Quiz: Electrical Engineering 10 Oct 2020

Quiz: Electrical Engineering
Exam: UPSSSC-JE
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. Two capacitor C1 and C2 have C1=20μ F and C2 = 30μ F, are connected in parallel
across a 100V source. The net capacitance of the circuit is?
(a) 50 μ F
(b) 10 μ F
(c) 12 μ F
(d) 60 μ F

Q2. A 100mA meter has accuracy of +2%. Its accuracy while reading 50mA will be:
(a) +1%
(b) +2%
(c) +4%
(d) +20%

Q3. An ammeter has a current range of 0-5 A, and its internal resistance is 0.2Ω. In order
to change the range of 0-25 A, what should be the value of resistance added and how it
would connect with meter (i.e. series/parallel)?
(a) (0.05Ω /series)
(b) (0.05Ω /parallel)
(c) (0.20Ω /parallel)
(d) (0.20Ω /series)

Q4. In ACSR conductors, steel is used for:
(a) compensating skin effect
(b) neutralizing proximity effect
(c) reducing line inductance
(d) increasing the tensile strength

Q5. A synchronous motor working at leading power factor can be used as
(a) Current booster
(b) Voltage booster
(c) Power factor correction
(d) All of the above

Q6. Earth wire on EHV overhead transmission line is provided to protect the line against:
(a) Lightning surge
(b) Switching surge
(c) Excessive fault voltage
(d) Corona effect

Q7. Three equal resistors, connected in series across a source of emf, dissipated 10W of
power. What would be the power dissipated in the same resistor when they are
connected in parallel across the same source?
(a) 10 W
(b) 30 W
(c) 90 W
(d) 270 W

Q8. If a 3-phase, 40V, 50Hz, 4 pole induction motor is running at a slip of 5% then the
relative speed of rotor field with respect to stator filed is:
(a) Zero
(b) 75 rpm
(c) 142.5 rpm
(d) 1500 rpm

Q9. A 3-phase induction motor is running at slip ‘s’. If its two supply leads are
interchanged, then the operating slip at that instant will be:
(a) 2s
(b) (1-s)
(c) (2-s)
(d) Zero

Q10. A 10Ω resistor is connected in parallel with a 15Ω resistor and combination in series
with a 12Ω resistor. The equivalent resistance of the circuit is:
(a) 37 Ω
(b) 27 Ω
(c) 18 Ω
(d) None of these

SOLUTIONS
S1. Ans.(a)
Sol. Ceq=C1+C2
= 30μF + 20μF
= 50 μF

S2. Ans.(c)
Sol. Error = 100 mA ×(±2)/100= ±2mA
% error = (±2mA)/50mA×100= ±4%

S3. Ans.(b)
Sol. Shunt resistance =G/(n-1)
G = Resistance of ammeter
n = times to increase the range=I/Im =25/5=5
R=0.2/(5-1)=0.2/4 = 0.05 (Parallel)

S4. Ans.(d)
Sol. ACSR stands for aluminum conductor steel reinforced and steel is used for increasing the tensile strength.

S5. Ans.(c)
Sol. An over-excited synchronous motor has a leading power factor. This makes it useful for power factor correction of industrial loads. If DC field excitation of a synchronous motor is such that back EMF Eb is greater than applied voltage V, then the motor is said to be over excited. An over excited synchronous motor draws leading current.

S6. Ans.(a)
Sol. Earth wire on EHV overhead transmission line is provided to protect the line against
lightning surge.

S7. Ans.(c)
Sol. P∝1/R
P1/P2 =R2/R1
P2=P1 (R1/R2 )
=10(3R/(R/3))=90W

S8. Ans.(a)
Sol. Each field rotates at synchronous speed. So, they are stationary w.r.t each other.

S9. Ans.(c)
Sol. Slip = (2- s)

S10. Ans.(c)
Sol.

