Quiz: Electrical Engineering 19 Oct 2020

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

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

Q1. Determine the self-inductance (in mH) of a 3m long-air cored solenoid, when the coil has 300 turns and the diameter of the coil is 12 cm.
(a) 0.24
(b) 0.32
(c) 0.35
(d) 0.42

Q2. Determine the reluctance (in Amp-turn/Wb) of a coil when the flux through the coil is 25 Wb and the value of produced mmf is 50 Amp-turns.
(a) 2
(b) 4
(c) 6
(d) 8

Q3. A current of 3A through a coil sets flux linkage of 15 Wb-turn. The inductance of the coil is?
(a) 1 H
(b) 3 H
(c) 5 H
(d) 15 H

Q4. Which of the following is bilateral element?
(a) Constant voltage source
(b) Constant current source
(c) Capacitance
(d) All of the above

Q5. In the following circuit, the equivalent capacitance between terminals A and B is

(a) C
(b) C/2
(c) 2C/3
(d) 3C/2

Q6. Which of the following materials has susceptibility independent of temperature?
(a) Ferromagnetic
(b) Ferrimagnetic
(c) Paramagnetic
(d) Diamagnetic

Q7. Determine the capacitive susceptance (in siemens) of a circuit if the capacitor of the circuit is 0.08 mF and supplied with a 50 Hz frequency.
(a) O.025
(b) 0.064
(c) 0.046
(d) 0.034

Q8. Which of the following is an effect of non-uniform current distribution in a conductor?
(a) Skin effect
(b) Ferranti effect
(c) Proximity effect
(d) Skin effect or proximity effect

Q9. A transformer has iron loss of 900 W and copper loss of 1600 W at full load. At what percentage of load will the efficiency be maximum?
(a) 133
(b) 125
(c) 75
(d) 66.66

Q10. A transformer is working at full load with maximum efficiency. Its iron loss is 1000 W. What will be its copper loss at half full load?
(a) 2000 W
(b) 1000 W
(c) 500 W
(d) 250 W

SOLUTIONS
S1. Ans.(d)
Sol. Given that, N=300 turns
d= 12 cm ⇒r=d/2=6 cm=0.06 m
l= 3m
for solenoid, L=(µ0 µr N^2 A)/l=(4π×10^(-7)×1×300×300×π×〖(0.06)〗^2)/3=0.42 mH
NOTE: for air µ_r=1
And µ_0=4π×10^(-7)H/m ≈ 12.57×10−7 H/m

S2. Ans.(a)
Sol. MMF produced in a coil=NI=Reluctance×ϕ
∴NI=Reluctance×ϕ
⇒reluctance=NI/ϕ=50/25=2 AT/Wb

S3. Ans.(c)
Sol. inductance of coil=L=Nϕ/I=15/3=5 H

S4. Ans.(c)
Sol. Bilateral elements are defined as the elements through which magnitude of current is independent of polarity of supply voltage.
e.g. of bilateral elements: resistors, inductors, capacitors

S5. Ans.(a)
Sol.

⇒C_AB=((C×C)/(C+C))+C/2=C/2+C/2=C

S6. Ans.(d)
Sol. Diamagnetic material: Magnetic materials which align against the magnetic field are known as diamagnetic materials. Magnetic susceptibility is χ<0 which means it is always a negative value for diamagnetic material.
These materials are independent of temperature. As these materials magnetize in the opposite direction, they do have a small amount of magnetization intensity. Gold, tin, mercury, water, etc are examples of diamagnetic materials.
For a diamagnetic material, the relative permeability lies between 0 ≤ μr < 1 and its susceptibility lies between −1< χ< 0. 𝑥=𝑢𝑟−1
Where 𝑥 is a dimensionless quantity defines as the magnetic susceptibility of the medium.

S7. Ans.(a)
Sol. capacitive susceptance(B)=1/XC =1/((1/2πfC) )
∴B==2πfC=2×3.14×50×0.08×10^(-3)=0.025 siemens

S8. Ans.(d)
Sol. Skin effect: When an Alternating Current flows through a conductor, it is not distributed uniformly throughout the conductor cross-section. AC current has a tendency to concentrate near the surface of the conductor. This phenomenon in alternating currents is called as the skin effect.
Proximity effect:
When two or more conductors carrying alternating current are close to each other, then distribution of current in each conductor is affected due to the varying magnetic field of each other. when the nearby conductors carrying current in the same direction, the current is concentrated at the farthest side of the conductors. When the nearby conductors are carrying current in opposite direction to each other, the current is concentrated at the nearest parts of the conductors. This effect is called as Proximity effect.

S9. Ans.(c)
Sol. maximum efficiency in %=√((iron loss(Pi))/(full load cu-loss(Pcu)) )×100=√(900/1600) ×100=75 % of full-load

S10. Ans.(d)
Sol. For maximum efficiency, Copper loss (Pc) = Iron losses (Pi) = 1000 W
And P_cu (x)=x^2×Pcu (full-load)
∴P_(cu(half-load))= (1/2)^2 P(cu(full-load))=1/4×1000=250 W

Quiz: Electrical Engineering 07 Oct 2020

Quiz: Electrical Engineering
Exam: SSC JE
Topic: Magnetic circuit

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

Q1. Magnetic field intensity at Centre of circular coil, of diameter 2 m and carrying a current of 2 A is:
(a) 1 A/m
(b) 2 A/m
(c) 4 A/m
(d) 8 A/m

Q2. A coil of 100 turns is rotated at 1200 rpm in a magnetic field having uniform density of 0.04 T. therefore, the value of frequency is……………….
(a) 20 Hz
(b) 25 Hz
(c) 30 Hz
(d) 50 Hz

Q3. Which of the following statements is true about magnetic lines of force?
(a) Magnetic lines of force are always closed.
(b) Magnetic lines of force always intersect each other.
(c) Magnetic lines of force tend to crowd far away from the poles of the magnet
(d) Magnetic lines of force do not pass through the vacuum.

Q4. The qualities aspired to obtain a good permanent magnet is/are ____________
(a) high residual flux
(b) low coercivity
(c) high coercivity
(d) high residual flux and high coercivity

Q5. For which of the following is magnetic susceptibility negative?
(a) Paramagnetic and Ferromagnetic materials
(b) Paramagnetic Materials only
(c) Ferromagnetic Materials only
(d) Diamagnetic Materials

Q6. Which of the following statements are true about the magnetic susceptibility xm of paramagnetic substance?
(a) Value of xm is inversely proportional to the absolute temperature of the sample
(b) xm is negative at all temperature
(c) xm is does not depend on the temperature of the sample.
(d) All of the above

Q7. …………materials having relative permeability much greater than unity
(a) Ferromagnetic
(b) Paramagnetic
(c) Diamagnetic
(d) None of the above

Q8. The materials having high retentivity and high coercivity are suitable for making
(a) Weak magnets
(b) Temporary magnets
(c) Permanent magnets
(d) None of the above

Q9. Both the number of turns and the core length of an inductive coil are doubled. Its self-inductance will be:
(a) Doubled
(b) Halved
(c) Tripled
(d) Quadrupled

Q10. Curie temperature is the temperature above which a ferro-magnetic material becomes……
(a) Paramagnetic
(b) Diamagnetic
(c) Either of the above
(d) None of the above

SOLUTIONS
S1. Ans.(a)
Sol. magnetic field intensity at Centre of circular coil is given by:H=I/2R=2/(2×1)=1 A/m

S2. Ans.(a)
Sol. speed of rotation =1200 rotations per minute=1200/60=20 rotations per second
Frequency is defined as rotations per second=20 Hz

S3. Ans.(a)
Sol. Magnetic lines of force are always closed.

S4. Ans.(d)
Sol. It’s both high residual flux as well as coercivity is desired for a good permanent magnet.

S5. Ans.(d)
Sol. Diamagnetic material: Magnetic materials which align against the magnetic field are known as diamagnetic materials. Magnetic susceptibility is χ<0 which means it is always a negative value for diamagnetic material.
These materials are independent of temperature. As these materials magnetize in the opposite direction, they do have a small amount of magnetization intensity. Gold, tin, mercury, water, etc are examples of diamagnetic materials.
For a diamagnetic material, the relative permeability lies between 0 ≤ μr < 1 and its susceptibility lies between −1< χ< 0. 𝑥=𝑢𝑟−1
Where 𝑥 is a dimensionless quantity defines as the magnetic susceptibility of the medium.

