# Power Systems

 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).

 A 585 B 632 C 225 D 415
GATE EE 2023      Switch Gear and Protection
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}
 Question 2
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).

 A 7.22 B 5.47 C 6.98 D 9.82
GATE EE 2023      Fault Analysis
Question 2 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, $I_{f}=\frac{\sqrt{3} E}{\left(X_{1}+X_{2}\right)}$ $=\frac{\sqrt{3} \times 1}{0.12+0.12} =7.217 \mathrm{pu}$

 Question 3
The three-bus power system shown in the figure has one alternator connected to bus 2 which supplies $200 \mathrm{MW}$ and $40 MVAR$ power. Bus 3 is infinite bus having a voltage of magnitude $\left|\mathrm{V}_{3}\right|=1.0$ p.u. and angle of $-15^{\circ}$. A variable current source, $|1| \angle \phi$ is connected at bus 1 and controlled such that the magnitude of the bus 1 voltage is maintained at 1.05 p.u. and the phase angle of the source current, $\phi=\theta_{1} \pm \frac{\pi}{2}$, where $\theta_{1}$ is the phase angle of the bus 1 voltage. The three buses can be categorized for load flow analysis as

 A Bus 1 Slack busBus 2 $P-|V|$ busBus 3 $P-Q$ bus B Bus 1 $P-|V|$ busBus 2 $P-|V|$ busBus 3 Slack bus C Bus 1 $P-Q$ busBus 2 $P-Q$ busBus 3 Slack bus D Bus 1 $P-|V|$ busBus 2 $P-Q$ busBus 3 Slack bus
GATE EE 2023      Load Flow Studies
Question 3 Explanation:
Concept : Classification of Bus system :
$\begin{array}{|c|c|c|} \hline \\ \textbf{Bus Type} & \textbf{Specified} & \textbf{Unspecified} \\ &\textbf{Value} &\textbf{Value} \\ \hline \text{Slack Bus} & |\mathrm{V}|, \delta &\mathrm{P}, \mathrm{Q} \\ \text{PV Bus} &\mathrm{P},|\mathrm{V}| & \mathrm{Q}, \delta \\ \text{PQ (load) Bus} & \mathrm{P}, \mathrm{Q} & |\mathrm{V}|, \delta \\ \hline \end{array}$
Now, from given power system network.
Bus-1 $\Rightarrow$ PV bus
Bus-2 $\Rightarrow \mathrm{PQ}$ (load) bus
Bus-3 $\Rightarrow$ Slack bus
 Question 4
A $50 \mathrm{~Hz}, 275 \mathrm{kV}$ line of length $400 \mathrm{~km}$ has the following parameters:

Resistance, $R=0.035 \Omega / \mathrm{km}$;
Inductance, $\mathrm{L}=1 \mathrm{mH} / \mathrm{km}$;
Capacitance, $\mathrm{C}=0.01 \mu \mathrm{F} / \mathrm{km}$;

The line is represented by the nominal $-\pi$ model. With the magnitudes of the sending end and the receiving end voltages of the line (denoted by $V_{S}$ and $V_{R}$, respectively) maintained at 275 $\mathrm{kV}$, the phase angle difference $(\theta)$ between $V_{S}$ and $V_{R}$ required for maximum possible active power to be delivered to the receiving end, in degree is ____ (Round off to 2 decimal places).
 A 42.36 B 64.88 C 83.64 D 98.25
GATE EE 2023      Performance of Transmission Lines, Line Parameters and Corona
Question 4 Explanation:
We have,
$\mathrm{P}_{\mathrm{R}}=\frac{\mathrm{V}_{\mathrm{S}} \mathrm{V}_{\mathrm{R}}}{\mathrm{B}} \cos (\beta-\delta)-\frac{\mathrm{AV}_{\mathrm{R}}^{2}}{\mathrm{~B}} \cos (\beta-\alpha)$

At max. power,
$\delta=\beta$

where, $\beta=$ angle of T-parameter of $B$.

$\pi$-Model :

$[T]=\begin{bmatrix} 1+\frac{yz}{2} &z \\ y\left ( 1+ \frac{yz}{4}\right )& 1+ \frac{yz}{2}\\ \end{bmatrix}=\begin{bmatrix} A & B\\ C&D \end{bmatrix}$
$\therefore \quad B=z$
$=(0.035 \times 400)+j(2 \pi \times 50 \times 10^{-3} \times 400) = 126.44\angle 83.643^{\circ}\Omega$
 Question 5
The bus admittance $\left(Y_{\text {bus }}\right)$ matrix of a 3-bus power system is given below.

$\begin{bmatrix} -j15 &j10 &j5 \\ j10& -j13.5 &j4 \\ j5& j4 & -j8 \end{bmatrix}$

Considering that there is no shunt inductor connected to any of the buses, which of the following can NOT be true?
 A Line charging capacitor of finite value is present in all three lines B Line charging capacitor of finite value is present in line 2-3 only C Line charging capacitor of finite value is present in line 2-3 only and shunt capacitor of finite value is present in bus 1 only D Line charging capacitor of finite value is present in line 2-3 only and shunt capacitor of finite value is present in bus 3 only
GATE EE 2023      Performance of Transmission Lines, Line Parameters and Corona
Question 5 Explanation:
From $Y_{\text {Bus }}$ matrix
\begin{aligned} & y_{10}=-j 15+j 10+j 5=0 \\ & y_{20}=j 10-j 13.5+j 4=j 0.5 \\ & y_{30}=j 5+j 4-j 8=j 1 \end{aligned}

Power system network :

Hence, option (A) and (C) will not be correct.

There are 5 questions to complete.