Q2GATE 2023MSQ

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?

🚀 Solve in Practice Mode

📖 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.

Q3GATE 2023NAT

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).

🚀 Solve in Practice Mode

📖 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}

$\therefore \quad B=z$ $=(0.035 \times 400)+j(2 \pi \times 50 \times 10^{-3} \times 400) = 126.44\angle 83.643^{\circ}\Omega$

Q4GATE 2022MCQ

The geometric mean radius of a conductor, having four equal strands with each strand of radius $'r'$ , as shown in the figure below, is

🚀 Solve in Practice Mode

📖 Explanation

Redraw the configuration: $\therefore \; GMR=(r' \times 2r\times 2r\times 2\sqrt{2}r)^{1/4}$ Where, $r'=0.7788r$ Hence, $GMR=1.723r$

Q5GATE 2020MCQ

A lossless transmission line with 0.2 pu reactance per phase uniformly distributed along the length of the line, connecting a generator bus to a load bus, is protected up to 80% of its length by a distance relay placed at the generator bus. The generator terminal voltage is 1 pu. There is no generation at the load bus. The threshold pu current for operation of the distance relay for a solid three phase-to-ground fault on the transmission line is closest to:

🚀 Solve in Practice Mode

📖 Explanation

$I_{f}=\frac{1}{Z_{Th}}=\frac{1}{0.2}$ =5 pu for 100% of line Relay is operated for 80% $Z_{f}=0.8\, Z_{t}\Rightarrow 0.8\times 0.2=0.16\, p.u.$ For 80% of line, $I_{f}=\frac{1}{0.16}=6.25\: p.u.$

Q6GATE 2020MCQ

Two buses, i and j, are connected with a transmission line of admittance Y, at the two ends of which there are ideal transformers with turns ratios as shown. Bus admittance matrix for the system is:

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} I&=Y(t_{i}V_{i}-V_{j}t_{j}) \\ I_{i}&=t_{i}I \\ &=t_{i}^{2}YV_{i}-t_{i}t_{j}YV_{j} \\ I_{j}&=-t_{j}I \\ &=-I_{i}t_{j}YV_{i}+t_{i}^{2}YV_{j} \\ \begin{bmatrix} I_{i}\\ I_{j} \end{bmatrix}&=\begin{bmatrix} t_{i}^{2}Y &-t_{i}t_{j}Y \\ -t_{i}t_{j}Y &t_{j}^{2}Y \end{bmatrix}\begin{bmatrix} V_{i}\\ v_{j} \end{bmatrix} \end{aligned}

Q7GATE 2019NAT

A three-phase 50 Hz, 400 kV transmission line is 300 km long. The line inductance is 1 mH/km per phase, and the capacitance is 0.01 $\mu F/km$ per phase. The line is under open circuit condition at the receiving end and energized with 400 kV at the sending end, the receiving end line voltage in kV (round off to two decimal places) will be ___________.

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} V_s&= 400kV\\ l&=300km\\ L_1&=1mH/km/phase\\ C_1&=0.01 \mu F/km/phase\\ v&=\frac{1}{\sqrt{L_1C_1}}\\ &=\frac{1}{\sqrt{1 \times 10^{-3} \times 0.01 \times 10^{-6}}}\\ &=3.16 \times 10^5 km/sec\\ \beta '&=\frac{2 \pi f l}{v}\\ &=\frac{2 \pi \times 50 \times 300}{3.16 \times 10^5}=0.29\\ A&=1-\frac{\beta ^2}{2}\\ &=1-\frac{(0.29)^2}{2}=0.955\\ V_R&=\frac{V_S}{A}=\frac{400}{0.955}=418.85kV \end{aligned}

Q8GATE 2018NAT

A three-phase load is connected to a three-phase balanced supply as shown in the figure. If $V_{an}=100\angle 0^{\circ} V$ , $V_{bn}=100\angle -120^{\circ}V$ and $V_{cn}=100\angle -240^{\circ} V$ (angles are considered positive in the anti-clockwise direction), the value of R for zero current in the neutral wire is ___________ $\Omega$ (up to 2 decimal places).

🚀 Solve in Practice Mode

📖 Explanation

From the given voltages,

\begin{aligned} I_R&=\frac{V_{RN}}{R}=\frac{100\angle 0^{\circ}}{R}\\ I_Y&=\frac{V_{YN}}{jX_L}=\frac{100\angle 120^{\circ}}{j10}\\ &= 10\angle -210^{\circ}\\ I_B&=\frac{V_{BN}}{-jX_C}=\frac{100\angle -240^{\circ}}{-j10}\\ &=10\angle -150^{\circ}\\ \text{For }\; I_N&=0\\ I_R+I_Y+I_B&=0\\ \frac{100}{R}+&10\angle -210^{\circ}+10\angle -150^{\circ}=0\\ R&=5.77\Omega \end{aligned}

Q9GATE 2017MCQ

For the balanced Y-Y connected 3-Phase circuit shown in the figure below, the line-line voltage is 208 V rms and the total power absorbed by the load is 432 W at a power factor of 0.6 leading. The approximate value of the impedance Z is

