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, are

75 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 |

Question 6 |

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 \muF and the capacitance between
each line conductor and the sheath is 0.4 \muF . The per-phase charging current is

2.0A | |

2.4A | |

2.7A | |

3.5A |

Question 6 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}

Question 7 |

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

e_{1}=3.74 kV,e_{2}=2.61 kV | |

e_{1}=3.46 kV,e_{2}=2.89 kV | |

e_{1}=6.0 kV,e_{2}=4.23 kV | |

e_{1}=5.5 kV,e_{2}=5.5 kV |

Question 7 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}

Question 8 |

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

Distribution company's policy requires radial system operation with minimum loss. This can be achieved by opening of the branch

e_{1} | |

e_{2} | |

e_{3} | |

e_{4} |

Question 8 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.

(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.

Question 9 |

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

5 | |

31.7 | |

37.8 | |

189 |

Question 9 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}

Question 10 |

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

RYB | |

RBY | |

BRY | |

YBR |

Question 10 Explanation:

The phase sequence of the given figure is RBY.

RYB, BRY, and YBR represent the same phase sequence of the figure (given below).

RYB, BRY, and YBR represent the same phase sequence of the figure (given below).

There are 10 questions to complete.