Req =(10×15)/25+12
= 6 + 12
= 18 Ω

Quiz: Electrical Engineering 26 Sep 2020

Quiz: Electrical Engineering
Exam: UPSSSC- JE
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. When an AC power is applied to a circuit having reactive load, then the voltage is_______
(a) 90 degree out of phase with current
(b) 180 degree out of phase with the current
(c) In phase with the current
(d) 270 degree out of phase with the current

Q2. If two supply terminals of a 3φ induction motor are interchanged, then the:
(a) Speed of the motor will become zero
(b) Speed of the motor will increase
(c) Motor will continue to run in the same direction with less speed
(d) Motor will rotate in the opposite direction

Q3. What is the relationship between the line voltage and the phase voltage in a delta -connected load?
(a) Line voltage = Phase current
(b) Line voltage > Phase voltage
(c) Line voltage < Phase voltage
(d) Line voltage = Phase voltage

Q4. An eight-pole wound rotor induction motor operating on 60 Hz supply is driven at 1800 rpm by a prime mover in the opposite direction of revolving magnetic field. The frequency of rotor current is:
(a) 60 Hz
(b) 200 Hz
(c) 120 Hz
(d) 180 Hz

Q5. The SI unit of conductivity is:
(a) Ohm/meter
(b) Ohm-meter
(c) Siemens/meter
(d) Siemens – meter

Q6. The SI unit of conductance is:
(a) Coulomb
(b) Ohm
(c) Siemens
(d) Newton

Q7. Select the incorrect statement.
A single-phase induction-motor
1. Requires only one winding
2. Can rotate in one direction only
3. Is not self-starting
(a) Only 1 and 2
(b) Only 2 and 3
(c) 0nly 3 only
(d) Only 1 and 3

Q8. In a single-phase induction motor, how are the windings placed?
(a) Both main and auxiliary windings on stator
(b) Both windings are on rotor
(c) Auxiliary on stator and main on rotor
(d) Main winding on rotor

Q9. Which of the following statements is correct for a single-phase hysteresis motor?
(a) It can run at synchronous and sub-synchronous speed
(b) It can run at sub synchronous speed only
(c) It can run at synchronous and super synchronous speed
(d) It can run at synchronous speed only

Q10. If a 500 KVA, 200HZ transformer is operated at 50HZ, its KVA(s) rating will be–
(a) 250 KVA
(b) 125 KVA
(c) 1000 KVA
(d) 2000 KVA

SOLUTIONS
S1. Ans.(a)
Sol. In the case of inductor or capacitor the voltage or current are out of phase. In the case of an inductor, the voltage leads the current by 90°, whereas in the capacitor the current lead voltage by 90°.

S2. Ans.(d)
Sol. The direction of rotation of a 3-phase motor can be changed by reversing two of its Stator leads.

S3. Ans.(d)
Sol. In delta connected system the line voltage is equal to the phase voltage and the Line current is √3 times of Phase Current.

S4. Ans.(d)
Sol. N_s=120f/P=(120×60)/8
= 900 rpm
Now it is rotating in opposite direction so slip (S) = (900-(-1800))/900=3
Rotter frequency (Fr) = Sf
= 3×60
= 180 HZ

S5. Ans.(c)
Sol. Conductivity (σ) = 1/(Resistivity (ρ) )

Unit =1/Ωm (mho/m)
Or
Siemen/meter

S6. Ans.(c)
Sol. Conductance (G) = 1/(Resistivity (R) )

Unit =1/ohm (mho or siemen)

S7. Ans.(a)
Sol. A single-phase induction motor is not self-starting and has two types of winding: main and auxiliary winding and direction of rotation can be changed by changing polarity of either main or auxiliary winding.

S8. Ans.(a)
Sol. The main parts of a single -phase induction motor are the Stator, Rotor, Windings. The stator is the fixed part of the motor to which A.C. is supplied. The stator contains two types of windings. One is the main winding and the other is the Auxiliary winding. These windings are placed perpendicular to each other.

S9. Ans.(d)
Sol. Hysteresis Motor is a synchronous motor with a cylindrical rotor and does not require any dc excitation to the rotor and it uses non-projected poles. It is a single-phase motor with rotor made up of ferromagnetic material.
initially, when hysteresis motor is started it behaves as a single-phase induction motor and while running it behaves as a synchronous motor.

