Power Systems Test - 4

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GATE
EEE

Power
Systems

Test-04Solutions

POWER SYSTEMS

1. In order to have a lower cost of energy generation,

a) The load factor and diversity factor should be low

b) The load factor should be low but diversity factor should be

high

c) The load factor should be high but diversity factor should be

low

d) The load factor and diversity factor should be high.

Answer: (d)

Solution:

Load factor = Average load (< 1.0)
Maximum demand

Average load is always less than the maximum demand. If the

average load is nearer to maximum demand then the effective

utilization of steam will occur, which will result in saving in

fuel. Hence running cost will be reduced.

Diversity factor = Sum of individual maximum demands (>1.0)
maximum demand on the system

Installed capacity of a power system is based on the

simultaneous maximum demand on the system. If the

simultaneous maximum demand is less than installed capacity is

less and hence fixed cost is reduced.

Hence load factor and diversity factors must be high.

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2. When the load on a transmission line is equal to the surge

impedance loading

a) the receiving end voltage is less than the sending end voltage

b) the sending end voltage is less than the receiving end voltage

c) the receiving end voltage is equal to the sending end voltage

d) none of these

Answer: (d)
3. For enhancing the power transmission in a long EHV

transmission line, the most preferred is to connect a

a) Series inductive compensator in the line

b) Shunt inductive compensator at the receiving end

c) Series capacitive compensator in the line

d) Shunt capacitive compensator at the sending end

Answer: (c)
Solution:
To enhance the power transmission in a long EHV transmission

line, a series capacity is used.

= 2 series capacitor will reduce the reactance.

4. Shunt reactors are sometimes used in high voltage transmission

systems to

a) limit the short circuit current through the line

b) compensate for the series reactance of the line under heavily

loaded condition

c) limit over-voltages at the load side under lightly loaded

condition

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d) compensate for the voltage drop in the line under heavily
loaded condition.

Answer: (c)
Solution:
Limit over voltages at the load side under lightly loaded
condition.
5. Power dispatch through a line can be increased by
a) installing series capacitors
b) installing shunt capacitors
c) installing series reactor
d) installing shunt reactor
Answer: (a)
6. The rated voltage of a 3-phase power system is given as
a) RMS phase voltage
b) PEAK phase voltage
c) RMS line to line voltage
d) PEAK line to line voltage
Answer: (c)
7. A load power factor of 0.95 lagging implies reactive power
demand of
a) 0.05 kVAR per kW
b) 0.10 kVAR per kW
c) 0.33 kVAR per kW
d) 0.95 kVAR per kW
Answer: (c)

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Solution:
From power triangle:

tan = = Reactive power (kVAR)
Active power (kW)

So, tan in (kVAR) is tan (cos−1(0.95))
(kW)

= tan18

tanϕ ≈ 0.33kVAR/kW

8. The most common type of unsymmetrical faults in three-phase

system is

a) single line to ground.

b) line to line.

c) double line to ground.

d) three-phase

Answer: (a)

Solution:

SLG (single line to ground) fault is most common fault. It is an

unsymmetrical fault.

9. As per IE rules the permissible variation of voltage at the

consumer end is:

a) ± 6%

b) ± 10%

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c) ± 12%
d) ± 2%
Answer: (a)
Solution:
As per IE rules “The Indian Electricity rules” the permissible
variation of voltage at the consumer end is ± 3%. Since it is not
in option, hence ± 6% is the correct option
10. To meet the reactive power requirements at load centres
usually
a) shunt capacitors are used
b) series capacitors are used
c) shunt reactors are used
d) tap changing transformers are used
Answer: (a)
Solution:
Shunt compensation by using shunt capacitors is generally, used
to meet the load reactive power demands hence to regulate the
load voltage.
11. The material commonly used for sheaths of underground cable
is
a) lead
b) steel
c) rubber
d) copper
Answer: (a)