S6. Ans.(a)
Sol. The mathematical definition of magnetic susceptibility is the ratio of magnetization to applied magnetizing field intensity. This is a dimensionless quantity.
x=M/H
Since it is the ratio of two magnetic fields, susceptibility is a dimensionless number. paramagnetic, superparamagnetic, and ferromagnetic substances have positive susceptibilities (χ>0).
For paramagnetic substance value of χm is inversely proportional to the absolute temperature of the sample
These materials (Paramagnetic)are temperature dependent and are weekly attracted by magnets with relative permeability 1.00001 to 1.003.

S7. Ans.(a)
Sol. for ferromagnetic materials, relative permeability is much greater than unity.
For a paramagnetic material, its susceptibility is positive. This is because its relative permeability is slightly greater than unity. For a diamagnetic material, the relative permeability lies between 0≤μr<1 and its susceptibility lie between −1<x <0.

S8. Ans.(c)
Sol. The permanent magnets are made from hard ferromagnetic materials (steel, carbon steel, cobalt steel etc). since, these materials have high retentivity, the magnet is quite strong. Due to their high coercivity, they are unlikely to be demagnetised by stray magnetic fields.

S9. Ans.(a)
Sol. self- inductance L=(µ0 µr AN^2)/L
Where, A=Area
N= number of turns
L=length of coil
If, both number of turns and core length of an inductive coil are doubled. Its self-inductance will be doubled.

S10. Ans.(a)
Sol. Curie temperature (TC) is the temperature at which certain materials lose their permanent magnetic properties, to be replaced by induced magnetism.
Below the Curie temperature, the atoms are aligned and parallel, causing spontaneous magnetism; the material is ferromagnetic.
Above the Curie temperature the material is paramagnetic, as the atoms lose their ordered magnetic moments when the material undergoes a phase transition.

Quiz: Electrical Engineering 17 Aug 2020

Quiz: Electrical Engineering
Exam: NLC-GET
Topic: Miscellaneous

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

Q1. In ……………, the susceptibility will decrease with increase in temperature and they have relatively small susceptibility at all temperatures.
(a)Anti-ferromagnetic materials
(b)Ferrimagnetic materials
(c)Superparamagnetic materials
(d)Ferromagnetic materials

Q2. In which of the following motors is torque produced due to the tendency of the rotor to align itself in the minimum reluctance position when the speed of the motor is close to the synchronous speed?
(a)Hysteresis motor
(b)Repulsion motor
(c)Reluctance motor
(d)Universal motor

Q3. For ………., the output C(s) of the system is equal to transfer function of the system.
(a)Impulse response
(b)Step response
(c)Parabolic response
(d)Ramp response

Q4. The function of the ………… is to facilitate collection of current from the armature conductors.
(a)Yoke
(b)Commutator
(c)Pole
(d)Brush

Q5. High resistances are provided with a guard terminal. This guard terminal used to:
(a)To eliminate the effect of leads
(b)Guard the resistance against stray electrostatic fields
(c)Bypass the leakage current
(d)Guard the resistance against overloads

Q6. If A is the number of parallel paths and P is the number of poles, then the number of parallel paths in lap winding and in wave winding is
(a) A = P, A = 2
(b) A = 2P, A = P
(c) A = 2, A = P
(d)A = P, A = 2P

Q7. In a d.c. machine, the current rating and voltage rating of wave winding is
(a) Low, high
(b) High, low
(c) High, high
(d) Low, low

Q8. A forward bias PN junction will act as a/an:
(a)Amplifier
(b)Open switch
(c)Closed Switch
(d)Attenuator

Q9. A single phase 230V energy meter has a constant load of 5 A passing through it for 6 hours at unity p.f. If the meter disc makes 2070 revolutions during this period, what is the meter constant in revolutions per unit?
(a) 100 rev/unit
(b) 200rev/unit
(c) 300rev/unit
(d) 400rev/unit

Q10. Resistance of an ammeter having range 0–5A is 1.8 Ω. It is shunted by a resistor of
0.2 Ω. What is the effective current when the pointer reads 2 A?
(a) 10 A
(b) 30 A
(c) 15 A
(d) 20 A

SOLUTIONS

S1. Ans.(a)
Sol. Antiferromagnetic materials have small positive susceptibilities at all temperatures. A critical temperature in this case is called Neel temperature. The material’s susceptibility changes around the Neel temperature as shown in fig. given below. The susceptibility increases inversely with temperature above TN and decreases inversely below this temperature.

S2. Ans.(c)
Sol. The reluctance motor has basically two main parts called stator and rotor. the stator has a laminated construction, made up of stampings. in the reluctance motor, rotor tries to align itself with the axis of rotating magnetic field in a minimum reluctance position. But due to rotor inertia, it is not possible, when rotor is in standstill condition.
So, rotor starts rotating near synchronous speed as a squirrel cage induction motor. When the rotor speed is about synchronous, stator magnetic field pulls rotor into synchronism i.e. minimum reluctance position and keeps it magnetically locked. Then rotor continues to rotate with a speed equal to synchronous speed. Such a torque exerted on the rotor is called the reluctance torque.

S3. Ans.(a)
Sol. For impulse response, the output C(s) of the system is equal to transfer function of the system.

S4. Ans.(b)
Sol. Functions of commutator in d.c. machines are:
(A) To facilitate the collection of current from armature conductors
(B) To convert internally developed induced emf to unidirectional emf
(C) To produce unidirectional torque in case of motors

S5. Ans.(c)
Sol. The guard terminal used to bypass the leakage current. Current flowing into the guard terminal is not measured by the instrument and so is ignored by the insulation resistance measurement.

S6. Ans.(a)
Sol. number of parallel paths for LAP winding: A = P
Number of parallel paths for wave winding: A = 2

S7. Ans.(a)
Sol. wave winding is used for HIGH VOLTAGE and low current rating applications.

S8. Ans.(c)
Sol. If the positive terminal of an external battery is connected to p-type and its negative terminal to n-type of PN junction then such a biasing called forward biasing.
An ideal diode acts as a short circuit or closed switch under forward bias condition and open circuit or open switch under reverse biased condition.

S9. Ans.(c)
Sol. Meter constant (k) =(No.of revolution)/(Energy in given period)
=2070/VIcosϕ(t) and (t)→ in hours
=2070/(230×5×1×6)
=2070/6.9=300 rev\/kwh
= 300 rev/unit

S10. Ans.(d)
Sol. Multiplying power of shunt, m=Rm/Rsh +1=1.8/0.2+1=10
Therefore, the range of ammeter increases 10 times. When the ammeter indicates 2 A, the circuit current = 2 ×10 = 20 A.

Quiz: Electrical Engineering 30 July 2020

Quiz: Electrical Engineering
Exam: UPPSC-AE
Topic: Miscellaneous

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

Q1. The amount of feedback applied to an amplifier reduces the gain by a factor of 10. The bandwidth
(a)Decreases by factor of 10
(b)Increases by factor of 10
(c)Remains the same
(d)None of the above

Q2. Transformer utilization factor of TUF = K signifies:
(a) That transformer for rectifier should be 1/K times larger than that for ac source
(b) That transformer for rectifier should be K-1 times larger than that for ac source
(c) That transformer for rectifier should be 1-K times larger than that for ac source
(d) That transformer for rectifier should be K time larger than that for ac source

Q3. A synchronous generator is feeding power to infinite bus bars at unity power factor. Its excitation is now increased. It will feed
(a)The same power but at leading power factor
(b)The same power but at lagging power factor
(c)More power at unity power factor
(d)Less power at unity power factor

Q4. A 400/200 volts transformer has pu impedance of 0.05. the HV side voltage required to circulate full load current during short circuit test is
(a)20 V
(b)40 V
(c)10 V
(d)5 V

Q5. In a D.C. machine, fractional pitch winding is used
(a)to increase the generated voltage
(b)to reduce sparking
(c)to save the copper because of shorter end connections
(d)due to (b) and (c) above

Q6. Why are shunt reactors connected at the receiving end of long transmission line system?
(a)To increase the terminal voltage
(b)To compensate voltage rise caused by capacitive charging at light load
(c)To improve power factor
(d)None of these

Q7. Which of the following transmission line have more initial cost?
(a)Overhead Transmission
(b)Underground transmission
(c)Both have almost the same initial cost
(d)None of the above

Q8. Which one of the following is the CORRECT relation between the peak and average value of an alternating current?
(a) Iavg=0.637IPeak
(b) Iavg=1.414IPeak
(c) Iavg=0.7071IPeak
(d) Iavg=0.8241IPeak

Q9. With regard to filtering property, lead compensator is
(a)Low pass filter
(b)High pass filter
(c)Band pass filter
(d)Band reject filter

Q10. The maximum phase occurs at the ………. of the two corner frequencies?
(a) arithmetic mean
(b) geometric mean
(c)Both a& b
(d)None of the above

SOLUTIONS

S1. Ans.(b)
Sol. for an amplifier, the product of gain-bandwidth is always constant. So, if gain reduces by a factor of 10 then the bandwidth increases by a factor of 10.