🚀 Solve in Practice Mode

📖 Explanation

For star connection,

\begin{aligned} P_{3-\phi }&=\left | \frac{V_L^2}{z} \right | \cos \phi \\ Z&=\frac{V_L^2}{P_{3-\phi }} \cos \phi \\ &=\frac{208^2 }{430} \times 0.6 \\ \cos \phi &=0.6 \text{ lead}\\ \phi &=53.13^{\circ}\\ \therefore \; Z&=60\angle -53.13^{\circ}\Omega \end{aligned}

Q10GATE 2017NAT

The nominal- $\pi$ circuit of a transmission line is shown in the figure. Impedance $Z=100\angle 80^{\circ}\Omega$ and reactance X=3300 $\Omega$ . The magnitude of the characteristic impedance of the transmission line, in $\Omega$ , is _______________. (Give the answer up to one decimal place.)

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} Z&=100\angle 80^{\circ}\Omega \\ X&=3300\Omega \\ \frac{y}{2}&=\frac{1}{X}=\frac{1}{3300} \end{aligned}

\begin{aligned} y&=\frac{2}{3300}=6.06 \times 10^{-4}\\ Z_C&=\sqrt{\frac{Z}{y}}=\sqrt{\frac{100}{6.06 \times 10^{-4}}}\\ &=406.2\Omega \end{aligned}

Q11GATE 2017MCQ

A source is supplying a load through a 2-phase, 3-wire transmission system as shown in figure below. The instantaneous voltage and current in phase-a are $V_{an}=220sin(100\pi t)$ V and $i_{a}=10sin (100\pi t)$ A, respectively. Similarly for phase-b the instantaneous voltage and current are $V_{bn}=220cos (100\pi t)$ V and $i_{b}=10cos( 100\pi t$ A, respectively The total instantaneous power flowing form the source to the load is

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} V_{an}&=220 \sin (100 \pi t)V\\ i_a&=10 \sin (100 \pi t)A\\ V_{bn}&=220 \cos (100 \pi t)V\\ i_b&=10 \cos (100 \pi t)A\\ p&=V_{an}i_a+V_{bn}i_b\\ &=2200 W \end{aligned}

Q12GATE 2017NAT

Consider an overhead transmission line with 3-phase, 50 Hz balanced system with conductors located at the vertices of an equilateral triangle of length $D_{ab}=D_{bc}=D_{ca}=1m$ as shown in figure below. The resistance of the conductors are neglected. The geometric mean radius (GMR) of each conductor is 0.01m. Neglecting the effect of ground, the magnitude of positive sequence reactance in $\Omega$ / km (rounded off to three decimal places) is ________

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} X_L&=2 \pi f L\\&=2 \pi f \times 2 \times 10^{-7} \ln \left ( \frac{D_m}{D_s} \right )\\ &=2 \pi \times 50 \times 2 \times 10^{-7} \ln \left ( \frac{1}{0.01} \right )\\ &=2.89 \times 10^{-4}\Omega /m\\ X_L&=0.289 \Omega /km \end{aligned}

Q13GATE 2016NAT

At no load condition, a 3-phase, 50 Hz, lossless power transmission line has sending-end and receiving-end voltages of 400 kV and 420 kV respectively. Assuming the velocity of traveling wave to be the velocity of light, the length of the line, in km, is ____________.

🚀 Solve in Practice Mode

📖 Explanation

At no load,

\begin{aligned} V_s&=AV_R\\ 400 &=A420\\ A&=\frac{400}{420}=0.9524\\ A&=1+\frac{YZ}{2}\\ &=1+\frac{(r+j\omega L)()g+j\omega C)}{2}\\ &\text{For lossless line,}\\ r&=0,g=0\\ \text{then,}\\ A&=1-\frac{(\omega C)(\omega L)}{2}\\ \beta l&=\sqrt{\omega L\omega C}\\ A&=0.9524=1-\frac{\beta ^2l^2}{2}\\ \beta l&=0.3085\\ \beta &=\frac{0.3085}{l}\\ \frac{V}{f}&=\frac{2\pi}{\beta }\\ \frac{3 \times 10^5}{50}&=\frac{2 \pi}{\left ( \frac{0.3085}{l} \right )}\\ l&=294.52km \end{aligned}

Q14GATE 2016NAT

A single-phase transmission line has two conductors each of 10 mm radius. These are fixed at a center-to-center distance of 1 m in a horizontal plane. This is now converted to a three-phase transmission line by introducing a third conductor of the same radius. This conductor is fixed at an equal distance D from the two single-phase conductors. The three-phase line is fully transposed. The positive sequence inductance per phase of the three-phase system is to be 5% more than that of the inductance per conductor of the single-phase system. The distance D, in meters, is _______.

