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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