Quiz: Electrical Engineering 22 Oct 2020

Quiz: Electrical Engineering
Exam: SSC JE
Topic: Miscellaneous

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

Q1. Owing to skin effect
(a) current flows through the half cross-section of the conductor
(b) portion of the conductor near the surface carries more current and core of the
conductor carries less current
(c) portion of the conductor near the surface carries less current and core of the
conductor carries more current
(d) any of the above

Q2. The speed regulation of a synchronous motor is always
(a) positive
(b) zero
(c) negative
(d) 100%

Q3. An induction motor and a synchronous motor are connected to a common feeder line;
To operate the feeder line at unity p.f., the synchronous motor should be
(a) normally excited
(b) under-excited
(c) over-excited
(d) none of the above

Q4. A synchronous condenser (or capacitor) means
(a) synchronous motor operating at no-load
(b) over-excited synchronous motor with capacitor
(c) under-excited synchronous motor
(d) none of the above

Q5. A synchronous capacitor is used to
(a) improve p.f. of the system
(b) reduce losses in the system
(c) improve voltage regulation
(d) achieve all of above

Q6. n similar resistors each of resistance r when connected in parallel have the total
resistance R. When these resistances are connected in series, the total resistance is
(a) nR
(b) R/n²
(c) n²R
(d) R/n

Q7. Fig. shows part of a closed circuit. What is the value of VA–VB?

(a) 12 V
(b) 9 V
(c) 18 V
(d) 6 V

Q8. A wire of resistance 0.1 Ω /cm is bent to form a square ABCD of side 10 cm. A similar
wire is connected between B and D to form the diagonal BD. If a 2 V battery of negligible
internal resistance is connected between A and C, then total power dissipated is
(a) 2 W
(b) 3 W
(c) 4 W
(d) 6 W

Q9. In a power station,
(a) Reserve capacity = Plant capacity + Max. demand
(b) Reserve capacity = Plant capacity – Average demand
(c) Reserve capacity = Plant capacity – Max. demand
(d) Reserve capacity = Plant capacity + Average demand

Q10. In a 3-phase system, the line losses are
(a) directly proportional to cos ϕ
(b) inversely proportional to cos ϕ
(c) inversely proportional to cos² ϕ
(d) none of the above

SOLUTIONS
S1. Ans.(b)
Sol. The distribution of current over the cross section of the conductor is uniform in DC
only. In alternating current flow through a conductor does not distribute uniformly but it
will concentrate near the surface of the conductor. In fact, in AC system no current flow
through the core because the whole current will flow through the surface region. Due to
this reason the effective area of conductor is reduced causing an increase in AC resistance
. The effective AC resistance is usually referred as the effective resistance of the conductor
.This phenomenon is called the Skin effect as it will cause concentration of current at the
skin of conductor.

S2. Ans.(b)
Sol. The speed of a synchronous motor is constant (= synchronous speed, N_s) from no
load to full-load. Therefore, speed regulation of a synchronous motor is always zero

S3. Ans.(c)
Sol. An over-excited synchronous motor (i.e.E_b>V) takes power with a leading power
factor and behaves like a capacitor,

S4. Ans.(a)
Sol. A synchronous motor operating at no-load is called a synchronous capacitor or
condenser. It is specially designed for power factor control and has no external shaft.

S5. Ans.(d)
Sol.

S6. Ans.(c)
Sol. In parallel,
R=r/n→r=nR
In series,
R_eq=nr=n²R

S7. Ans.(b)
Sol.

V_AB=3+3×2=9V

S8. Ans.(c)
Sol.

This is a wheat stone bridge in balanced form.
∴ R_eq between A and C
R_eq=(2‖2)=1Ω
∴ Total power =V^2/R= 4/1=4 W

S9. Ans.(c)
Sol. Reserve capacity = Plant capacity – Max. demand

S10. Ans.(c)
Sol. Line current, IL=P/(√3 〖 V〗L cos⁡ϕ )
Since line losses are proportional to the square of I_L, it follows that line losses are proportional to 1/cos² ϕ

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 15 Oct 2020

Quiz: Electrical Engineering
Exam: SSC-JE
Topic: Measurement and measuring instrument

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

Q1. Watt-hour meter is classified as a
(a) Recording instrument
(b) Deflecting instrument
(c) Integrating instrument
(d) Indicating instrument

Q2. A PMMC meter rated at 200 µA is used in a rectifier type instrument which uses full wave rectification. What is the sensitivity on sinusoidal AC?
(a) 9 KΩ/V
(b) 4.5 KΩ/V
(c) 18 KΩ/V
(d) 10 KΩ/V