S2. Ans.(a)
Sol. The transformer utilization factor (TUF) of a rectifier circuit is defined as the ratio of the DC power available at the load resistor to the AC rating of the secondary coil of a transformer.
TUF = (DC POWER output (pdc))/(Effective VA rating of transformer)
TUF indicate how much is the utilization of the transformer in a circuit.
▭(TUF=Pdc/Pac(rated) )

▭(■(TUF for full wave rectifier=0.812@TUF for half wave rectifier=0.287))

For ideal condition, TUF=1 and apart from this, TUF is always less than 1.

S3. Ans.(b)
Sol. from the V-curve of an alternator, it is clear that when excitation is increased, the alternator feeds a lagging power factor.

S4. Ans.(a)
Sol. pu quantity will be same on both side.
Given Z_pu=0.05
And we know 〖Vsc〗pu=〖Isc〗pu×Zpu and 〖Isc)pu=I_rated/(I_base)
As rated current on HV side is same as base current on HV side. Then, 〖I_sc〗_pu=1
And 〖Vsc〗pu=〖Isc〗pu×Zpu=1×0.05=0.05
∴Vsc (HV)=〖Vsc〗pu×Vbase=0.05×400=20 V

S5. Ans.(d)
Sol. Some advantages of short pitch winding are:
Due to shortening span, the copper required is less.
Low copper losses
Improve waveform due to reduction in harmonic
Fractional Pitch winding reduces sparking in DC machines

S6. Ans.(b)
Sol. Shunt Reactor compensation at the receiving end help to reduce the effect of capacitance thus reducing the Ferranti effect.
Shunt Reactor absorbs the excess reactive power during no load or light load condition and thus helps in stabilizing the voltage of Transmission Line.

S7. Ans.(b)
Sol. As the voltage level increases the cost of insulation is increased therefore the underground cable is restricted to low and medium voltages.

S8. Ans.(a)
Sol. For AC, Iavg=(2Ip)/π=0.637 Ip

S9. Ans.(b)
Sol. In terms of filtering property, lead compensator acts as high pass filter.
Following are the properties of lead compensator….
For sinusoidal input, output is leading
Acts as high pass filter
Reduces rise time and peak overshoot
Increases bandwidth & stability of the system.

S10. Ans.(b)
Sol. The maximum phase occurs at the GM (geometric mean) of the two corners
frequencies.

Quiz: Electrical Engineering 29 July 2020

Quiz: Electrical Engineering
Exam: SSC JE
Topic: Miscellaneous

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

Q1. The inductance of certain moving-iron ammeter is expressed as
L=10+3Ө-Ө^ 2/4 µH,
Where Ө is deflection in radians from the zero position. The control spring torque is 25 ×10^(-6) Nm/radian. The deflection of the pointer in radian when the meter carries a current of 5 A, is
(a) 2.4
(b) 2.0
(c) 1.2
(d) 1.0

Q2. The relative permeability of the diamagnetic material is
(a) Greater than 1
(b) Greater than 10
(c) Less than 1
(d) Greater than 100

Q3. Which of the following provides support to the insulators used in overhead lines?
(a) Cross-Arm
(b) Support/base
(c) Phase Plate
(d) Conductor

Q4. Assertion A: Fractional pitch winding is used in DC machines
Reason B: Fractional Pitch winding reduces sparking in DC machines
Which of the following is correct?
(a) B is true, but A is false
(b) A and B are true, but B is not the correct explanation of A
(c) A and B are true, and B is the correct explanation of A
(d) A is true but B is false

Q5. The essential requirement of good heating elements is
(a) High Specific resistance
(b) Free from oxidation
(c) Low-temperature coefficient of resistance
(d) All of the above

Q6. In order to switch-off and EHV circuit for maintenance, the following sequence is adopted:
(a) Open the circuit breaker, open the isolator, operate the earth switch
(b) Operate the earth switch, open the isolator, open the circuit breaker
(c) Open the isolator, operate the earth switch, open the circuit breaker
(d) Open the isolator, open the circuit breaker, operate the earth switch

Q7. The yoke of small DC machine is made up of–
(a) Cast iron
(b) Aluminium
(c) Stainless steel
(d) Copper

Q8. A 200/400 V, 10 kVA, 50 Hz single-phase transformer has, at full-load, a copper loss
of 120 W. If it has an efficiency of 98% at full-load, determine the iron losses.
(a) 84 W
(b) 117 W
(c) 92 W
(d) 106 W

Q9. An isolation transformer has primary to secondary turns ratio of
(a) 1:1
(b) 1:2
(c) 2:1
(d) can be any ratio

Q10. An autotransformer having voltage transformation ratio 0.8 supplies a load of 10
kW. The power transferred inductively from the primary to the secondary is
(a) 10 kW
(b) 8 kW
(c) zero
(d) 2 kW

solutions

S1. Ans.(c)
Sol. Given that, L=10+3Ө-Ө^2/4µH
Then dL/dӨ=(3-Ө/2)×10^(-6)
K=25×10^(-6)
And we know that the control torque for moving iron ammeter is given by :
T_c=kӨ=1/2 I^2 dL/dӨ
∴(25×10^(-6))Ө=1/2×5^2×(3-Ө/2)×10^(-6)
∴2Ө=(3-Ө/2)
⇒Ө=6/5=1.2 rad

S2. Ans.(c)
Sol. Diamagnetic materials are materials with relative permeabilities slightly smaller than 1 (μr < 1). Permeability is a measure of how easy it is to establish the flux in a material. Ferromagnetic materials have high permeability and hence low Reluctance, while non-magnetic materials have low permeability and high Reluctance.

S3. Ans.(a)
Sol. Cross arms are installed at the top of the pole for holding for holding the insulator on which the conductors are fastened. Cross arms are either made up of wood or steel angle sections.

S4. Ans.(c)
Sol. Some advantages of short pitch winding are:
(a)Due to shortening span, the copper required is less.
(b)Low copper losses
(c)Improve waveform due to reduction in harmonic
(d)Fractional Pitch winding reduces sparking in DC machines

S5. Ans.(d)
Sol. The required properties in material used for heating elements-
#High melting point.
#Free from oxidation in open atmosphere.
#High tensile strength.
#Sufficient ductility to draw the metal or alloy in the form of wire.
#High resistivity.
#Low temperature coefficient of resistance.
Following material are used for manufacturing heating element-
Nichrome
Kanthal
Cupronickel (CuNi): an alloy of copper that contains nickel and strengthening elements, such as iron and manganese.
Platinum

S6. Ans.(a)
Sol. In order to switch-off and EHV circuit for maintenance, the following sequence is adopted:
S1: Open the circuit breaker: CB is a device which can make and break the circuit under normal as well as under faulty condition by manually or automatically.
S2: Open the isolator: Isolator is a device which always operates under no load condition.
S3: Operate the Earth switch: Its function is to discharge any remnant charge due to residual magnetism which might be lethal to human and expensive testing equipment.

S7. Ans.(a)
Sol. The yoke of small Dc machine is made up of cast iron because it has a low reluctance so they will support a strong magnetic field with a high density.

S8. Ans.(a)
Sol. Output at full load at unity pf = 10×1 = 10kW
Input = output / η = 10 / 0.98 = 10.204 kW
Total F.L = 10.204-10=0.204 Kw=204W
Iron loss = 204 – 120 = 84 W

S9. Ans.(a)
Sol. For isolation transformer = 1:1

S10. Ans.(d)
Sol. Power transferred inductively
= (1-0.8) × 10 = 2kW

Quiz: Electrical Engineering 13 July 2020

Quiz: Electrical Engineering

Exam: SSC JE

Topic: Miscellaneous

Each question carries 1 mark.