S10. Ans.(b)
Sol. For constant load;
KVA i.e. S is directly proportional to induced emf and emf is directly proportional to frequency
⇒ S α f
So, S1/f1 =S2/f2
⇒ 500/200=S2/50
⇒ S₂ = 125 KVA

Quiz: Electrical Engineering 14 Sep 2020

Quiz: Electrical Engineering
Exam: UPSSSC JE
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. An ammeter is convertible to a voltmeter by
(a) Changing the scale
(b) Putting a large resistance in parallel with the actual measuring part of the instrument
(c) Putting a large resistance in series with the actual measuring part of the instrument
(d) Simply installing the instrument in parallel with the circuit

Q2. What should be the size of the slide wire of the potentiometer to make it to achieve high accuracy?
(a) As long as possible
(b) As short as possible
(c) 1 meter
(d) Neither too thin nor too thick

Q3. Power transformers are designed to have maximum efficiency at
(a) Full load
(b) 50% load
(c) 80% load
(d) no load

Q4. If the field of a DC shunt motor gets opened while the motor is running, then the
(a) Motor will become slow
(b) The motor will attain the dangerously high speed
(c) Armature current will drop
(d) Armature will oscillate about original speed as the mean speed

Q5. In an electric heater, the metal case is connected to…………
(a) Phase wire
(b) earth wire
(c) neutral wire
(d) none of these

Q6. Which of the following is used for heating of non-conducting materials?
(a) Eddy current heating
(b) Arc heating
(c) Dielectric heating
(d) Induction heating

Q7. A high starting torque can be obtained in a 3-phase induction motor by
(a) Increasing Rotor Reactance
(b) Decreasing Rotor Resistance
(c) Increasing Rotor Resistance
(d) Any of the above

Q8. At starting, rotor reactance of a 3-phase induction motor is ………… as compared to rotor resistance.
(a) Small
(b) Equal to
(c) Large
(d) None of the above

Q9. The form factor is the ratio of
(a) Peak value to rms value
(b) Rms value to average value
(c) Average value to rms value
(d) None of the above

Q10. The element which is capable of delivering energy on its own is known as:
(a) Non-Linear elements
(b) Unilateral elements
(c) Active element
(d) Passive element

SOLUTIONS
S1. Ans.(c)
Sol. An ammeter is converted into voltmeter by putting a large resistance in series with the actual measuring part of the instrument.

S2. Ans.(a)
Sol. in case of potentiometer, E ∝ L ⇒ E = φL
Where φ is the potential gradient
So, In the case of longer wire, the fall of potential per unit length is small. In other words, the potential gradient is small. Lesser the potential gradient, more accurate is the potentiometer.

S3. Ans.(a)
Sol. Power transformer are used for transmission as a step-up device hence they are not directly connected to consumer therefore, load fluctuation is very less. So, the power transformer can operate on full load.

S4. Ans.(b)
Sol. N ∝ Eb/φ
Now if the field winding gets open than flux will become zero i.e. φ = 0
∴ N ∝ Eb/0
Or N = ∞
Hence the speed of the DC shunt Motor will attain the dangerous High Seed.

S5. Ans.(b)
Sol. In an electric heater, the metal case is connected to ‘earth wire’ for protection against fault.

S6. Ans.(c)
Sol. Induction heating → for magnetic and conducting materials
Dielectric heating → for non-conducting materials.

S7. Ans.(c)
Sol. in case of induction motor, the starting torque(T) α Rotor resistance.
So, A high starting torque can be obtained in a 3-phase induction motor by increasing rotor resistance.

S8. Ans.(c)
Sol. The rotor resistance does not depend on the slip of the motor. The rotor reactance depends on the slip. At start, the rotor reactance is large because slip of the motor at start is equal to unity. The large rotor reactance at start makes the starting torque of the motor poor. The motor draws about 5 to 6 times current of its full load current and produce very less torque. The most of the current is reactive in nature and does not contribute in production of torque. That is why the starting torque of the squirrel cage induction motor is poor.

S9. Ans.(b)
Sol. The form factor of an alternating current waveform (signal) is the ratio of the RMS (root mean square) value to the average value.

S10. Ans.(c)
Sol. The active elements generate energy. Batteries, generators, operational amplifiers etc are active elements.