Q3. The bridge suitable for the measurement of an unknown inductance in terms of known capacitance would include
(a) Maxwell and Schering
(b) Maxwell and Hay
(c) Hay and Schering
(d) Maxwell, Hay and Schering

Q4. If an energy meter disc makes 40 revolutions in 50 sec when a load of 5 KW is connected to it, the meter constant (in rev/kw-h) is:
(a) 500
(b) 1000
(c) 480
(d) 576

Q5. The current and potential coils of a dynamometer type wattmeter were accidentally interchanged while connecting. After energizing the circuit, it was observed that the wattmeter did not show the reading. This could be due to the
(a) Damage to current coil
(b) Damage to potential coil
(c) Loose contacts
(d) Damage to both the potential and current coil

Q6. The voltage coil of a single-phase house service energy meter
(a) Is highly capacitive
(b) Is highly resistive
(c) Is highly inductive
(d) None of the above

Q7. Consider the following types of damping:
1. Air-friction damping
2. Fluid-friction damping
3. Eddy current damping
PMMC types instruments use which of the above?
(a) 1 only
(b) 2 only
(c) 3 only
(d) 1, 2 and 3

Q8. The correct statement about Electrodynamometer instrument amongst the following is:
(a) Its scale in linear
(b) It measures only DC
(c) It is a transfer instrument
(d) Its sensitivity is less than M.I. type instruments

Q9. The type of instruments used mainly for standardizing instruments in laboratories is
(a) Indicating instrument
(b) Integrating instrument
(c) Absolute instrument
(d) Recording instrument

Q10. A 0-200 V voltmeter has an accuracy of 0.75% of full-scale reading. If voltage measured is 100 V, the error is:
(a) 3%
(b) 2%
(c) 1.5%
(d) .75

SOLUTIONS
S1. Ans.(c)
Sol.
Indicating Instrument – The instrument which indicates the magnitude of the measured quantity is known as the indicating instrument.
e.g. ammeter, voltmeter
Integrating Instrument – The instrument which measures the total energy supplied at a particular interval of time is known as the integrating instrument. The total energy measured by the instrument is the product of the time and the measures electrical quantities.
e.g. energy meter, watt-hour meter
Recording Instrument – The instrument records the circuit condition at a particular interval of time is known as the recording instrument.
Deflecting instrument: The instrument in which the deflection provides the basis for measuring the electrical quantity is known as the deflection type instrument.
E.g. PMMC type ammeter

S2. Ans.(b)
Sol. sensitivity is reciprocal of full-scale deflection (I_FSD).
∴S_DC=1/I_FSD =1/(200×10^(-6) )=5 KΩ/V
For full-wave rectification, AC sensitivity=0.9×DC sensitivity
∴S_AC=0.9×5=4.5 KΩ/V

S3. Ans.(b)
Sol. maxwell and hay’s bridge are used to find the value of self-inductance.
Schering bridge is used to measure the capacitance.

S4. Ans.(d)
Sol. No. of revolutions in one hour=(40×3600)/50=2880
∴meter constant(k)=(no of revolutions)/(kili-watt hours)=2880/5=576 rev/kw-h

S5. Ans.(a)
Sol. current coil is connected in series and carries the load current. If both coils are interchanged then current coils gets directly connected across the supply and large current flows through the coil resulting into damage to current coil.

S6. Ans.(c)
Sol. The voltage coil of a single-phase house service energy meter is highly inductive.

S7. Ans.(c)
Sol. PMMC types instruments use eddy current damping.

S8. Ans.(c)
Sol. Transfer instrument: – Any instrument which is calibrated with DC source and used without any modification for AC measurements. Such a transfer instrument having same accuracy for both DC and AC.
E.g.: – Dynamometer.
(Td ⍺ I^2 orV^2 )
So, for dynamometer instruments: –
I. Scale is non-uniform
II. It is a transfer instrument
III. Coils are air-cored

S9. Ans.(c)
Sol. Standardizing instruments: – use for scrutinizing the accuracy of a measuring instrument by comparing it with a standard instrument.