Negative marking: 1/4 mark

Time: 10 Minute

 

Q1. Increase in percentage of carbon in carbon steel reduces

(a) resistivity

(b) coercive force

(c) permeability

(d) retentivity

 

 

Q2. Uniaxial anisotropy can be induced in bulk material by

(a) magnetic annealing

(b) magnetic quenching

(c) cold working

(d) any of the above method

 

 

Q3. The relative permeability is less than 1 in

(a) ferromagnetic material

(b) diamagnetic materials

(c) ferrites

(d) paramagnetic materials

 

 

Q4. The errors introduced by an instrument fall in which category?

(a) systematic error

(b) Random error

(c) Gross error

(d) Environmental error

 

 

Q5. Torque/weight ratio of an instrument indicates

(a) selectivity

(b) accuracy

(c) fidelity

(d) sensitivity

 

 

Q6. Sensitivity of a voltmeter is given as

(a) Ω/V

(b) reciprocal of full-scale deflection current

(c) both

(d) none

 

 

Q7. For defining a standard meter, wavelength of which material is used

(a) Neon

(b) Helium

(c) Krypton

(d) Xenon

 

 

Q8. The most suitable primary standard for frequency is

(a) Rubidium vapour standard

(b) Cesium beam standard

(c) Quartz standard

(d) Hydrogen maser standard

 

 

Q9. The following is not essential for the working of an indicating instrument

(a) deflecting torque

(b) braking torque

(c) damping torque

(d) controlling torque

 

 

Q10. Preferred material for permanent magnet is

(a) stainless steel

(b) alnico

(c) tungsten steel

(d) soft iron

 

Solutions

S1. Ans. (d)

Sol. Retentivity

 

S2. Ans. (d)

Sol. Uniaxial anisotropy can be induced in bulk material by all of the following method

  • magnetic annealing
  • magnetic quenching
  • cold working

 

S3. Ans. (b)

Sol. diamagnetic materials

 

S4. Ans. (a)

Sol. systematic error

 

S5. Ans. (d)

Sol. Sensitivity

 

S6. Ans. (c)

Sol. Both

 

S7. Ans. (c)

Sol. Krypton

 

S8. Ans. (b)

Sol. Cesium beam standard

 

S9. Ans. (b)

Sol. braking torque

 

S10. Ans. (b)

Sol. alnico

 

 

 

Quiz: Electrical Engineering 9 July 2020

Quiz: Electrical Engineering
Exam: UPPSC-AE
Topic: Miscellaneous

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

Q1. Form factor is the ratio of
(a) Average value to rms value
(b) Rms value to average value
(c) Peak value to average value
(d) Peak value to rms value

Q2. Peak factor is the ratio of
(a) Average value to rms value
(b) rms value to average value
(c) maximum value to average value
(d) maximum value to rms value

Q3. The transformer used for AC welding sets is
(a) Booster type
(b) Step up type
(c) Step down type
(d) Equal turn ratio type

Q4. The commutator of a DC generator acts as:
(a) Amplifier
(b) Rectifier
(c) Load
(d) Multiplier

Q5. For RLC ac series circuits at resonance the current is:
(a) Minimum at leading p.f
(b) Minimum at lagging p.f
(c) Maximum at unity p.f
(d) Maximum at leading p.f

Q6. In star delta starting of three phase induction motor the starting voltage reduced by
(a) 1.73 times of normal voltage
(b) 3 times of normal voltage
(c) 1/3 times of normal voltage
(d) 1/1.73 times of normal voltage

Q7. An electric iron is rated at 230V, 400W, 50Hz. The voltage rating 230V refers to
(a) Rms value
(b) Peak to peak value
(c) Average value
(d) Peak value

Q8. A star delta transformer has a phase to phase voltage transformation ratio of a:1[delta phase: star phase]. The line to line voltage ratio of star- delta is given by:
(a) a/1
(b)a/1.73
(c) 1.73a
(d) 1.73/a

Q9. Transformer core laminations are made from
(a) Low carbon steel
(b) Silicon steel
(c) Nickel steel
(d) Chrome steel

Q10. If dc voltage is applied to the primary of a transformer it may
(a) Work
(b) Not work
(c) Burn the winding
(d) Give lower voltage on the secondary side

Solution

S1. Ans(b)
Sol. Form factor is the ratio of Rms value to average value

S2. Ans(d)
Sol. Peak factor is the ratio of maximum value to rms value

S3. Ans(c)
Sol. Step down type

S4. Ans(b)
Sol. Rectifier

S5. Ans(c)
Sol. Maximum at unity power factor

S6. Ans(d)
Sol. In star delta starting of three phase induction motor the starting voltage reduced by 1/1.73 times of
its normal voltage

S7. Ans(a)
Sol. Rms value

S8. Ans(d)
Sol. 1.73/a

S9. Ans(b)
Sol. Silicon steel

S10. Ans(c)
Sol. Burn the winding

Quiz: Electrical Engineering 8 July 2020

Quiz: Electrical Engineering
Exam: SSC JE
Topic: Miscellaneous

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

Q1. Which of the following function represents parabolic characteristics?
a) ƒ(t) = t
b) ƒ(t) = t^2
c) ƒ(t) = e^(-at)
d) ƒ(t) = Ut

Q2. The power factor of a purely resistive circuit will be
a) unity
b) zero
c) infinity
d) 0.5

Q 3. The quality factor of RCL circuit will increase if the
a) R decreases
b) R increases
c) Voltage increases
d) Impedance increases

Q 4. If the diameter of a current carrying conductor is doubled, the resistance will
a) Be reduced to half
b) To reduce to one-fourth
c) Remain same
d) Be doubled

Q5. A moving coil instrument can be used to measure
a) Low frequency alternating current
b) High frequency alternating current
c) Direct current
d) Direct current and alternating current both

Q6. The cost of a capacitor increases as
a) The size of a capacitor increase
b) Voltage rating of the capacitor increases
c) Microfarad rating of the capacitor increases
d) None of the above

Q7. The thermal time constant is the time
a) To reach the final steady temperature if the initial rate of increase of temperature were maintained constant
b) To reach 63% of fined steady temperature
c) To reach 86.6% of final steady temperature
d) To reach 37% of final steady temperature

Q8. The auto-starters (using three auto-transformers) can be used to start cage induction motor of the following type:
a) Star connected only
b) Delta connected only
c) (a) and (b) both
d) None of the above

Q9. The crawling in the induction motor is caused by
a) Improper design of the machine
b) Low voltage supply
c) High Loads
d) Harmonics developed in the motor

Q10. When the equivalent circuit diagram of double squirrel cage induction motor is constructed the two rotor cages can be considered
a) In parallel
b) In series parallel
c) In series
d) In parallel with stator

Solution
Sol.1
Ans: (b)
ƒ(t) = t^2 (parabolic characteristics)

Sol.2
Ans: (a)
Unity

Sol.3
Ans: (a)
Value of resistance, R decreases

Sol.4
Ans: (b)
To reduce to one-fourth

Sol.5
Ans: (c)
Direct current

Sol.6
Ans: (b)
Voltage rating of capacitor increases

Sol.7
Ans: (b)
To reach 63% of fined steady temperature

Sol.8
Ans: (c)
Star connected
Delta connected

Sol.9
Ans: (d)
Harmonics developed in the motor

Sol.10
Ans: (a)
In parallel

Quiz: Electrical Engineering 6 July 2020

Quiz: Electrical Engineering
Exam: SSC JE
Topic: Utilization of electrical energy

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

Q1. Which of following method is not employed for lighting Calculation?
(a) Inverse square law
(b) Stefan Tube method
(c) Watts per square method
(d) Lumen method

Q2. Which of the following will need lowest level of illumination?
(a) Display
(b) fine engraving
(c) Railway platform
(d) Auditorium

Q3. A mercury lamp vapour lamp gives
(a) pink light
(b) Yellow light
(c) greenish blue light
(d) White light

Q4. Principles are used in street lighting is
(a) diffusion and specular reflection
(b) induction
(c) electromagnetic
(d) electrostatic

Q5. High pressure mercury discharge lamps are not used in
(a) park
(b) in the street light of play ground
(c) stage lighting
(d) drawing room

Q6. Which of the following lamp gives nearly monochromatic light?
(a) Sodium vapour lamp
(b) GLS lamp
(c) Tube light
(d) Mercury vapour lamp

Q7. Power factor is highest in case of
(a) mercury arc lamp
(b) sodium vapour lamp
(c) Tube lights
(d) GLS lamp

Q8. To prevent excessive brightness which type of lighting is used
(a) direct
(b) indirect
(c) general
(d) local

Q9. The most important factor with respect to lighting quality is
(a) color
(b) glare
(c) illumination
(d) uniform lighting

Q10. Arc can be produced by
(a) AC current only
(b) DC current only
(c) Either AC & DC current
(d) none of the above

Solution

S1. Ans. (b)
Sol. Stefan Tube method is not used for lighting calculation method.