Quiz: Electrical Engineering 02 Sep 2020

Quiz: Electrical Engineering
Exam: UPSSSC JE
Topic: Measurement and measuring instrument

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. Swamping resistance is used to compensate error due to
(a)Stray magnetic field
(b)Temperature variations
(c)Large supply variations
(d)None of the above

Q2. Which of the following is an Integrating instrument?
(a) Watt-hour meter
(b) Ammeter
(c) Voltmeter
(d) Wattmeter

Q3. Induction type energy meter measures energy in _______:
(a) Joules
(b) kW
(c) kWh
(d) W

Q4. Creeping in a singly-phase induction type energy meter may be due to-
(a) Overcompensation for friction
(b) Over voltages
(c) Vibrations
(d) All of these

Q5. In two wattmeter method of power measurement method, the power factor is given as:
(a) sin⁡{tan^(–1)⁡(±√3 (W1 – W2)/(W1 + W_2 )) }
(b) cos⁡{tan^(–1)⁡(±√3 (W1+ W2)/(W1 – W2 )) }
(c) sin⁡{tan^(–1)⁡(±√3 (W1+ W2)/(W1 – W2 )) }
(d) cos⁡{tan^(–1)⁡(±√3 (W1– W2)/(W1+ W2 )) }

Q6. The minimum number of wattmeter(s) required to measure 3-phase. 3-wire balanced or unbalanced power is:
(a) 4
(b) 1
(c) 2
(d) 3

Q7. For increasing the range of voltmeter, connect a
(a) Low value resistance in series with voltmeter
(b) High value resistance in parallel with voltmeter
(c) High value resistance in series with voltmeter
(d) Low value resistance in parallel with voltmeter

Q8. A voltmeter must have very high internal resistance so that
(a) Accuracy is high
(b) Resolution is high
(c) Draws small amount of current
(d) Creates high loading effect of the circuit

Q9. Electrostatic instruments are suitable for the measurement of:
(a) ac and dc voltages
(b) ac voltage and current
(c) dc voltage and current
(d) ac and dc current

Q10. …….. voltmeter has the least power consumption.
(a)Electrostatic type
(b)Hot wire type
(c)Induction type
(d)Moving iron attraction type

SOLUTIONS
S1. Ans.(b)
Sol. The ammeter is a sensitive device which is easily affected by the surrounding temperature. The variation in temperature causes the error in the reading. This can reduce by swamping resistance. The resistance having zero temperature coefficient is known as the swamping resistance. It connects in series with the ammeter.

S2. Ans.(a)
Sol. integrating instrument are those instruments which measures the electrical quantity over a period.

S3. Ans.(c)
Sol. induction type works on AC. Due to series and shunt magnet eddy current induced on ‘Al’ disc, which interacts and provides rotation. The commercial unit of electrical energy is kilowatt-hour (KWh). For measurement of energy in a.c. circuit, the meter used is based on “electro-magnetic induction” principle. They are known as induction type instruments.

S4. Ans.(d)
Sol. Creeping in Energy meter: – Creeping is phenomena by which Al disc rotates continuously.
This is due to following reasons-
a) Pressure coil energized and current coil having no current.
b) due to over voltage and friction.

S5. Ans.(d)
Sol. power measurement by two wattmeter method –
tanθ = √3 ((W1-W2)/(W1+W2 ))
So, power factor =cosθ=cos⁡{tan^(-1)±√3 ((W_1-W_2)/(W_1+W_2 ))}
And two wattmeter method used for-
I. balanced and unbalanced load
II. star and delta connected load.

S6. Ans.(c)
Sol. for n-wire system, minimum number of wattmeter required = (n-1)

S7. Ans.(c)
Sol. Voltmeter range increment: – High value resistance in series with it
I.e.(Rs=Rm (m-1) )
Rm = meter resistance
m=V/Vm≃I/Im

S8. Ans.(c)
Sol. voltmeter having high internal resistance so it draws low current and correctly measure the potential difference connected across the point a-b.

S9. Ans.(a)
Sol. Electrostatics instruments are suitable for only voltage (AC or DC) measurement.
Electrostatics instruments do not measure current, power energy etc.

S10. Ans.(a)
Sol. advantages of electrostatic type voltmeter:
Power consumption is quite low in these types of instruments as the current drawn by these instruments is quite low.
We can measure high value of voltage.
Suitable for voltage (ac or dc) measurement.