S10. Ans.(c)
Sol. ▭( limiting error=(%Accuracy×FSD)/(True value (AT) ))
=(0.75×200)/100
=1.5%

Quiz: Electrical Engineering 12 Oct 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. According to inverse square law illumination, what is the relation between illumination(E) and the distance between the light source and surface(d)?
(a) E ⍺ d
(b) E ⍺1/d
(c) E ⍺ d^2
(d) E ⍺1/d^2

Q2. According to lambert’s law, the illumination on a surface is proportional to
(a) cos⁡Ө
(b) cos^2⁡Ө
(c) 1/cos^2⁡Ө
(d) 1/cos^3⁡Ө

Q3. Direct lighting method of illumination is used in
(a) Houses
(b) Industries
(c) Outdoor lightning
(d) All of the above

Q4. In the indirect method of lighting about how much percentage of light falls on the surface?
(a) 10 %
(b) 50 %
(c) 75 %
(d) 90 %

Q5. Lumen/ watt is the unit of
(a) Light flux
(b) Luminous intensity
(c) Brightness
(d) Luminous efficiency

Q6. Luminous flux is
(a) Rate of energy radiation in the form of light waves
(b) Light energy radiated by sun
(c) Part of light energy radiated by sun which is received on earth
(d) None of the above

Q7. In dielectric heating current flows through
(a) Metallic conductor
(b) Air
(c) Dielectric medium
(d) Ionic discharge between dielectric medium and metallic conductor

Q8. In dielectric heating, if the capacitance is loss free, the heat produced will be
(a) Proportional to the value of capacitance
(b) Proportional to the frequency
(c) Infinity
(d) Zero

Q9. Plant capacity factor of a power plant may be calculated by the formula:
(a) (Average demand)/(Plant capacity)
(b) (Plant capacity)/(Average demand)
(c) (Sum of individual max. ⁡demand)/(Max demand of plant)
(d) (Station output)/(Plant Hours of use)

Q10. Demand factor of a power plant is:
(a) (maximum demand)/(connected load)
(b) (average demand)/(plant capacity)
(c) (station energy output)/(plant capacity×hrs of use)
(d) (average demand)/(maximum demand)

SOLUTIONS
S1. Ans.(d)
Sol. laws of illumination:
Illumination is directly proportional to the luminous intensity of the source.
Inverse square law – The illumination of a surface receiving its flux from a point source is inversely proportional to the square of the distance between the surface and the source.
i.e. E=I/d^2
where, E=illumination
I= intensity
d =distance

S2. Ans.(a)
Sol. laws of illumination:
Illumination is directly proportional to the luminous intensity of the source.
Inverse square law – The illumination of a surface receiving its flux from a point source is inversely proportional to the square of the distance between the surface and the source.
Lambert’s cosine law – The illumination of a surface at any point is proportional to the cosine of the angle between the normal at the point and the direction of the luminous flux.
i.e. E ⍺1/(d^2 )⍺1/d^2 cos⁡Ө
Illumination(E)=I/d^2 cos⁡Ө
where, E=illumination
I= intensity
d =distance

S3. Ans.(d)
Sol. In direct lighting method of lighting, an opaque reflector directs 90 % of the light vertically downwards. Only 10 % of light will be absorbed in this system by the reflector. It is uses in houses, industries and general outdoor lighting.

S4. Ans.(a)
Sol. indirect lighting method of lighting, 90 % of the light is reflected upwards and only 10 % falls on the surface. They are used for mainly decoration purposes in cinemas, hotels, club etc.

S5. Ans.(d)
Sol. Luminous efficiency is the ratio of the total luminous flux radiated by any source to the total radiant flux from the source commonly expressed in lumens per watt.

S6. Ans.(a)
Sol. Luminous flux is the measure of brightness of a light source in terms of energy being emitted. Luminous flux, in SI units, is measured in the lumen (lm). It is a measurement of energy released in the form of visible light from a light-producing source.

S7. Ans.(c)
Sol. In dielectric heating current flows through dielectric materials. The dielectric acts as capacitance.

S8. Ans.(d)
Sol. Dielectric loss(P)=2πfCV^2 cos⁡Ө
For loss free capacitor, Ө=90⁰⇒P=0
Hence, heat produced will be zero.

S9. Ans.(a)
Sol. (plant capacity factor=(peak load)/(plant capacity)×load factor)
=(Average demand)/(Plant capacity)

NOTE: If peak load equals to plant capacity then load factor = capacity factor.

S10. Ans.(a)
Sol. Demand factor = (Maximum demand)/(connected load)
Demand factor is less than 1.

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 05 Oct 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 is the dimensional formula of conductance?
(a) M^1 L^2 T^(-3) A^(-1)
(b) M^1 L^2 T^(-3) A^(-2)
(c) M^(-1) L^(-2) T^3 A^2
(d) M^1 L^2 T^(-3) A^1

Q2. Which one of the following statements is TRUE about the resistance of a conductor?
(a) the resistance of a conductor is inversely proportional to the length of the conductor.
(b) the resistance of a conductor is directly proportional to the area of the conductor,
(c) the resistance of a conductor is inversely proportional to the pressure applied on the conductor,
(d) the resistance of a conductor is inversely proportional to the area of the conductor.

