# Fault Analysis

 Question 1
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 ______
 A 2.2 B 1.8 C 3.6 D 4.2
GATE EE 2019   Power Systems
Question 1 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 2
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_________
 A 50 B 75 C 100 D 150
GATE EE 2019   Power Systems
Question 2 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}
 Question 3
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
 A 60 B -45 C 90 D -30
GATE EE 2019   Power Systems
Question 3 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}
 Question 4
The positive, negative and zero sequence impedances of a three phase generator are $Z_{1},Z_{2}$ and $Z_{0}$ respectively. For a line-to-line fault with fault impedance $Z_{f}$ ,the fault current is $I_{f1}=kI_{f}$ , where $I_{f}$ is the fault current with zero fault impedance. The relation between $Z_{f}$ and k is
 A $Z_{f}=\frac{(Z_{1}+Z_{2})(1-k)}{k}$ B $Z_{f}=\frac{(Z_{1}+Z_{2})(1+k)}{k}$ C $Z_{f}=\frac{(Z_{1}+Z_{2})k}{1-k}$ D $Z_{f}=\frac{(Z_{1}+Z_{2})k}{1+k}$
GATE EE 2018   Power Systems
Question 4 Explanation:
For LL fault:
Without $Z_f$

\begin{aligned} \text{Given, } I_{f1}&=k \times I_f \\ \frac{\sqrt{3} E_g}{Z_1+Z_2+Z_f} &=\left ( \frac{\sqrt{3} E_g}{Z_1+Z_2} \right )k \\ Z_1+Z_2&=k(Z_1+Z_2+Z_f) \\ Z_f &=\frac{(Z_1+Z_2)(1-k)}{k} \end{aligned}
 Question 5
The positive, negative and zero sequence impedances of a 125 MVA, three-phase, 15.5 kV, star-grounded, 50 Hz generator are j0.1 pu, j0.05 pu and j0.01 pu respectively on the machine rating base. The machine is unloaded and working at the rated terminal voltage. If the grounding impedance of the generator is j0.01 pu, then the magnitude of fault current for a b-phase to ground fault (in kA) is __________ (up to 2 decimal places).
 A 35.82 B 73.52 C 87.23 D 97.66
GATE EE 2018   Power Systems
Question 5 Explanation:
For LG fault,
\begin{aligned} I_{f\;p.u.}&=\frac{3V_{Th}}{Z_1+Z_2+Z_0+3Z_n}\\ &=\frac{3 \times 1 }{0.1+0.05+0.01+3(0.01)}\\ &=15.789\; p.u.\\ I_{base}&=\frac{P}{\sqrt{3}\cdot V_L}\\ &=\frac{125 \times 10^6}{\sqrt{3} \times 15.5 \times 10^3}\\ &=4656.05A\\ I_f&=I_{f\;p.u.}\cdot I_{base}\\ &=15.789 \times 4656.05\\ &=73516.538=73.52kA \end{aligned}
 Question 6
The series impedance matrix of a short three-phase transmission line in phase coordinates is
$\begin{bmatrix} Z_{s} & Z_{m} &Z_{m} \\ Z_{m}& Z_{s}& Z_{m}\\ Z_{m}& Z_{m}& Z_{s} \end{bmatrix}$.
If the positive sequence impedance is $(1 + j10)\Omega$, and the zero sequence is $(4 + j31)\Omega$, then the imaginary part of $Z_{m}$ (in $\Omega$) is ______(up to 2 decimal places).
 A 3 B 5 C 7 D 9
GATE EE 2018   Power Systems
Question 6 Explanation:
\begin{aligned} Z_1&=(1+j10)\Omega \\ Z_0&=(4+j31)\Omega \\ \text{we know,}\\ Z_1&=Z_s-Z_m\\ Z_0&=Z_s+2Z_m\\ Z_1-Z_0&=-3Z_m\\ Z_m&=\frac{Z_0-Z_1}{3}\\ &=\frac{4+j31-1-j10}{3}\\ &=\frac{3+j21}{3}\\ &=1+j7 \end{aligned}
The imaginary part of $Z_m$ is 7.00.
 Question 7
The positive, negative and zero sequence reactances of a wye-connected synchronous generator are 0.2 pu, 0.2 pu, and 0.1 pu, respectively. The generator is on open circuit with a terminal voltage of 1 pu. The minimum value of the inductive reactance, in pu, required to be connected between neutral and ground so that the fault current does not exceed 3.75 pu if a single line to ground fault occurs at the terminals is _______ (assume fault impedance to be zero).
 A 0.033 B 0.05 C 0.1 D 0.2
GATE EE 2017-SET-1   Power Systems
Question 7 Explanation:
\begin{aligned} Z_1&=0.2\\ Z_2&=0.2\\ Z_0&=0.1\\ V_{Th}&=1\\ I_f&=3.75\\ &\text{For LG fault,}\\ I_f&=\frac{3V_{Th}}{Z_1+Z_2+Z_0+3Z_n}\\ 3.75&=\frac{3 \times 1}{0.2+0.2+0.1+3Z_n}\\ Z_n&=0.1\;p.u. \end{aligned}
 Question 8
Two identical unloaded generators are connected in parallel as shown in the figure. Both the generators are having positive, negative and zero sequence impedances of j0.4 p.u., j0.3 p.u. and j0.15 p.u., respectively. If the pre-fault voltage is 1 p.u., for a line-to-ground (L-G) fault at the terminals of the generators, the fault current, in p.u., is ___________.
 A 4.5 B 8.9 C 6 D 9.6
GATE EE 2016-SET-2   Power Systems
Question 8 Explanation:

