Q2GATE 2021NAT

Two single-core power cables have total conductor resistances of $0.7\;\Omega$ and $0.5\;\Omega$ , respectively, and their insulation resistances (between core and sheath) are $600\;M\Omega$ and $900\;M\Omega$ , respectively. When the two cables are joined in series, the ratio of insulation resistance to conductor resistance is ___________ $\times 10^{6}$ .

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📖 Explanation

\begin{aligned} \left(R_{\text {eq }}\right)_{\text {conductor }} &=0.7+0.5=1.2 \Omega \\ \left(R_{\text {eq }}\right)_{\text {\in }} &=\frac{1}{\left(R_{e q}\right)_{\text {\in }}}=\frac{1}{\left(R_{\text {\in }}\right)_{1}}+\frac{1}{\left(R_{\text {\in }}\right)_{2}}=\frac{1}{900 \times 10^{6}}+\frac{1}{600 \times 10^{6}} \\ \left(R_{\mathrm{eq}}\right)_{\text {\in }} &=360 \times 10^{6} \Omega\\ \frac{\left(R_{\text {eq }}\right)_{\text {\in }}}{\left(R_{\text {eq }}\right)_{\text {conductor }}}&=\frac{360 \times 10^{6}}{1.2}=300 \times 10^{6} \end{aligned}

Q3GATE 2016NAT

A three-phase cable is supplying 800 kW and 600 kVAr to an inductive load. It is intended to supply an additional resistive load of 100 kW through the same cable without increasing the heat dissipation in the cable, by providing a three-phase bank of capacitors connected in star across the load. Given the line voltage is 3.3 kV, 50 Hz, the capacitance per phase of the bank, expressed in microfarads, is ________.

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📖 Explanation

$\text{KVA}_1=\sqrt{800^{2+}600^2}=100 \text{ KVA}$ Without excessive heat dissipation means current should be constant (i.e.) KVA erating must be constatn. In second case active power, P=800+100=900 KW Reactive power in second case, $Q_2=\sqrt{1000^{2-}900^2}=435.889 \text{ KVAR}$ Reactive power supplied by the three phase bank = 600-435.889 =164.44 KVAR

\begin{aligned} Q_{bank}/ph&=\frac{164.11}{3}=54.7 \text{ KVAR} \\ V/ph&= \frac{3.3}{\sqrt{3}}=1.9051 \text{ KV}\\ Q_c/ph&=\frac{(V/ph)^2}{X_C} \\ X_C&=\frac{(1.9052 \times 10^3)^2}{54.7 \times 10^3} \\ &= 66.36\; W\\ C&= \frac{1}{2 \pi f X_C}\\ &=\frac{1}{2 \pi \times 50 \times 66.36}\\ &=47.96 \mu F \end{aligned}

Q4GATE 2014MCQ

The undesirable property of an electrical insulating material is

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Q5GATE 2014MCQ

A distribution feeder of 1 km length having resistance, but negligible reactance, is fed from both the ends by 400 V, 50 Hz balanced sources. Both voltage sources $S_1 \; and \; S_2$ are in phase. The feeder supplies concentrated loads of unity power factor as shown in the figure. The contributions of $S_1 \; and \; S_2$ in 100 A current supplied at location P respectively, are

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📖 Explanation

Let the resistance of whole length of feeder be $R\;\Omega$ , length of feeder =1000m. $\therefore \;$ Resistance per unit length $=\frac{R}{1000}\Omega /meter$ $\therefore \;$ Resistance of 400 m length $=\frac{R}{1000} \times 400 =\frac{2R}{5} \Omega$ $\therefore \;$ Resistance of 200 m length $=\frac{R}{5} \Omega$ Let the current supplied by the sources be $I_1 \text{ and } I_2$ . Applying KVL from source $S_1 \text{ and } S_2$ , we have:

\begin{aligned} 400 &= \frac{2R}{5}I_1+\frac{R}{5}(I_1-200)\\ & -\frac{R}{5}(I_2-200)-\frac{R}{5}I_2+400\\ 0&= \frac{2}{5}I_1 +\frac{I_1-200}{5}+\frac{200-I_2}{5}-\frac{I_2}{5}\\ 0&=2I_1+I_1-200+200-I_2-I_2 \\ 0&=3I_1-2I_2\\ I_1&=\frac{2}{3}I_2\;\;\;...(i)\\ \text{also, }& I_1 +I_2-400=100\\ \text{or, }&I_1+I_2=500 \;\;\;...(ii)\\ &\text{Solvinf eq. (i) and (ii), we have}\\ I_1&=200A \text{ and }I_2=300A\\ &\text{Contribution of }S_1 \text{\in 100 A at location}\\ P&=I_1-200=0A \\ &\text{Contribution of }S_2 \text{\in 100 A at location}\\ P&=I_2-200=300-200=100A \end{aligned}

Therefore, $S_2$ alone supplies the total load at location P.

