three-phase fault at the generator terminal are equal. The value of grounding reactance in ohms (round off to one decimal place) is ______

2.2 | |

1.8 | |

3.6 | |

4.2 |

Question 3 Explanation:

\begin{aligned} X_1&=0.25 \; p.u.\\ X_2&=0.15 \; p.u.\\ X_0&=0.05 \; p.u.\\ I_{f(LG)}&=I_{f(3-\phi )}\\ \frac{3V_{pu}}{(X_1+X_2+X_0+3X_n)}&=\frac{V_{pu}}{X_1}\\ \frac{3 \times 1}{(0.25+0.15+0.05+3X_n)}&=\frac{1}{0.25}\\ \frac{3}{0.46+3X_n}&=\frac{1}{0.25}\\ \Rightarrow \;\; X_n&=0.1\; p.u.\\ X_n&=0.1 \times Z_B\\ &=0.1 \times \frac{30^2}{50}\\ &=1.8\Omega \end{aligned}

Question 4 |

Five alternators each rated 5 MVA, 13.2 kV with 25% of reactance on its own base are connected in parallel to a busbar. The short-circuit level in MVA at the busbar is_________

50 | |

75 | |

100 | |

150 |

Question 4 Explanation:

Net reactance of parallel connection,

\begin{aligned} X&=\frac{0.25}{5}=0.05\;p.u.\\ I_{SC}&=\frac{1}{X}=\frac{1}{0.05}=20\; p.u.\\ \text{SC MVA}&=20 \times 5\\ &=100 \text{ MVA} \end{aligned}

\begin{aligned} X&=\frac{0.25}{5}=0.05\;p.u.\\ I_{SC}&=\frac{1}{X}=\frac{1}{0.05}=20\; p.u.\\ \text{SC MVA}&=20 \times 5\\ &=100 \text{ MVA} \end{aligned}

Question 5 |

In the circuit shown below, the switch is closed at t=0. The value of \theta in degrees which will give the maximum value of DC offset of the current at the time of switching is

60 | |

-45 | |

90 | |

-30 |

Question 5 Explanation:

If the switch is closed at t=0 in series R-L circuit. Then the circuit current i(t) expression is

\begin{aligned} i(t)&=\left \{ \frac{-V_m}{\sqrt{R^2+X_L^2}}\sin (\theta -\phi ) \right \}e^{-t/\tau }\\ &+\frac{V_m}{\sqrt{R^2+X_L^2}}\sin (\omega t -\phi ) \end{aligned}

The first term of expression indicates DC offset current.

For maximum value of DC offset current, the angle should be 90^{\circ}

\begin{aligned} -\left ( \theta -\tan ^{-1}\left ( \frac{\omega L}{R} \right ) \right )&=90^{\circ}\\ -\left ( \theta -\tan ^{-1}\left ( \frac{377 \times 10 \times 10^{-3}}{3.77} \right ) \right )&=90^{\circ}\\ (\theta -45^{\circ})&=-90^{\circ}\\ \theta &=-45^{\circ} \end{aligned}

\begin{aligned} i(t)&=\left \{ \frac{-V_m}{\sqrt{R^2+X_L^2}}\sin (\theta -\phi ) \right \}e^{-t/\tau }\\ &+\frac{V_m}{\sqrt{R^2+X_L^2}}\sin (\omega t -\phi ) \end{aligned}

The first term of expression indicates DC offset current.

For maximum value of DC offset current, the angle should be 90^{\circ}

\begin{aligned} -\left ( \theta -\tan ^{-1}\left ( \frac{\omega L}{R} \right ) \right )&=90^{\circ}\\ -\left ( \theta -\tan ^{-1}\left ( \frac{377 \times 10 \times 10^{-3}}{3.77} \right ) \right )&=90^{\circ}\\ (\theta -45^{\circ})&=-90^{\circ}\\ \theta &=-45^{\circ} \end{aligned}

There are 5 questions to complete.