Q3. In parallel combination of resistance, the voltage is _________.
(a) lower across largest resistance
(b) higher across largest resistance
(c) same across each resistance
(d) higher across smaller resistance

Q4. Electrical conductivity of a conductor is measure in __________.
(a) Siemens
(b) Ohms
(c) Siemens/meter
(d) Ohms/meter

Q5. What will be the equivalent capacitance of a parallel combination of four capacitors having equal value of capacitance ‘C’?
(a) C/4
(b) 4C
(c) C/2
(d) 2C

Q6. Determine the Thevenin’s equivalent resistance (in ohms) across the terminal a and b for the electrical circuit given below.

(a) 1
(b) 0.5
(c) 0.3
(d) 0.2

Q7. The no-load ratio of a 50 Hz single phase transformer is 6000/250 V. The maximum flux in the core is 0.06 Wb. What is the number of primary turns?
(a) 450
(b) 900
(c) 350
(d) 210

Q8. A 3-phase induction motor is ………
(a) essentially a constant-speed motor
(b) a variable speed motors
(c) very costly
(d) not easily maintainable

Q9. If a 3-phase induction motor is running at slip s (in decimal), then, rotor copper loss is equal to …….
(a) (1–s)× rotor input
(b) (1+s) × rotor input
(c) s× rotor input
(d) none of the above

Q10. In a squirrel cage motor, the number of stator slots is ……. Rotor slots.
(a) always equal to the number of
(b) always greater than the number of
(c) always less than the number of
(d) either more or less than the number of

SOLUTIONS
S1. Ans.(c)
Sol. conductance = [1/((Resistance) )]=1/R
R = (V/I)
& V = (W/Q) = (W / It) …. [as V = (work / charge) = {(work) / (current × time)}]
∴ V = ([M^1 L^2 T^(–2)] / [A^1 T]) = [M^1 L^2 T^(–3) A^(–1)]
Hence R = [M^1 L^2 T^(–3) A^(–1) ]/[A] =[M^1 L^2 T^(–3) A^(–2)]
∴ conductance = 1/([M^1 L^2 T^(–3) A^(–2)] )==[M^(-1) L^(-2) T^3 A^2]

S2. Ans.(d)
Sol. R=ρ l/A
So, the resistance of a conductor is inversely proportional to the area of the conductor.

S3. Ans.(c)
Sol. In parallel combination of resistance, the voltage is same across each resistance.
i.e. V1=V2=⋯=V

S4. Ans.(c)
Sol. The electric conductivity is the measure the ability of a conductor to conduct electricity.
The conductivity formula is the inverse of the resistivity that is:
σ=1/ρ
Here
σ = refers to the electrical conductivity
ρ = refers to the resistivity
The conductivity unit is Siemens per meter.

S5. Ans.(b)
Sol. for parallel combination: Ceqv=C+C+C+C=4C

S6. Ans.(b)
Sol. The value of the equivalent resistance, Rab is found by calculating the total resistance looking back from the terminals a and b with all the voltage sources shorted.
∴Rab=1 ‖(3 ‖3 ‖3)
∴Rab=1 ‖ 1=1/2=0.5

S7. Ans.(a)
Sol. E1=4.44 f N1 ϕ_m or 6000=4.44×50×N1×0.06
∴N1=450

S8. Ans.(a)
Sol. At no-load, the rotor lags behind the stator flux only a small amount since the only torque required is that needed to overcome small no-load losses. As mechanical load is added, the rotor speed decreases. A decrease in rotor speed allows the constant speed rotating field to sweep across the rotor conductors at a faster rate, thereby inducing larger rotor current (since rotor impedance is low). This results in a large increase in torque which tends to bring the speed to the original value. Although the motor speed does decrease is classed as a constant-speed motor

S9. Ans.(c)
Sol. Rotor copper loss = s× rotor input

S10. Ans.(c)
Sol. The rotor has a larger number of slots than the stator and should be a non-integer multiple of the number of stator slots so as to prevent magnetic interlocking of rotor and stator teeth at the starting instant.
Also, it leads to good starting characteristics, high efficiency and power factor, low stator current, and less vibrations and mechanical oscillations at nominal speed.