S2. Ans. (c)
Sol. Railway platform

S3. Ans. (c)
Sol. greenish blue light

S4. Ans. (a)
Sol. diffusion and specular reflection

S5. Ans. (d)
Sol. drawing room

S6. Ans. (a)
Sol. Sodium vapour lamp

S7. Ans. (d)
Sol. Power factor is highest in case of GLS lamp.

S8. Ans. (b)
Sol. Indirect

S9. Ans. (b)
Sol. Glare

S10. Ans. (c)
Sol. Either AC or DC

Quiz: Electrical Engineering 4 July 2020

Quiz: Electrical Engineering
Exam: NLC
Topic: Switchgear and protection

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

Q1. A balanced 3-phase system consist of
(a) zero sequence current only
(b) positive sequence current only
(c) negative and zero sequence currents
(d) zero, negative and positive sequence currents

Q2. The positive sequence current of a transmission line is:
(a) always zero
(b) 1/3 of negative sequence current
(c) equal to negative sequence current
(d) 3 times negative sequence current

Q3. In case of an unbalanced star-connected load supplied from an unbalanced 3-phase, 3-wire system, load currents will consist of
(a) positive-sequence component
(b) negative-sequence component
(c) zero-sequence component
(d) only (a) and (b)

Q4. For an unbalanced fault, with paths of zero sequence currents, at the point of faults
(a) the negative and zero sequence voltage are minimum
(b) the negative and zero sequence voltage are maximum
(c) the negative sequence voltage is minimum and zero sequence voltage is maximum
(d) the negative sequence voltage is maximum and zero sequence voltage is minimum

Q5. Zero-sequence fault current is absent when fault is
(a) single-line-to-ground fault
(b) line-to-line-ground fault
(c) double-line-ground fault
(d) line-to-line

Q6. In case of single-line to ground fault
(a) all sequence networks are connected in parallel
(b) all sequence networks are connected in series
(c) positive and negative sequence networks are connected in parallel
(d) zero and negative sequence networks are connected in series

Q7. Zero sequence current can flow from a line to transformer bank if the windings are in
(a) grounded star/delta
(b) delta/star
(c) star/grounded star
(d) delta/delta

Q8. Negative-sequence reactance of a transformer is
(a) equal to the positive sequence reactance
(b) larger than the positive sequence reactance
(c) smaller than the positive sequence reactance
(d) None of the above

Q9. If all the sequence voltages at the fault point in a power system are equal, then the fault is a
(a) three phase faults
(b) line to ground fault
(c) line to line fault
(d) double line to ground fault

Q10. In an un-transposed three-phase transmission line
(a) the sequence network does not have mutual coupling
(b) a positive sequence currents may cause a negative sequence voltage drop
(c) the sequence impedance matrix is diagonal
(d) none of the above

Solutions
S1. Ans. (b)
Sol. positive sequence current only

S2. Ans. (c)
Sol. equal to negative sequence current

S3. Ans. (d)
Sol. only (a) and (b)

S4. Ans. (b)
Sol. the negative and zero sequence voltage are maximum

S5. Ans. (d)
Sol. Zero-sequence fault current is absent when fault is line-to-line.

S6. Ans. (b)
Sol. all sequence networks are connected in series

S7. Ans. (a)
Sol. grounded star/delta

S8. Ans. (a)
Sol. equal to the positive sequence reactance

S9. Ans. (d)
Sol. double line to ground fault

S10. Ans. (b)
Sol. a positive sequence currents may cause a negative sequence voltage drop

Quiz: Electrical Engineering 2 july 2020

Quiz: Electrical Engineering

Exam: UPSSSC JE

Topic: Miscellaneous

 

Each question carries 1 mark.

Negative marking: 1/4 mark

Time: 10 Minute

 

 

  1. Potential transformers are used to measure _________
    a) high voltages
    b) low voltages
    c) high currents
    d) low currents

 

  1. Potential transformer is similar in design to a
    a) C.T.
    b) Step up transformer
    c) Power transformer
    d) Step down transformer

 

  1. The primary current in a C.T. is _________
    a) independent of secondary circuit
    b) dependent on the secondary circuit
    c) depends on the transformation ratio
    d) depends on the nominal ratio

 

  1. The excitation current of a C.T.
    a) varies over a fixed range of operation
    b) varies over a wide range of normal operation
    c) is fixed over a range of operation
    d) is fixed always

 

  1. Instrument transformers provide _________
    a) electrical isolation from low rated winding
    b) electrical isolation from high rated winding
    c) electrical isolation from medium rated winding
    d) no electrical isolation at all

 

  1. A 5A ammeter can measure a current of up to 1000 A using a _________
    a) 5/1000A C.T.
    b) 1000A C.T.
    c) 5A C.T.
    d) 1000/5A C.T.

 

  1. How can the meter circuit be isolated from power circuit?
    a) by grounding
    b) through electrical isolation
    c) by physical separation
    d) through mechanical isolation

 

  1. In a DC machine, how coil-side emf varies towards the outer side of poles?
    a) Decreases
    b) Remains same
    c) Increases
    d) First increases the decreases

 

  1. What is the effect of armature coils at points where brushes are located?
    a) Induces positive emf
    b) Induces negative emf
    c) Induces zero emf
    d) Depends on the speed of rotor

 

  1. Coil span for 4-pole, 12-slot armature winding is_______
    a) 24
    b) 48
    c) 8
    d) 3

 

Solution

Sol. 1 Ans (a)

Potential transformers are also known as P.T. and are used in the measurement of high magnitude of voltages.

 

Sol. 2 Ans (c)

In terms of design, the potential transformer resembles a power transformer. Potential transformers have a very low loading capacity of the order of a few volt amperes.

 

Sol. 3 Ans (a)

A C.T. is used for the measurement of high magnitude of currents in a circuit, while a P.T. is used for the measurement of high magnitude of voltages in a circuit. Primary current in a C.T. is independent of the secondary circuit conditions.

 

Sol. 4 Ans (b)

In a potential transformer, the excitation current remains constant under normal operation. While in a current transformer, the excitation current varies over a wide range of operation.

Sol. 5 Ans (b)

In an instrument transformer, the low rated secondary windings provide electrical isolation from the high rated primary winding.

 

Sol. 6 Ans (d)

A 1000/5A current transformer can be used for measuring a current of up to 1000A by making use of an ammeter with 5A current reading.

 

Sol. 7 Ans (b)

Leads of the secondary winding transformer are brought to the switch board thus separating them from high voltage windings. In this way, the meter circuit is isolated from the high voltage power circuit.

 

Sol. 8 Ans (a)

Coil side current pattern is the same as the emf pattern. Only difference is that while the coil-side emf reduces towards the outer side of poles, the current remains the same in all the coil-sides except for alterations from pole to pole, while the coil side current reverses, the current exchanged with external circuit must be unidirectional.

 

Sol. 9 Ans (c)

Brushes are at magnetically neutral region hence, induced emf due to armature coils at brushes will be equal to zero. As in the magnetically neutral region change in flux will be equal to the zero, emf will not be induced (Faraday’s law).

 

Sol. 10 Ans (d)

Coil span is defined as a ratio of number of slots in the armature winding which are also equal to the number of commutator segments to the number of poles. Here, Slots in the armature winding= 12, Number of poles= 4.
YCS= 12/4= 3.