Quiz: Electrical Engineering 26 Aug 2020

Quiz: Electrical Engineering
Exam: UPSSSC-JE
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. While conducting short circuit test on a transformer the following side is short circuited:
(a)High voltage side
(b)Low voltage side
(c)Primary side
(d)Secondary side

Q2. The main disadvantages of using short pitch winding in alternators is that it…….
(a)Reduces harmonics in the generated voltage
(b)Reduces the total voltage around the armature coils
(c)Produces asymmetry in the three phase windings
(d)Increase Cu of end connections

Q3. Out of the following methods of heating the one which is independent of supply frequency is
(a)Electric arc heating
(b)Induction heating
(c)Electric resistance
(d)Dielectric heating

Q4. An RLC series circuit of R=4Ω inductive reactance =6Ω and capacitive =3Ω is connected across a 220 V, 50 Hz single phase AC supply. Power factor of the circuit is:
(a) 0.6 leading
(b) 0.8 lagging
(c) 0.6 lagging
(d) 0.8 leading

Q5. Potentiometer is basically a device for:
(a) measuring a current
(b) comparing two voltages
(c) comparing two currents
(d) measuring a voltage

Q6. Which of the following generators have two field windings?
(a)Series wound generator
(b)Shunt wound generator
(c)Compound wound generator
(d)All of the above

Q7. The equalizer rings are used in
(a)Lap winding
(b)Wave winding
(c)Multilayer wave winding
(d)None of these

Q8. In an RLC series circuit the condition below the resonant frequency is:
(a)Xc > XL
(b)Xc + XL
(c)Xc < XL
(d)Xc = XL

Q9. What is the main drawback of paper as an insulating material?
(a)Has poor dielectric strength
(b)Has low insulation resistivity
(c)Has high capacitance
(d)It is hygroscopic

Q10. The unit of luminous flux is:
(a)Candela
(b)Lumen
(c)Weber
(d)Lux

SOLUTIONS
S1. Ans.(b)
Sol. During short-circuit test of a transformer, low voltage side is short-circuited and during open circuit test, high voltage side is open circuited.

S2. Ans.(b)
Sol. if the coil span is less than the pole pitch, then the winding is referred as short pitched. Short-pitches coils reduce the amount of copper needed and it eliminates the harmonics and improves the waveform of generated emf.
The disadvantage of using short-pitch winding is that the total voltage around the coils is somewhat reduced.

S3. Ans.(c)
Sol. Heating by resistance methods is independent of frequency.

S4. Ans.(b)
Sol. When XL>XC: the phase angle Ф is positive and the circuit is inductive. The circuit behaves like a series RL circuit. i.e. lagging power factor.
For given problem: Z=√(R^2+(XL-XC )^2 )
∴Z=√(4^2+(6-3)^2 )=5 Ω
And power factor=R/Z=4/5=0.8 lagging

S5. Ans.(b)
Sol. A potentiometer is an instrument for measuring voltage or ‘potential difference’ by comparison of an unknown voltage with a known reference voltage.

S6. Ans.(c)
Sol. In a Compound Wound Generator, there are two sets of the field winding on each pole. One of them is connected in series having few turns of thick wire, and the other is connected in parallel having many turns of fine wire with the armature windings.

S7. Ans.(a)
Sol. To overcome the detrimental effects resulting from the circulating currents, it is customary to employ equalizer connections in all lap wound armatures.
Therefore, the function of equalizer rings is to avoid unequal current distribution at the brushes and to reduce sparking between brushes and commutator. The equalizer rings are low resistance copper conductor which connects those points in the armature winding which have potential difference between them.

S8. Ans.(a)
Sol. In an RLC series circuit, below the resonant frequency: Xc > XL (capacitive nature)

S9. Ans.(d)
Sol. The only disadvantage is that paper is hygroscopic and even if it is impregnated with suitable compound, it absorbs moisture and thus lowers the insulation resistance of the cable.

S10. Ans.(b)
Sol. The SI unit of luminous flux is the lumen.

