Previous Year Questions- Distribution Systems, Cables and Insulators | Power Systems - Electrical Engineering (EE) PDF Download

Q1: 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)       (2022)
(a) 8
(b) 13
(c) 17
(d) 15
Ans: (b)
Sol: 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 (95/5) = 12.78.

Q2: Two single-core power cables have total conductor resistances of 0.7Ω and 0.5Ω, respectively, and their insulation resistances (between core and sheath) are 600MΩ and 900MΩ, respectively. When the two cables are joined in series, the ratio of insulation resistance to conductor resistance is ___________ × 10⁶.       (2021)
(a) 300
(b) 100
(c) 150
(d) 600
Ans: (a)
Sol: 
Q3: 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 ________.      (SET-1 (2016))
(a) 500.42
(b) 19.25
(c) 47.96
(d) 98.32
Ans: (c)
Sol: 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,
Reactive power supplied by the three phase bank = 600 - 435.889 = 164.44 KVAR

Q4: 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 S₁ and S₂ are in phase. The feeder supplies concentrated loads of unity power factor as shown in the figure.
The contributions of S₁ and S₂ in 100 A current supplied at location P respectively, are       (SET-1 (2014))
(a) 75 A and 25 A
(b) 50 A and 50 A
(c) 25 A and 75 A
(d) 0 A and 100 A
Ans: (d)
Sol: Let the resistance of whole length of feeder be RΩ, length of feeder =1000m.
∴ Resistance per unit length
∴ Resistance of 400 m length
∴ Resistance of 200 m length = (R/5) Ω
Let the current supplied by the sources be I₁ and I₂.
Applying KVL from source S₁ and S₂, we have:  
Therefore, S₂ alone supplies the total load at location P.

Q5: The undesirable property of an electrical insulating material is      (SET-1 (2014))
(a) high dielectric strength
(b) high relative permittivity
(c) high thermal conductivity
(d) high insulation resistivity
Ans: (b)

Q6: 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 μF and the capacitance between each line conductor and the sheath is 0.4 μF . The per-phase charging current is      (2010)
(a) 2.0 A
(b) 2.4 A
(c) 2.7 A
(d) 3.5 A
Ans: (a)
Sol: Equivalent capacitance (C_eq) between a phase and ground

Q7: 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       (2010)(a) $e1=3.74kV,e2=2.61kV$e₁ = 3.74kV, e₂ = 2.61kV  
(b) $e1=3.46kV,e2=2.89kV$e₁ = 3.46kV, e₂ = 2.89kV
(c) $e1=6.0kV,e2=4.23kV$e₁ = 6.0kV, e₂ = 4.23kV
(d) $e1=5.5kV,e2=5.5kV$e₁ = 5.5kV, e₂ = 5.5kV
Ans: (b)
Sol: Line to line voltage = V_l-l = 11 kV
V_P = Phase to ground voltage

Q8: Single line diagram of a 4-bus single source distribution system is shown below. Branches e₁, e₂, e₃ and e₄ 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      (2008)
(a) e₁
(b) e₂
(c) e₃
(d) e₄
Ans: (d)
Sol: Assuming impedance of each branch R
(i) if e₁ is opened
Total losse = 8²R + 3²R + 1²R = 74R
(ii) if e₂ is opened
Total losses = 8²R + 7²R + 5²R = 138R
(i) if e₃ is opened  
Total losses =1²R + 7²R + 2²R = 54R
(i) if e₄ is opened
Total losses = 5²R + 3²R + 1²R = 38R
Operation with minimum loss can be achieved by opening line e₄.  

Q9: 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 × 10⁻⁴, then dielectric power loss in this cable in W/km is       (2004)
(a) 5
(b) 31.7
(c) 37.8
(d) 189
Ans: (b)
Sol: Dielectric power loss in cable

Q10: The phase sequences of the 3-phase system shown in figure is        (2004)
(a) RYB
(b) RBY
(c) BRY
(d) YBR
Ans: (b)
Sol: The phase sequence of the given figure is RBY.
RYB, BRY, and YBR represent the same phase sequence of the figure (given below).

Q11: The rated voltage of a 3-phase power system is given as      (2004)
(a) rms phase voltage
(b) peak phase voltage
(c) rms line to line voltage
(d) peak line to line voltage
Ans: (c)
Sol: NOTE:
Generator are specified in 3 − ϕ MVA, line to line voltage and per-phase reactance (equivalent star).
Tranformers are specified in 3 − ϕ MVA, line to line transformation ratio and per phase (equivalent star) impedance on one side.
Load are specified in 3 − ϕ MVA, line to line voltage and power factor.

Q12: 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 = 3V. The value of the voltage V_P for a minimum voltage of 220 V at any point along the feeder is     (2003)
(a) 225.89 V
(b) 222.89 V
(c) 220.0 V
(d) 228.58 V
Ans: (a)
Sol: Let current injected at point P = I_P 
direction of current will be in opposite in section SQ, so, points will have minimum voltage,