 

Quiz: Electrical Engineering 1 July 2020

Quiz: Electrical Engineering
Exam: UPPSC AE
Topic: Measurement and measuring instrument

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

 

1.Range of an electrical instrument depends on __________

a) current

b) voltage

c) power

d) resistance

 

 

2. Potential terminals have a __________

a) high current capacity

b) low voltage capacity

c) low current capacity

d) high voltage capacity

 

3.What is the condition for using a multiplier in A.C. voltmeters?

a) by using ac supply

b) by maintaining a uniform impedance

c) by maintaining a uniform frequency

d) by using a galvanometer

 

4. What is the relation between the balance equation and the magnitude of input voltage?

a) directly proportional

b) independent

c) inversely proportional

d) depends on the null indicator

 

5. Accuracy of bridge circuit depends on _________

a) component values

b) null detector

c) voltage source

d) current source

 

6.The bridge circuit can be used in _________

a) high voltage circuits

b) low power circuits

c) control circuits

d) digital integrated circuits

 

7. What are the physical parameters that are to be controlled when a bridge is used in control applications?

a) area and volume

b) mass and weight

c) pressure and temperature

d) current and voltage

 

8.Induction type instruments are used for ____________

a) A.C. measurements

b) D.C. measurements

c) Resistance measurements

d) Voltage measurements

 

9.What is the effect of eddy currents in the aluminum disc?

a) varies by a factor of twice the disc length

b) independent of the disc speed

c) varies by a factor of four times the disc size

d) proportional to the disc speed

 

10.Resistance of pressure coil in a low power factor dynamometer type wattmeter is

a) once time

b) three times

c) hundred times

d) ten times

 

Solution

 

Sol.1 Ans (a)

The amount of current safely passing through the coil of the instrument and the spiral springs. This acts as the leads of the current to the instrument. As a result, the range of an electrical instrument depends on the current.

 

Sol.2 Ans (c)

A shunt is normally a very low value of resistance, connected in parallel with the ammeter coil. In a shunt, the potential terminals have a low current carrying capacity. As a result, a low range ammeter is used to measure the large current.

 

Sol.3 Ans (c)

A multiplier can be used for A.C. voltmeters. The condition to be satisfied is that the total impedance of the voltmeter and the multiplier circuit must be constant for a wide range of frequencies.

 

Sol.4 Ans (c)

The input voltage does not appear in the expression for the balance equation. Thus, balance equation is independent of the magnitude of input voltage.

 

Sol.5 Ans (a)

The accuracy of measurement of a bridge circuit depends on the values of the components used in it. Voltage source supplies dc bias to the circuit while the detector is used for balance condition.

 

Sol.6 Ans (c)

The bridge circuit is generally used in control applications. Control systems make use of bridge circuits for industrial applications.

 

Sol.7 Ans (c)

In control applications, when one of the arms of the bridge circuit consists of a resistance element, sensitive physical parameters such as pressure and temperature are to be controlled.

 

Sol.8 Ans (a)

A.C. measurements are made using Induction type instruments. Induction type energy meter is used to measure the energy that is consumed in A.C. circuits only.

 

Sol.9 Ans (d)

The eddy currents induced in an aluminum disc vary in proportion to the speed of the disc. As a result, the braking torque exerted on the disc varies in proportion to the speed.

 

Sol.10 Ans (d)

In a low power factor dynamometer type wattmeter, pressure coil has a resistance value that is one tenth of the actual with respect to unity power wattmeter. This is done in order to ensure a reasonable amount of torque at low power factors.

 

Quiz: Electrical Engineering 29 June 2020

Quiz: Electrical Engineering
Exam: UPSSSC JE
Topic: Miscellaneous

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

Q1. In an induction motor, the torque T is related to the supply voltage V as
(a) T α V
(b) T α V²
(c) T α√V
(d) T α 1/V

 

Q2. As load angle is increased, degree of stability __________
a) reduces
b) increases
c) remains same
d) no change in stability

 

Q3. Synchronizing power is
a) transient in nature
b) steady state in nature
c) pulsating at small frequencies
d) constant for a machine

 

Q4. Which of the following is/are used in synchronous machines to maintain mechanical stability?
a) Damper winding
b) Interpole winding
c) Compensating winding
d) Equalizer rings

 

Q5. For a 4-pole 3 phase, 400V alternator has synchronizing power of 300 units. Then the synchronizing power per mechanical degree is
a) 300
b) 1200
c) 900
d) 600

 

Q6. The operation of a synchronous motor operating on constant excitation across infinite bus will not be stable if power angle δ
a) is less than θ
b) exceeds internal angle θ
c) is more than θ/2
d) is less than θ/2

 

Q7. Variation in the dc excitation of a synchronous motor causes variation in
a) speed
b) power factor
c) armature current
d) armature current and power factor

 

Q8. Which motor can conveniently operate at lagging as well as leading power factor?
a) Squirrel cage induction motor
b) Slip ring induction motor
c) Synchronous motor
d) Stepper motor

 

Q9. In a rotating electrical machine with 2 poles on the stator and 4 poles on the rotor, spaced equally, the net electromagnetic torque developed is ____________
a) maximum
b) zero or no torque is developed
c) minimum
d) none of the mentioned

 

Q10. The basic torque and EMF expression of rotating electrical machines are ____________
a) applicable to DC machines only
b) applicable to AC machines only
c) applicable to both AC and DC machines
d) none of the mentioned

 

 

Solution
Sol 1. (b)
Relation between torque and supply voltage of an induction motor
T α V²

 

Sol 2. (a)
Synchronizing power = EV×cosδ/X.

 

Sol 3. (a)
It is transient in nature and determines the stability while disturbances.

 

Sol 4. (a)
Damper windings are the dummy windings which help to stabilize the machine during transient instability.

 

Sol 5. (a)

P(sy,mech) = Poles* P(sy,ele)/2
P(sy,ele)=300 units.

Sol 6. (b)
Power angle should not exceed θ, to operate under normal conditions.

Sol 7. (d)
Armature current and power factor both will vary by varying the dc excitation.

Sol 8. (c)
It’s the synchronous motor which can favorably operate both at leading and lagging power factor.

Sol 9. (b)
In a machine with 2 stator poles and 4 rotor poles spaced equally, the force of attraction will be cancelled due to the force of repulsion, and hence the net electromagnetic torque will be zero.

Sol 10. (c)
Basic torque and EMF expressions are same for both AC and DC machines as the fundamental principles underlying the operation of AC and DC machines are same.

Quiz: Electrical Engineering 27 June 2020

Quiz: Electrical Engineering
Exam: NLC
Topic: Control system

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

1. Effect of feedback on the plant is to
a) Control system transient response
b) Reduce the sensitivity to plant parameter variations
c) Both (a) and (b)
d) None of these

 

2. Transfer function of a system is defined as the ratio of output to input in
a) Z-transformer
b) Fourier transform
c) Laplace transform
d) All of these

3. Electrical resistance is analogous to
a) Inertia
b) Dampers
c) Spring
d) Fluid capacity

 

4. Relation between Fourier integral and Laplace transformer is through
a) Time domain
b) Frequency domain
c) Both (a) and (b)
d) None of these

 

5. At resonance peak, ratio of output to input is
a) Zero
b) Lowest
c) Highest
d) None of these

6. If gain of the system is zero, then the roots
a) Coincide with the poles
b) Move away from the zeros
c) Move away from the poles
d) None of these

 

7. Settling time is inversely proportional to product of the damping ratio and
a) Time constant
b) Maximum overshoot
c) Peak time
d) Undamped natural frequency of the roots

 

8. If gain of the critically damped system is increased, the system will behave as
a) Under damped
b) Over damped
c) Critically damped
d) Oscillatory

 

9. If gain of the system is increased, then
a) Roots move away from the zeros
b) Roots move towards the origin of the S-plot
c) Roots move away from the poles
d) None of these

 

10. Physical meaning of zero initial condition is that the
a) System is at rest and stores no energy
b) System is at rest but stores energy
c) Reference input to working system is zero
d) System is working but stores no energy

 

Solution

Sol 1. (c)
Both (a) and (b)

Sol 2. (c)
Laplace transform

Sol 3. (b)
Inertia

Sol 4. (c)
Both Time domain and frequency.

Sol 5. (c)
At resonance peak, ratio of output to input is highest.