Quiz: Electrical Engineering 19 Aug 2020

Quiz: Electrical Engineering
Exam: UPSSSC-JE
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. Holes are drilled on the opposite sides of the spindles of an energy meter to
(a)Avoid creep on load
(b)Balance the disc
(c)Dissipate heat generated due to eddy currents
(d)Increases the deflection torque

Q2. The input of an A.C. circuit having power factor of 0.8 lagging is 40 kVA. The power
drawn by the circuit is
(a) 12 kW
(b) 22 kW
(c) 32 kW
(d) 64 kW

Q3. Which of the following expressions is true for apparent power in an A.C. circuit?
(a) VI cos ∅
(b) Vav.×Iav.
(c) V(r.m.s.) ×I(r.m.s.)
(d) Vpeak×Ipeak

Q4. The efficiency for maximum power transfer to the load is
(a) 25%
(b) 50%
(c) 75%
(d) 100%

Q5. In a parallel R-L-C circuit Susceptance is equal to
(a) 1/X
(b) 1/R
(c) R/Z^2
(d) X/Z^2

Q6.A circuit component that opposes the change in circuit voltage is
(a) Resistance
(b) Capacitance
(c) Inductance
(d) All the above

Q7. A series resonant circuit implies
(a) Zero power factor and maximum current
(b) Unity power factor and maximum current
(c) Unity power factor and minimum current
(d) Zero power factor and minimum current

Q8. A 6 pole, 50 Hz, 3-phase induction motor is running at 950 rpm and has rotor cu loss of 5 Kw. Its rotor input is……….KW.
(a)100
(b)10
(c)95
(d)5.5

Q9. Reactance relay is normally preferred for protection against
(a)Over load currents only
(b)Phase faults only
(c)Earth faults only
(d)High voltage protection only

Q10. Which of the following relay is used for the protection of medium transmission line?
(a)Impedance Relay
(b)Mho’s Relay
(c)Differential Relay
(d)None of the above

SOLUTIONS

S1. Ans.(a)
Sol. The slow but continuous rotation of the disc when only the pressure coils are excited but no current is flowing in the circuit is called “creeping”.
To overcome this creeping effect on no-load, two holes are drilled in the disc on a diameter, i.e. on opposite sides of the spindle. This causes sufficient distortion of the field to prevent rotation when one of the holes comes under one of the poles of the shunt magnet.

S2. Ans.(c)
Sol. P.F.= P/S
Or
P = P.F. x S = 0.8 x 40 = 32 kW

S3. Ans.(c)
Sol. The product of voltage and current if and only if the phase angle differences between current and voltage are ignored. In an AC circuit, the product of the r.m.s voltage and the r.m.s current is called apparent power.

S4. Ans.(b)
Sol. For maximum power transfer to load
Z_L=Z_s or R_L=R_s
I=V/(R_s+R_L )=V/(2R_s ) (Since R_L=R_s)
P_in=V.I=(R_s+R_L ).I
= I^2 (R_s+R_L )=I^2 (2R_s)
P_out=P_load=I^2 R_L=I^2 R_s
η = P_out/P_in
= (I^2 (R_s))/(I^2 (2R_s ) )=1/2=0.5
% η=50%
So, the efficiency for maximum power transfer.

S5. Ans.(a)
Sol. For Parallel RLC Circuit
1/Z=1/R+J(1/X_L +1/X_C )
Y = 1/R+J(1/X)
For here Susceptance = 1/X

S6. Ans.(b)
Sol. A capacitance is a circuit component that opposes the change in circuit.

S7. Ans.(b)
Sol. At resonance, we know that
ω=ω_0=1/√LC& X_L=X_c then
Impedance Z = √(R^2+(X_L-X_c )^2 )
Here L = Inductance
C = Capacitance
⇒ Z = R so there is unity power factor and Current is maximum.

S8. Ans.(a)
Sol. rotor speed (N_r)=950 rpm
And synchronous speed of the motor, N_s= 120f/P=(120×50)/6=1000 rpm
∴slip(s)=(N_s-N_r)/N_s =(1000-950)/1000=0.05
Rotor cu loss is given as, P_L=sP_r
Where, P_r=rotor input
∴P_r=P_l/s=(5×10^3)/0.05=100 KW.

S9. Ans.(c)
Sol. A reactance relay is an overcurrent relay with directional limitations. So, it is normally used against earth faults only.

S10. Ans.(a)
Sol. Relay used for different configuration of transmission lines……
Short transmission line: reactance relay
Medium transmission line: impedance relay
Long transmission line: mho relay