$I_f=3I_{a_1}=\frac{3}{0.2+0.15+0.15}=6 \; p.u.$
 Question 9
The single line diagram of a balanced power system is shown in the figure. The voltage magnitude at the generator internal bus is constant and 1.0 p.u. The p.u. reactances of different components in the system are also shown in the figure. The infinite bus voltage magnitude is 1.0 p.u. A three phase fault occurs at the middle of line 2.

The ratio of the maximum real power that can be transferred during the pre-fault condition to the maximum real power that can be transferred under the faulted condition is _________.
 A 2.28 B 1.25 C 3.65 D 1.82
GATE EE 2016-SET-2   Power Systems
Question 9 Explanation:
The ratio$=\frac{P_1}{P_2}=\frac{X_1}{X_2}$

\begin{aligned} 1.\; \Rightarrow \; \;&\frac{j0.35 \times j0.6}{j1.2} &=j0.175 \\ 2.\; \Rightarrow \; \; &\frac{j0.6 \times j0.25}{j1.2} &=j0.125 \\ 3.\; \Rightarrow \; \; &\frac{j0.35 \times j0.25}{j1.2} &=j0.0729 \end{aligned}

\begin{aligned} X_{12}&=X_2\\ &=j0.375+j0.125+\frac{j0.375\times j0.125}{j0.0729}\\ &=j1.1430\;p.u.\\ \frac{P_1}{P_2}&=\frac{X_2}{X_1}=\frac{j1.1430}{j0.5}=2.286 \end{aligned}
 Question 10
A 50 MVA, 10 kV, 50 Hz, star-connected, unloaded three-phase alternator has a synchronous reactance of 1 p.u. and a sub-transient reactance of 0.2 p.u. If a 3-phase short circuit occurs close to the generator terminals, the ratio of initial and final values of the sinusoidal component of the short circuit current is ________.
 A 3 B 4 C 5 D 6
GATE EE 2016-SET-2   Power Systems
Question 10 Explanation:
\begin{aligned} I''&=\frac{E_g}{X_d''};\; I=\frac{E_g}{X}\\ \frac{I''}{I}&=\frac{X}{X_d''}=\frac{1.0}{0.2}=5.0\; p.u. \end{aligned}
There are 10 questions to complete.