🚀 Solve in Practice Mode

📖 Explanation

In first case,

\begin{aligned} L_1&=2 \times 10^{-7} \ln \frac{D}{r}\\ &=2 \times 10^{-7} \ln \left ( \frac{100}{0.7788} \right )\\ &=0.97\mu H/m\\ L_2&=1.05 \times 0.97=1.0185 \mu H/m\\ L_2&=2 \times 10^{-7} \ln \left ( \frac{\sqrt[3]{D^2 \times 100}}{0.7788} \right ) \end{aligned}

\begin{aligned} \ln \left ( \frac{\sqrt[3]{D^2 \times 100}}{0.7788} \right )&=\frac{1.0185 \times 10^{-6}}{2 \times 10^{-7}}\\ &=5.0925\\ e^{5.0925}&= \frac{\sqrt[3]{D^2 \times 100}}{0.7788} \\ D&=142.7cm=1.427m \end{aligned}

Q15GATE 2015NAT

A composite conductor consists of three conductors of radius R each. The conductors are arranged as shown below. The geometric mean radius (GMR) (in cm) of the composite conductor is kR . The value of k is ______

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \text{GMR}&=\$[(0.7788R \times 3R \times 3R)^3]^{1/3^2}\\ &=1.913 R=KR\\ K&=1.913 \end{aligned}

Q16GATE 2014NAT

For a 400 km long transmission line, the series impedance is $(0.0+j0.5)\Omega /km$ and the shunt admittance is $(0.0+j5.0)\mu mho /km$ . The magnitude of the series impedance (in $\Omega$ ) of the equivalent $\pi$ circuit of the transmission line is ____.

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} z &=(0+j0.5)\Omega /km \\ Z &=zl = (0+j200) \Omega\\ y&= (0+j5)\mu \Omega/km\\ Y&=yl=(0+j5) \times 10^{-6} \times 400\\ &=(0+j2 \times 10^{-3})\Omega \\ \gamma l &=\sqrt{ZY}=\sqrt{(200 \times 2 \times 10^{-3})} \\ \gamma l &= j0.6324 \\ &= \alpha l+j\beta l=0+j\beta l\\ Z_c&=\sqrt{\frac{Z}{Y}}=\sqrt{\frac{200}{2 \times 10^{-3}}} \\ &= 316.22\Omega \end{aligned}

Equivalent $\pi$ -network:

\begin{aligned} Z &=Z_C \sin \gamma l \\ &= Z_C \sinh (\alpha l+j\beta l)\\ &=Z_C ( \sinh \alpha l\cdot \cos \beta l+j \cos \alpha l \sin \beta l) \\ &=316.22[\sin (0) \times \cos \beta l+j \cosh(0)] \\ & \sin \left ( 0.6324 \times \frac{180}{\pi} \right ) \\ &=187\Omega \end{aligned}

Q17GATE 2014MCQ

The horizontally placed conductors of a single phase line operating at 50 Hz are having outside diameter of 1.6 cm, and the spacing between centers of the conductors is 6 m. The permittivity of free space is $8.854 \times 10^{-12}$ F/m. The capacitance to ground per kilometer of each line is

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} C &=\frac{2 \pi \in _0}{\ln \left ( \frac{d}{r} \right )} \\ &=\frac{2 \pi \in _0}{\ln \left ( \frac{2 \times 6}{1.6 \times 10^{-2}} \right )} \\ &= 8.4 \times 10^{-12} F/m\\ &= 8.4 \times 10^{-9} F/km \end{aligned}

Q18GATE 2014MCQ

In a long transmission line with $r , l , g$ and c are the resistance, inductance, shunt conductance and capacitance per unit length, respectively, the condition for distortionless transmission is

🚀 Solve in Practice Mode

📖 Explanation

The condition for a transmission line to be distortionless is $\frac{r}{g}=\frac{l}{c}$

Q19GATE 2010MCQ

A 50 Hz synchronous generator is initially connected to a long lossless transmission line which is open circuited at the receiving end. With the field voltage held constant, the generator is disconnected from the transmission line. Which of the following may be said about the steady state terminal voltage and field current of the generator ?

🚀 Solve in Practice Mode

📖 Explanation

As field voltage is held constant, so field current does not change. When the generator is connected with open-circuit transmission line, line draws charging current. Therefore, $V_t$ is higher than $E_g$ i.e. $V_t \gt E_g$ But when the generator is disconnected from the line, no charging current is delivered by the generator i.e. $I_C=0$ . In this $V_t=E_g$ So, terminal voltage decreases.

Q20GATE 2009MCQ

For a fixed value of complex power flow in a transmission line having a sending end voltage V, the real loss will be proportional to

🚀 Solve in Practice Mode

📖 Explanation

S= Complex power S is defined as, S=VI $I=\frac{S}{V}$ If R is the resistance of transmission line Real power loss $=I^2R=\left ( \frac{S}{V} \right )^2R=\frac{S^2R}{V^2}$ Since, S and R are fixed Real power loss $\propto \frac{1}{V^2}$

Performance Of Transmission Lines Line Parameters And Corona Practice Questions

Get the full Performance Of Transmission Lines Line Parameters And Corona experience — 100% free