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)
8 | |
13 | |
17 | |
15 |
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
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}.
300 | |
100 | |
150 | |
600 |
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 ________.
500.42 | |
19.25 | |
47.96 | |
98.32 |
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}
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
The contributions of S_1 \; and \; S_2 in 100 A current supplied at location P respectively, are75 A and 25 A | |
50 A and 50 A | |
25 A and 75 A | |
0 A and 100 A |
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.
\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
high dielectric strength | |
high relative permittivity | |
high thermal conductivity | |
high insulation resistivity |
There are 5 questions to complete.