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Solution:
Lead-sheathed cables are widely being used cables. Also, an
alloy-lead sheath is much like a pure lead sheath but this if
manufactured with 2% in lead in it.
12. In bay on wiring the cables are carried on seasoned teak wood
perfectly straight and well varnished teak wood button of
thickness not less then
a) 3 cm
b) 4 cm
c) 1 cm
d) 2 cm
Answer: (c)
Solution:
Thickness not less than 1 cm
13. Consider the problem of relay co-ordination for the distance
relays Rl and R2 on adjacent lines of a transmission system
figure. The zone 1

and zone 2 settings for both the relays are indicated on the
diagram. Which of the following indicates the correct time
setting for the zone 2 of relays R1 and R2?

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a) TZ2R1 = 0.6s, TZ2R2 = 0.3s
b) TZ2R1 = 0.3s, TZ2R2 = 0.6s
c) TZ2R1 = 0.3s, TZ2R2 = 0.3s
d) TZ2R1 = 0.1s, TZ2R2 = 0.3s
Answers: (a)
Solution:
From figure, it is clear that zone2 of relay1 and relay2 are
overlapped. If there is a fault in overlapped section (line2), the
fault should be clear by relay2. Hence zone2 operating time of
relay2 must be less than zone1 operating time (TZ2R1 > TZ2R2).
14. Advantage of transmitting power at high voltage is
a) magnitude of current will be small.
b) power loss will be less.
c) it will reduce the voltage drop in the line impedance.
d) all the above
Answer: (d)
Solution:
All the options are advantages of high voltage power
transmission.
15. The minimum area of cross section of a three and half core
cable should be
a) 60 cm2
b) 40 cm2
c) 30 cm2
d) 50 cm2

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Answer: (c)
Solution:
It is a standard that for 3.5 core cable the cross sectional area
must be greater than or equal to 25 cm2.
Here, appropriate answer is 30 cm2.
16. The sag of the conductors of a transmission line is 2.5 m when
the span is 250 m. Now if the height of the supporting towers is
increased by 25%, the sag will
a) reduce by 25%
b) increase by 25%
c) reduce by 12.5%
d) remain unchanged
Answer: (d)
Solution:
Sag of the conductors is directly proportional to the square of
the span length if other conditions, such as type of conductor,
working tension, temperature remain same. Here the height of
both the supporting towers is increased to 25% so the sag will
remain unchanged.
17. In arc welding, the voltage required to maintain the arc is in
the range of
a) 200 - 250 volts
b) 1000 - 1200 volts
c) 2 - 5 volts
d) 20 - 30 volts

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Answer: (d)
Solution:
In arc welding the moderate voltage and moderate current
electricity from the utility mains (230 V AC) is converted into a
high current and low voltage typically between 17 to 45 V (open
circuit voltage) and 55 to 590 A respectively
18. The inner surface of a fluorescent tube is coated with a
fluorescent material which
a) absorbs ultraviolet rays and radiates visible rays.
b) reduces glare.
c) improves life.
d) absorbs infrared rays and radiates visible rays.
Answer: (a)
Solution:
The inner surface of a fluorescent tube is coated with a
fluorescent material to absorb ultraviolet rays and to radiate
visible rays.
19. A relay performs the function of
a) fault isolation
b) fault detection
c) fault prevention
d) all of the above
Answer: (b)

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Solution:
A fault is sensed/detected by the relay which is further cleared
by the circuit breakers.
20. Which of the following surfaces has the lowest reflection
factor for white light?
a) Aluminium sheets
b) White plaster work
c) Blue curtains
d) White oil paint
Answer: (c)
Solution:
Blue curtain has the lowest reflection factor for white light.
21. In arc welding, arc is created between the electrode and work
by
a) flow of current
b) voltage
c) material characteristics
d) contact resistance
Answer: (d)
22. A two machine power system in shown below. Transmission
line XY has positive sequence impedance of Z1 Ω and zero
sequence impedance of Z0 Ω.