Sol 6. (a)
Coincide with the poles

Sol 7. (b)
Maximum overshoot

Sol 8. (a)
Under damped

Sol 9. (c)
Roots move away from the poles

Sol 10. (a)
System is at rest and stores no energy

Quiz: Electrical Engineering 24 June 2020

Quiz: Electrical Engineering
Exam: UPSSSC JE
Topic: Utilization of electrical energy

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

Q1. The ratio of horizontal distance between lamps and the mounting Height of lamp is referred to as
(a) space to height ratio
(b) space to intensity ratio
(c) Height to intensity ratio
(d) Luminosity ratio

Q2. Power factor of fluorescent lamp is about
(a) zero
(b) 0.5 lead
(c) 0.5 lag
(d) unity

Q3. Which of the following lamp gives nearly monochromatic light?
(a) Sodium Vapour Lamp
(b) GLS lamp
(c) Tube Light
(d) Mercury Vapour lamp

Q4. The function of inert gas in filament lamp is
(a) increase the illumination
(b) decrease the power consumption
(c) minimize the effect of vaporization during service
(d) decrease the Clare

Q5. To prevent the excessive brightness, which type of lighting scheme is used?
(a) direct
(b) indirect
(c) general
(d) local

Q6. What is the function of choke in a fluorescent lamp?
(a) starting purpose
(b) produce high voltage at starting
(c) improve power factor
(d) minimize the power consumption

Q7. Carbon arc lamp are commonly used in
(a) Domestic lighting
(b) Street lightning
(c) Cinema projectors
(d) Photography

Q8. Arc can be produced by?
(a) AC current only
(b) DC current only
(c) Both AC & DC current
(d) All options are incorrect

Q9. The illumination at a point 5 meter below a lamp is 6 lux. The candle power of the lamp is
(a) 160
(b) 180
(c) 150
(d) 260

Q10. Luminous intensity is expressed as
(a) radians
(b) candela
(c) lumens
(d) none of the above

Solutions
S1. Ans. (a)
Sol. space to height ratio

S2. Ans. (c)
Sol. 0.5 Lag

S3. Ans. (a)
Sol. Sodium Vapour Lamp

S4. Ans. (c)
Sol. minimize the effect of vaporization during service

S5. Ans. (b)
Sol. Indirect lighting scheme

S6. Ans. (b)
Sol. produce high voltage at starting

S7. Ans. (c)
Sol. Cinema projectors

S8. Ans. (c)
Sol. Both AC & DC current

S9. Ans. (c)
Sol. Candela power =? Hight = 5-meter, Illumination = 6 lux
Illumination = CP/h^2 ,
CP = 6×5^2 = 150 W

S10. Ans. (b)
Sol. Luminous Intensity (I)=(Luminous Flux)/(Solid angle)=Candela

Quiz: Electrical Engineering 22 June 2020

Quiz: Electrical Engineering
Exam: UPSSSC JE
Topic: Miscellaneous

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

Q1. The advantage of putting tapping at the phase ends of a Transformer is
(a) to obtain fine variation of voltage
(b) to operate with ease
(c) to reduce the no of bushing
(d) to obtain better regulation

Q2. The field of self-excited generator is excited by
(a) AC
(b) DC
(c) by its own current
(d) Either AC or DC

Q3. The main use of DC shunt motor for driving the
(a) train
(b) crane
(c) hoist
(d) machine tools

Q4. In DC shunt motors, as load is reduced then the speed will be
(a) increase abruptly
(b) increase in proportion to reduction in load
(c) remain almost constant
(d) reduce

Q5. The variable resister shunting the field of DC series motor is called
(a) Armature diverter
(b) voltage regulator
(c) potential diverter
(d) field diverter

Q6. A DC series motor is accidently connected to single phase ac supply. The torque produce will be
(a) of zero average value
(b) Oscillating
(c) steady and unidirectional
(d) pulsating and unidirectional

Q7. In case of DC series motor, it is possible to have finite no-load speed if a resistance is connected across its-
(a) Field terminals
(b) Armature terminals
(c) Field and armature together
(d) It is not possible

Q8. The current drawn by a 120 V DC motor of armature resistance 0.8 Ω and back emf 110 V is
(a) 12.5 A
(b) 24.5 A
(c) 22.4 A
(d) 10.5 A

Q9. If the variable losses in a DC motor is 500 W. then for maximum efficiency of the motor, the constant losses must be
(a) 250 W
(b) 450 W
(c) 650 W
(d) 500 W

Q10. Ward-Leonard system of speed control is NOT recommended for:
(a) frequent motor reversals
(b) very low speeds
(c) constant speed operation
(d) wide speed range

Solutions
S1. Ans. (c)
Sol. to reduce the no of bushing

S2. Ans. (c)
Sol. The field of self-excited generator is excited by its own current.

S3. Ans. (d)
Sol. The main use of DC shunt motor is driving machine tools.

S4. Ans. (c)
Sol. In DC shunt motors, as load is reduced then the speed will be remaining almost constant.

S5. Ans. (d)
Sol. Field diverter

S6. Ans. (d)
Sol. pulsating and unidirectional

S7. Ans. (b)
Sol. Armature terminals

S8. Ans. (a)
Sol. We have relation in case of DC motor as V=E_b+I_a R_a
I_a=(V-E_b)/R_a = (120-110)/0.8=12.5 A
S9. Ans. (d)
Sol. According to the condition of DC motor maximum efficiency, Variable losses = Constant losses

S10. Ans. (c)
Sol. constant speed operation

Quiz: Electrical Engineering 17 June 2020

Quiz: Electrical Engineering
Exam: UPSSSC JE
Topic: AC fundamentals

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

Q1. A Tank circuit consist of
(a) an inductor and a capacitor connected in series
(b) an inductor and a capacitor connected in parallel
(c) a pure inductor and a pure capacitor connected in series
(d) a pure inductor and a pure capacitor connected in parallel

Q2. In a series R-L-C circuit, “Q-factor” is given by
(a) Q= 1/R √(L/C)
(b) Q= R√(L/C)
(c) Q= 1/R √(C/L)
(d) Q= R√(C/L)

Q3. A non-sinusoidal periodic waveform is free from DC component, cosine components and even harmonics. The waveform has
(a) half wave and odd function symmetry
(b) half wave and even function symmetry
(c) only odd function symmetry
(d) only half wave symmetry

Q4. The main reason for using high voltage for long distance power transmission is
(a) reduction in transmission losses
(b) reduction in time of transmission
(c) increase in system reliability
(d) none of these

Q5. Capacitors used for improvement of power factor of a system
(a) Draws lagging power and supply leading power
(b) Draws lagging power and supply lagging power
(c) Draws leading power and supply lagging power
(d) Draws leading power and supply leading power

Q6. Materials having high dielectric constants, which is non-linear is called
(a) hard dielectrics
(b) super dielectrics
(c) Ferroelectric material
(d) paramagnetic material

Q7. For a sinusoidal waveform, the ratio of average value to rms value is
(a) π/(2√2)
(b) √2π/2
(c) 2/√2π
(d) (2√2)/π

Q8. The current in the circuit follows the relation i=200 sin⁡ωt. If frequency is 50 Hz, how long will it take for the current to rise to 100A?
(a) 2.89 ms
(b) 1.66 ms
(c) 6.84 ms
(d) 1.96 ms

Q9. The power of single-phase AC circuit is given by
(a) VI
(b) VI cos Ǿ
(c) VI sin Ǿ
(d) None of these

Q10. Which of the following distribution system is used for combined power and lighting load?
(a) single phase 2-wire AC system
(b) Three phase 3-wire AC system
(c) Three phase 4-wire AC system
(d) None of these

Solutions
S1. Ans. (d)
Sol. a pure inductor and a pure capacitor connected in parallel

S2. Ans. (a)
Sol. Q= 1/R √(L/C)

S3. Ans. (a)
Sol. half wave and odd function symmetry

S4. Ans. (a)
Sol. Reduction in transmission losses

S5. Ans. (c)
Sol. Draws leading power and supply lagging power

S6. Ans. (c)
Sol. Ferroelectric material

S7. Ans. (d)
Sol. (2√2)/π ; For sinusoidal waveform,I_av=I_m/(π/2)=〖2I〗_m/π
I_rms=I_m/√2
So, I_av/I_rms =(2√2)/π

S8. Ans. (b)
Sol. Given, i=200 sin⁡ωt
Time to reach 100 A. Putting i=100 A
100=200 sin⁡ωt
1/2=sin⁡ωt
ωt=π/6
t=1.66 ms

S9. Ans. (b)
Sol. VI cos Ǿ

S10. Ans. (c)
Sol. Three phase 4-wire AC system

Quiz: Electrical Engineering 15 June 2020

Quiz: Electrical Engineering
Exam: UPSSSC JE
Topic: Miscellaneous

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

Q1. 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

Q2. Total instantaneous power supplied by a 3-phase ac supply to a balanced R-C load as
(a) zero
(b) constant
(c) pulsating with zero average
(d) pulsating with non-zero average