Q6GATE 2010MCQ

Consider a three-core, three-phase, 50 Hz, 11 kV cable whose conductors are denoted as R,Y and B in the figure. The inter-phase capacitance(C1) between each pair of conductors is 0.2 $\mu$ F and the capacitance between each line conductor and the sheath is 0.4 $\mu$ F . The per-phase charging current is

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📖 Explanation

\begin{aligned} Z'&=Z/3\\ \frac{1}{\omega C'}&=\frac{1}{3\omega C}\\ C'&=3C \end{aligned}

Equivalent capacitance ( $C_{eq}$ ) between a phase and ground

\begin{aligned} C_{eq} &=C'+C_2=3C_1+C_2\\ C_{eq} &= 3 \times 0.2+0.4=1\mu F\\ &\text{Per phase voltage,}\\ V_P&=\frac{11}{\sqrt{3}}kV\\ &\text{Per phase charging current}\\ I&=j\omega C_{eq}V_P\\ |I_C|&=\omega C_{eq}V_P\\ &=2 \pi \times 50 \times 1 \times 1 \times 10^{-6}\\ & \times \frac{11}{\sqrt{3}} \times 10^3\\ &\approx 2A \end{aligned}

Q7GATE 2010MCQ

Consider a three-phase, 50 Hz, 11 kV distribution system. Each of the conductors is suspended by an insulator string having two identical porcelain insulators. The self capacitance of the insulator is 5 times the shunt capacitance between the link and the ground, as shown in the figures. The voltages across the two insulators are

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📖 Explanation

Line to line voltage $=V_{l-l}=11kV$

\begin{aligned} V_P&=\text{Phase to ground voltage} \\ V_P&=\frac{V_{l-l}}{\sqrt{3}}=\frac{11}{\sqrt{3}}kV \\ e_1+e_2&= \frac{11}{\sqrt{3}}kV=6.35kV \;\;...(i)\\ I&=I_1+I_2 \\ (j\omega 5C) \times e_1 &= (j\omega C) \times e_2 +(j\omega 5C) \times C_2 \;\;..(ii)\\ 5e_1&=6e_2\\ \text{Solving } & \text{equation (i) and (ii), we get,} \\ e_1&=3.46kV\\ e_2&=2.89kV \end{aligned}

Q8GATE 2008MCQ

Single line diagram of a 4-bus single source distribution system is shown below. Branches $e_1,e_2,e_3 \; and \; e_4$ have equal impedances. The load current values indicated in the figure are in per unit. Distribution company's policy requires radial system operation with minimum loss. This can be achieved by opening of the branch

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📖 Explanation

Assuming impedance of each branch R (i) if $e_1$ is opened Total losses $=8^2R+3^2R+1^2R=74R$ (ii) if $e_2$ is opened Total losses $=8^2R+7^2R+5^2R=138R$ (i) if $e_3$ is opened Total losses $=1^2R+7^2R+2^2R=54R$ (i) if $e_4$ is opened Total losses $=5^2R+3^2R+1^2R=38R$ Operation with minimum loss can be achieved by opening line $e_4$ .

Q9GATE 2004MCQ

The rated voltage of a 3-phase power system is given as

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📖 Explanation

NOTE: Generator are specified in $3-\phi$ MVA, line to line voltage and per-phase reactance (equivalent star). Tranformers are specified in $3-\phi$ MVA, line to line transformation ratio and per phase (equivalent star) impedance on one side. Load are specified in $3-\phi$ MVA, line to line voltage and power factor.

Q10GATE 2004MCQ

The phase sequences of the 3-phase system shown in figure is

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Q11GATE 2004MCQ

A 110 kV, single core coaxial, XLPE insulated power cable delivering power at 50 Hz, has a capacitance of 125 nF/km. If the dielectric loss tangent of XLPE is $2 \times 10^{-4}$ , then dielectric power loss in this cable in W/km is

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📖 Explanation

\begin{aligned} \text{V(Phase voltage)} &=\frac{110}{\sqrt{3}}kV \\ C&=125 nF/km \\ \tan \delta &= 2 \times 10^{-4} \end{aligned}

Dielectric power loss in cable

\begin{aligned} P&=V62 \omega C \tan \delta \\ P&=\left ( \frac{110 \times 10^3}{\sqrt{3}} \right )^2 \times 2 \\ & \times \pi \times 50 \times 125 \times 10^{-9} \times 2 \times 10^{-4}\\ &\approx 31.7 W/km \end{aligned}

Q12GATE 2003MCQ

A dc distribution system is shown in figure with load current as marked. The two ends of the feeder are fed by voltage sources such that $V_P - V_Q = 3 V$ . The value of the voltage $V_P$ for a minimum voltage of 220 V at any point along the feeder is

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📖 Explanation

Let current injected at point $P=I_P$

\begin{aligned} V_P-V_Q &=0.1(I_P-10)+0.15(I_P-30) \\ & +0.2(I_P-60) \\ 3&=0.45I_P-17.5 \\ I_P&=45.55A \end{aligned}

direction of current will be in opposite in section SQ, so, points will have minimum voltage,

\begin{aligned} V_s&=220V\\ V_p&=V_s+0.1(I_p-10)+0.15(I_P-30)\\ &=220+0.1 \times (45.55-10)\\ &+0.15(45.55-30)\\ &\approx 225.89V \end{aligned}

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