# Distribution Systems, Cables and Insulators

 Question 1
If only 5% of the supplied power to a cable reaches the output terminal, the power loss in the cable, in decibels, is _________. (round off to nearest integer)
 A 8 B 13 C 17 D 15
GATE EE 2022   Power Systems
Question 1 Explanation:
We have, the power loss (in decibel) in the cable is given power loss = 95%
Power output as a % of power is 5%
$P_L=10 \log \left ( \frac{95}{5} \right )=12.78$
 Question 2
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 ___________$\times 10^{6}$.
 A 300 B 100 C 150 D 600
GATE EE 2021   Power Systems
Question 2 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}

 Question 3
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 ________.
 A 500.42 B 19.25 C 47.96 D 98.32
GATE EE 2016-SET-1   Power Systems
Question 3 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}
 Question 4
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
 A 75 A and 25 A B 50 A and 50 A C 25 A and 75 A D 0 A and 100 A
GATE EE 2014-SET-1   Power Systems
Question 4 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.
 Question 5
The undesirable property of an electrical insulating material is
 A high dielectric strength B high relative permittivity C high thermal conductivity D high insulation resistivity
GATE EE 2014-SET-1   Power Systems

There are 5 questions to complete.