Q3. The phase voltage of a three-phase, star-connected alternator is V. By mistake the connection of R phase got reversed. The new line voltages will have a relationship
(a) V_RY=V_BR= V_YB/(√3)
(b) V_RY=V_YB= V_BR/(√3)
(c) V_YB=V_BR= V_RY/(√3)
(d) V_RY=V_YB= V_BR

Q4. A three-phase star-connected load is operating at a power factor angle ∅, with ∅ being the angle between
(a) line voltage and line current
(b) phase voltage and phase current
(c) line voltage and phase current
(d) phase voltage and line current

Q5. Which of the following is a polar dielectric?
(a) Teflon
(b) polyethylene
(c) Nylon
(d) Quartz

Q6. Orbital magnetic moment of an electron, in an atom, is of the order of
(a) 0.1 Bohr magneton
(b) 1.0 Bohr magneton
(c) 10 Bohr magnetons
(d) 100 Bohr magnetons

Q7. The magnetic material exhibits the property of magnetization due to
(a) spin of nucleus
(b) spin of electrons
(c) orbital motion of electrons
(d) all of the above

Q8. A dc-series generator is provided with divertor and is delivering its rated current. If the divertor switch is opened, the terminal voltage will
(a) remain constant
(b) decrease
(c) increase
(d) none of these

Q9. In a dc shunt generator, the voltage builds up is generally restricted due to
(a) armature heating
(b) insulation restriction
(c) saturation of iron
(d) speed limitation

Q10. Which of the following dc generator has rising V-I characteristic?
(a) series
(b) shunt
(c) compound
(d) None

Solutions
S1. Ans. (c)
Sol. rms line-to-line voltage

S2. Ans. (b)
Sol. Constant

S3. Ans. (a)
Sol. V_RY=V_BR= V_YB/(√3)

S4. Ans. (b)
Sol. phase voltage and phase current

S5. Ans. (c)
Sol. Nylon

S6. Ans. (b)
Sol. 1.0 Bohr magneton

S7. Ans. (d)
Sol. The magnetic material exhibits the property of magnetization due to
spin of nucleus
spin of electrons
orbital motion of electrons

S8. Ans. (c)
Sol. Increase

S9. Ans. (c)
Sol. Saturation of iron

S10. Ans. (a)
Sol. Series

Quiz: Electrical Engineering 10 June 2020

Quiz: Electrical Engineering
Exam: UPSSSC JE
Topic: Miscellaneous

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

Q1. A 3-phase 4- wire system is commonly used for
(a) primary distribution
(b) secondary distribution
(c) primary transmission
(d) secondary transmission

Q2. Which of the following is usually not the generating voltage?
(a) 6.6 kV
(b) 9.9 kV
(c) 11 kV
(d) 13.2 kV

Q3. The main drawbacks of overhead transmission system over underground system is/are
(a) underground system is more flexible than overhead system
(b) higher charging current
(c) surge problem
(d) high initial cost

Q4. Improving pf
(a) reduces current for a given output
(b) increases losses in line
(c) increase the cost of station equipment
(d) none of the above

Q5. DC motors are still preferred for use in
(a) electric excavators
(b) lathes and machine tools
(c) floor mills and jaw crusher
(d) paper industry

Q6. The least significant feature while selecting a motor for centrifugal pump is
(a) speed control
(b) power rating of motor
(c) operating speed
(d) starting characteristic

Q7. BCD code is
(a) a binary code
(b) unweighted code
(c) the same thing as binary number
(d) the same as gray code

Q8. A unit impulse response of a second order system is 1/6 e^(-0.8t) sin (0.6t). Then natural frequency and damping ratio of the system are
(a) 1 and 0.6
(b) 1 and 0.8
(c) 2 and 0.4
(d) 2 and 0.3

Q9. Damping ratio ξ and peak overshoot M_p are measure of
(a) relative stability
(b) absolute stability
(c) speed of response
(d) steady state error

Q10. For a critically damped system, the closed-loop poles are
(a) purely imaginary
(b) real, equal and negative
(c) complex conjugate with negative real part
(d) real unequal and negative

Solutions
S1. Ans. (b)
Sol. A 3-phase 4- wire system is commonly used for secondary distribution.

S2. Ans. (b)
Sol. 9.9 kV is usually not the generating voltage.

S3. Ans. (c)
Sol. Surge problem is only advantage of overhead transmission system.

S4. Ans. (a)
Sol. Since Power, P = VI cos φ , cos φ (power factor)
If cos φ will improve for a given output the value of current will decrease.
S5. Ans. (a)
Sol. electric excavators

S6. Ans. (a)
Sol. speed control

S7. Ans. (a)
Sol. BCD code is a binary code

S8. Ans. (b)
Sol. h (t) = 1/6 e^(-0.8t) sin (0.6t)
On comparing it with the standard equation,
i.e. Ke^(-ξω_n t)/√(1-ξ^2 ) sin⁡〖(ω_d t+ Ǿ)〗
we have, ξω_n=0.8 …………………………. (i)
ω_d= ω_n √(1-ξ^2 )
= 0.6 rad/sec ……………………. (ii)

Solving both equation
ω_n=1 rad/sec and 𝞷 = 0.8

S9. Ans. (c)
Sol. speed of response

S10. Ans. (b)
Sol. real, equal and negative

Quiz: Electrical Engineering 8 June 2020

Quiz: Electrical Engineering

Exam: UPSSSC JE

Topic: DC Generator

 

Each question carries 1 mark.

Negative marking: 1/4 mark

Time: 10 Minute

 

Q1. The voltmeter connected across a generator reads voltage same at no load and at full load (rated). The generator is of type

(a) shunt generator

(b) series generator

(c) Level compound

(d) Short-shunt compound

 

 

Q2. The residual magnetism of a dc shunt generator can be regained by

(a) connecting the shunt field to a battery

(b) running the generator on no load for some time

(c) ground the shunt field

(d) reversing the direction of rotation of the generator

 

 

Q3. Flashing of field of dc generator means

(a) neutralization of residual magnetism

(b) creation of residual magnetism by a dc source

(c) increasing flux density by providing extra ampere-turns in field

(d) none of the above

 

 

Q4. A dc series generator is employed

(a) as a booster to maintain constant voltage at the load end of the feeder.

(b)for supplying traction load

(c) for supplying industrial load at constant voltage

(d) for battery charging

 

 

Q5. The type of dc generator used for arc welding purpose is a

(a) series generator

(b) shunt generator

(c) cumulatively compound generator

(d) differentially compound generator

 

 

Q6. When two dc series generator are operating in parallel, an equalizer bar is used

(a) to reduce armature reaction

(b) to increase emf

(c) to increase the speed

(d) so that the two similar machines take approximately equal load current

 

 

Q7. The simplest way of shifting load from one shunt generator to the other operating in parallel is by

(a) adjustment of speed

(b) adjustment of armature resistance

(c) adjustment of field rheostats

(d) using equalizer connections

 

 

Q8. If field of one of two generators operating in parallel is made very weak, then it will

(a) not take any load

(b) take major share of load

(c) operate as a motor and run in the same direction

(d) operate as a motor and run in the opposite direction

 

 

 

Q9. For parallel operation, the dc generators generally preferred are

(a) shunt

(b) series

(c) under compound

(d) Both (a) and (c)

 

 

Q10. The output of any electrical motor is taken from the

(a) armature

(b) field

(c) coupling mounted on shaft

(d) motor frame

 

Solutions

S1. Ans. (c)

Sol. Level compound

 

S2. Ans. (a)

Sol. connecting the shunt field to a battery

 

S3. Ans. (b)

Sol. creation of residual magnetism by a dc source

 

S4. Ans. (a)

Sol. as a booster to maintain constant voltage at the load end of the feeder

 

S5. Ans. (d)

Sol. differentially compound generator

 

S6. Ans. (d)

Sol. so that the two similar machines take approximately equal load current

 

S7. Ans. (c)

Sol. adjustment of field rheostats

 

S8. Ans. (c)

Sol. operate as a motor and run in the same direction

 

S9. Ans. (d)

Sol. Both shunt and under compound

 

S10. Ans. (c)

Sol. coupling mounted on shaft