GATE EE 2026 | Load Flow Studies Question 2 | Power System Analysis

A three-phase two-winding transformer has a voltage transformation ratio $\frac{V_P}{V_S}=0.866+j 0.5$, where $V_P$ is the primary side voltage in p.u., and $V_S$ is the secondary side voltage in p.u. $I_P$ and $I_S$ represent the currents injected into the primary and secondary sides of the transformer, respectively. The admittance corresponding to the leakage impedance of the transformer referred to the secondary is $y_t$, p.u. Neglect the magnetizing branch. GATE EE 2026 Power System Analysis - Load Flow Studies Question 1 English

GATE EE 2026 Power System Analysis - Load Flow Studies Question 1 English

The $Y$ bus representation of this transformation is

$\left[\begin{array}{c}I_P \\ I_S\end{array}\right]=\left[\begin{array}{cc}\frac{y_t}{0.866+j 0.5} & \frac{y_t}{0.866+j 0.5} \\ -\frac{y_t}{0.866+j 0.5} & \frac{y_t}{0.866+j 0.5}\end{array}\right]\left[\begin{array}{l}V_P \\ V_S\end{array}\right]$

$\left[\begin{array}{l}I_P \\ I_S\end{array}\right]=\left[\begin{array}{cc}y_t & -y_t \\ -y_t & y_t\end{array}\right]\left[\begin{array}{l}V_P \\ V_S\end{array}\right]$

$\left[\begin{array}{c}I_P \\ I_S\end{array}\right]=\left[\begin{array}{cc}y_t & -\frac{y_t}{0.866+j 0.5} \\ -\frac{y_t}{0.866+j 0.5} & y_t\end{array}\right]\left[\begin{array}{l}V_P \\ V_S\end{array}\right]$

$\left[\begin{array}{c}I_P \\ I_S\end{array}\right]=\left[\begin{array}{cc}y_t & -\frac{y_t}{0.866-j 0.5} \\ -\frac{y_t}{0.866+j 0.5} & y_t\end{array}\right]\left[\begin{array}{l}V_P \\ V_S\end{array}\right]$

GATE EE 2024

MCQ (Single Correct Answer)

-1.33

For the three-bus lossless power network shown in the figure, the voltage magnitudes at all the buses are equal to 1 per unit (pu), and the differences of the voltage phase angles are very small. The line reactances are marked in the figure, where $\alpha$, $\beta$, $\gamma$, and $x$ are strictly positive. The bus injections $P_1$ and $P_2$ are in pu. If $P_1 = mP_2$, where $m > 0$, and the real power flow from bus 1 to bus 2 is 0 pu, then which one of the following options is correct?

GATE EE 2024 Power System Analysis - Load Flow Studies Question 6 English

$\gamma = m\beta$

$\beta = m\gamma$

$\alpha = m\gamma$

$\alpha = m\beta$

GATE EE 2021

MCQ (Single Correct Answer)

-0.67

A 3-bus network is shown. Consider generators as ideal voltage sources. If rows 1,2 and 3 of the $Y_{h s}$ matrix correspond to bus 1,2 and 3 respectively, then $Y_{h s}$ of the network is

GATE EE 2021 Power System Analysis - Load Flow Studies Question 3 English

$\left[\begin{array}{ccc}-4 j & j & j \\ j & -4 j & j \\ j & j & -4 j\end{array}\right]$

$\left[\begin{array}{ccc}-4 j & 2 j & 2 j \\ 2 j & -4 j & 2 j \\ 2 j & 2 j & -4 j\end{array}\right]$

$\left[\begin{array}{ccc}-\frac{3}{4} j & \frac{1}{4} j & \frac{1}{4} j \\ \frac{1}{4} j & -\frac{3}{4} j & \frac{1}{4} j \\ \frac{1}{4} j & \frac{1}{4} j & \frac{-3}{4} j\end{array}\right]$

$\left[\begin{array}{ccc}\frac{-1}{2} j & \frac{1}{4} j & \frac{1}{4} j \\ \frac{1}{4} j & -\frac{1}{2} j & \frac{1}{4} j \\ \frac{1}{4} j & \frac{1}{4} j & \frac{-1}{2} j\end{array}\right]$

GATE EE 2018

MCQ (Single Correct Answer)

-0.67

The per-unit power output of a salient-pole generator which is connected to an infinite bus, is given by the expression, P = 1.4 sin $$\delta $$ + 0.15 sin 2$$\delta $$, where $$\delta $$ is the load angle. Newton-Raphson method is used to calculate the value of $$\delta $$ for P = 0.8 pu. If the initial guess is $$30^\circ $$, then its value (in degree) at the end of the first iteration is

$$15^\circ $$

$$28.48^\circ $$

$$31.20^\circ $$

$$28.74^\circ $$