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An 'a' phase to ground fault with zero fault impedance occurs at
the centre of the transmission line. Bus voltage at X and line
current from X to F for the phase ‘a’, are given by Va Volts and
Ia Amperes, respectively. Then, the impedance measured by the
ground distance relay located at the terminal X of line XY will
be given by
a) Z1/2 Ω
b) Z0/2 Ω
c) (Z0 + Z1)/2 Ω
d) Va/Ia Ω
Answers: (d)
Solution:

Impedance seen by the relay =

23. An electric iron is rated at 230 V, 400 W, 50 Hz. The voltage

rating 230 V refers to

a) Rms value

b) Peak-to-peak value

c) Average value

d) Peak value

Answer: (a)

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Solution:

The voltage rating 230V refers to rms value (or) effective value
of voltage.
24. Isolators are capable of breaking
a) fault current
b) no load
c) load current
d) charging current
Answer: (d)
Solution:
Isolators are not capable of breaking any appreciable current and
their operation is normally inter locked to ensure that this does
not occur. They are, however capable of breaking the small
charging currents of bus bars.
25. A consumer is offered the following rate of tariff. He has to
pay a fixed charge of ₹ 1,000 per month and a running charge of
₹ 4.50 per unit consumed. If the consumer runs a motor load of
1 kW at 0.85 power factor lagging on an average of 15 hours per
day, his annual bill is
a) ₹ 25637.50
b) ₹ 36637.50

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c) ₹ 40985.29

d) ₹ 32941.88

Answer: (b)

Solution:

Total number of kWh per day = 1 kW × 15h

= 15 kWh

Cost per day = 15 × 4.50

= ₹ 67.5/day

Cost per annum = 67.5 × 365

= ₹ 24637.5

Fixed charge per annum = ₹ 12000

Total = ₹ 36637.50

26. Bundled conductors are mainly used in high voltage overhead

transmission lines to

a) Reduce transmission line losses

b) Increase mechanical strength of the line

c) Reduce corona

d) Reduce sag

Answer: (c)

Solution:

Electric field intensity at surface of each conductor

E = V
(self GMD)ln�seGlfMGDMD�

For bundle conductors increase in self GMD will be

predominant compared to decrease in ln �seGlfMGDMD�.

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Hence electric field intensity at surface of each sub conductor
will be reduced. Hence the corona loss will be reduced, because
chances for ionization of air are reduced.
27. How many watt-seconds are supplied by a motor developing
2hp (British) for 5 hours?
a) 2.6856 x 107watt-seconds
b) 4.476 x 107watt-seconds
c) 2.646 x 1 07 watt-seconds
d) 6.3943 x 107 watt-seconds
Answer: (a)
Solution:

2 × 746 = 1492 W is the load
Total number of hours = 5h

∴ Energy = 1492 × 5 = 7460 Wh
To convert the energy into W-sec, we have

Energy = 7460 × (60 × 60)
= 2.6856 × 107 Wsec

28. In a system, if the base load is the same as the maximum
demand, the load factor will be
a) 1.0
b) 0.5
c) zero
d) infinity
Answer: (a)

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Solution:

Load factor = Average load
Maximum demand

Average load can be said as base load. It is given that base load

is same as the maximum demand then, load factor = 1.0.

29. In an insulated cable having core diameter d and overall

diameter D, the ratio of maximum to minimum dielectric stress

is given by

a) (D/d)1/2

b) (D/d)2

c) D/d

d) d/D

Answer: (c)

Solution:

Maximum dielectric stress = gmax

=
ln

Minimum dielectric stress = gmin

=
ln

∴ The ratio of maximum to minimum dielectric stress =

ln
Or, = = =

ln

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30. Keeping in view the cost and overall effectiveness, the
following circuit breaker is best suited for capacitor bank
switching
a) Vacuum
b) Air blast
c) SF6
d) Oil
Answer: (a)
Solution:
Vacuum circuit breakers are specifically used for low cost
switch having low fault interrupting capacity, but capable of
large number of load switching operations without maintenance
and in some applications capable of interrupting line charging
(or) capacitor current without restrike.

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