A surge of 20 kV magnitude travels along a lossless cable towards its junction with two identical lossless overhead transmission lines. The inductance and the capacitance of the cable are 0.4 mH and 0.5 F per km. The inductance and capacitance of the overhead transmission lines are 1.5 mH and 0.015 F per km. The magnitude of the voltage at the junction due to surge is

where, Parameter of cable: Inducatance mH/km and Capacitance Surge impedance of the cable Parameters of OH Transmission lines Inductance mH/km and Capacitance Surge impedance of the line E= 20 kV Junction voltage

GATE EE · Power Systems

Power System Transients – GATE EE Power SystemsPractice Questions & PYQs

Generate GATE-level questions on Power System Transients. Focus on: 1. Lightning and Switching surges. 2. Traveling waves on transmission lines: Reflection and Refraction coefficients. 3. Protection against surges: Lightning arresters and Insulation coordination.

5 questions · 5 PYQs · 0 AI practice · GATE EE 2027

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Q1GATE 2025MCQ

In the system shown below, the generator was initially supplying power to the grid. A temporary LLLG bolted fault occurs at F very close to circuit breaker 1. The circuit breakers open to isolate the line. The fault self-clears. The circuit breakers reclose and restore the line. Which one of the following diagrams best indicates the rotor accelerating and decelerating areas?

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Q2GATE 2010MCQ

Consider a step voltage of magnitude 1 pu travelling along a lossless transmission line that terminates in a reactor. The voltage magnitude across the reactor at the instant travelling wave reaches the reactor is

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📖 Explanation

If a step voltage of magnitude 'V' travels through transmission line which is terminated with a inductive load $Z_L=L_1s$ then voltage transmitted (Induced) across the $Z_L=L_1s$

\begin{aligned} V'(t)&=2Ve^{-Z_st/L_1} \\ \text{At time }&t=0 \\ V'(t)&=2V \\ \text{If }V&=1 \\ V'&=2p.u. \end{aligned}

Q3GATE 2005MCQ

The insulation strength of an EHV transmission line is mainly governed by

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📖 Explanation

NOTE: At transmission line voltages upto around 230 kV, the insulation level is dictated by the requirement of protection against lighting. For voltages from 230 kV to 700 kV, both switching transients and lightning overlotages must be accounted for in deciding the insulation levels. In EHV ( $\gt$ 700 kV) switching surges cause overvoltages than lightning and are therefore mainly responsible for insulation level decision.

Q4GATE 2004MCQ

Total instantaneous power supplied by a 3-phase ac supply to a balanced R-L load is

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📖 Explanation

Impedance of the load $=Z_L=R+j\omega L=|Z|\angle \theta _L$ where, $\theta _L=\tan ^{-1}\left ( \frac{\omega L}{R} \right )$ Voltages of $3-\phi$ supply

\begin{aligned} V_a&=V_m \sin \omega t \\ V_b&=V_m \sin (\omega t-120^{\circ}) \\ V_c &=V_m \sin (\omega t+120^{\circ}) \\ I_a&=\frac{V_a}{Z_L}=\frac{V_m \sin \omega t}{|Z|\angle \theta _L} \\ &= I_M \sin (\omega t-\theta _L)\\ \text{where, } I_m&=\frac{V_m}{|Z|}\\ \text{Similarly,}&\\ I_b&=I_m \sin (\omega t-120-\theta _L)\\ I_c&=I_m \sin (\omega t+120-\theta _L)\\ & \text{Instantaneous power}\\ &=P=V_aI_a+V_bI_b+V_cI_c\\ P&=V_mI_m\$[\sin \omega t\cdot \sin (\omega t-\theta _L)\\ &+\sin (\omega t-120^{\circ}) \cdot \sin (\omega t-120-\theta _L)\\ &+\sin (\omega t+120^{\circ})\cdot \sin (\omega t+120-\theta _L)]\$\\ &=\frac{V_mI_m}{2}[(\cos \theta _L -\cos(2\omega t-\theta _L) ) \\ &+(\cos \theta _L -\cos(2\omega t-240-\theta _L) )\\ &+(\cos \theta _L -\cos(2\omega t+240-\theta _L) )]\\ P&=\frac{3V_mI_m}{2}\cos \phi = \text{constant} \end{aligned}

Q5GATE 2003MCQ

A surge of 20 kV magnitude travels along a lossless cable towards its junction with two identical lossless overhead transmission lines. The inductance and the capacitance of the cable are 0.4 mH and 0.5 $\mu$ F per km. The inductance and capacitance of the overhead transmission lines are 1.5 mH and 0.015 $\mu$ F per km. The magnitude of the voltage at the junction due to surge is

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📖 Explanation

where, $z_1=z_2$ Parameter of cable: Inducatance $=L_C=0.4$ mH/km and Capacitance $=C_C=0.5\mu F/km$ Surge impedance of the cable $=Z_C=\sqrt{\frac{L_C}{C_C}}=\sqrt{\frac{0.4 \times 10^{-3}}{0.5 \times 10^{-6}}}=28.28\Omega$ Parameters of OH Transmission lines Inductance $L_1=1.5$ mH/km and Capacitance $=C_1==0.015\mu F/km$ Surge impedance of the line $=Z_1=Z_2=\sqrt{\frac{L_1}{C_1}}=\sqrt{\frac{1.5 \times 10^{-3}}{0.015 \times 10^{-6}}}=316.22\Omega$ E= 20 kV Junction voltage

\begin{aligned} E_j&=\frac{2E \times \frac{1}{Z_C}}{\frac{1}{Z_C}+\frac{1}{Z_1}+\frac{1}{Z_2}} \\ &= \frac{2 \times 20 \times \frac{1}{28.28}}{\frac{1}{28.28}+\frac{2}{316.22}}\\ &=33.93kV \end{aligned}

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Power System Transients – GATE EE Power SystemsPractice Questions & PYQs

5 questions · 5 PYQs · 0 AI practice · GATE EE 2027