Quiz: Electrical Engineering 28 Sep 2020

Quiz: Electrical Engineering
Exam: SSC JE
Topic: Miscellaneous

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

Q1. What will be the potential difference (in v) between the ends of a conductor when the current flowing through the conductor is 3 A and the value of conductance is 0.3 mho?
(a) 10
(b) 100
(c) 20
(d) 0.1

Q2. Determine the value of charge stored (in mC) in a capacitor, when the value of capacitance is 0.01 mF and the potential difference between the ends of the capacitor is 20 V.
(a) 0.2
(b) 2
(c) 20
(d) 200

Q3. What will be the value of load impedance (in ohms) for transmitting maximum power from the source to load when the source impedance is 8 + j4 ohms?
(a) 8 – j4
(b) 8 + j4
(c) 4 + j8
(d) 4 – j8

Q4. Determine the value of current (in A) drawn from the voltage source for the electric circuit given below.

(a) 4
(b) 3.5
(c) 2.5
(d) 1.6

Q5. Which of the following is the dimensional formula for inductance?
(a) ML^2 T^2 A^(-2)
(b) ML²T²A²
(c) ML^2 T^(-2) A^(-2)
(d) ML^(-2) T²A²

Q6. Which of the following is the CORRECT expression for hysteresis loss occurring in a material?
(a) η×Bm^2×f^2×V
(b) η×Bm^2×f^2×V^2.5
(c) η×Bm^1.6×f×V
(d) η×Bm^2×f^1.6×V

Q7. Which of the following expression satisfies the Faraday’s law of electromagnetic induction?
(a) e= -Ndϕ/dt
(b) e=N\/ dϕ/dt
(c) e= -N⎰dϕdt
(d) e=(Nd^2 ϕ)/(dt^2 )

Q8. Which property of a material opposes the passage of magnetic flux through it?
(a) Permeance
(b) Capacitance
(c) Inductance
(d) Reluctance

Q9. Which of the following statement is TRUE?
(a) A galvanometer can be converted into ammeter by connecting a low value of resistance in series with the galvanometer.
(b) a galvanometer can be converted into ammeter by connecting low value of resistance in parallel with the galvanometer.
(c) A galvanometer can be converted into ammeter by connecting a high value of resistance in series with the galvanometer.
(d) A galvanometer can be converted into ammeter by connecting a high value of resistance in parallel with the galvanometer.

Q10. Which of the statement is TRUE about megger?
(a) megger is used for the measurement of voltage.
(b) megger is used for the measurement of current.
(c) Megger is used for the measurement of insulation resistance.
(d) Megger is used for the measurement of breakdown voltage of insulation.

SOLUTIONS
S1. Ans.(a)
Sol. conductance(G)=1/R=0.3 mho
∴R=1/G=1/(0.3 ) Ω
∴potential difference(V)=IR=3×1/0.3=10 V

S2. Ans.(a)
Sol. from the expression, Q=CV
⇒Q=0.01×20=0.2 mC

S3. Ans.(a)
Sol. for maximum power transfer, ZL=ZS^* i.e. conjugate of source impedance
Given that, Z_S=8+j4
∴ZL=ZS^*=8-j4 ohm

S4. Ans.(c)
Sol.

Reqv=(20‖20)+(20‖20)=20 Ω
∴I=V/R=50/20=2.5 A

S5. Ans.(c)
Sol. We have the formula for inductance as,L=V/(dI/dt)

Hence, dimensional formula for inductance is [L]= [[ML^2 T^(-3) A^(-1) ]/[AT^(-1) ] ]=[ML^2 T^(-2) A^(-2)]

S6. Ans.(c)
Sol. The equation for hysteresis loss is given as:
Pb = η * Bmaxn * f * V
Pb = hysteresis loss (W)
η = Steinmetz hysteresis coefficient, depending on material (J/m3)
Bmax = maximum flux density (Wb/m2)
n = Steinmetz exponent, ranges from 1.5 to 2.5, depending on material
f = frequency of magnetic reversals per second (Hz)
V = volume of magnetic material (m3)

S7. Ans.(a)
Sol. expression for the Faraday’s law of electromagnetic induction: e= -Ndϕ/dt

S8. Ans.(d)
Sol. The obstruction offered by a magnetic circuit to the magnetic flux is known as reluctance. As in electric circuit, there is resistance similarly in the magnetic circuit, there is a reluctance, but resistance in an electrical circuit dissipates the electric energy and the reluctance in magnetic circuit stores the magnetic energy.

The differences between reluctance and permeance have been discussed in the table below.

Reluctance Permeance
Reluctance opposes the production of
magnetic flux in a magnetic circuit.
Permeance is a measure of the ease with
which magnetic flux can be set up in the magnetic circuit.
It is denoted by S. It is denoted by P.
Its unit is AT/Wb or 1/Henry or H-1. Its unit is Wb/AT or Henry.
It is analogous to resistance in an electric circuit. It is analogous to conductance in an electric
circuit.
Reluctance adds in a series of the
magnetic circuit.
Permeance adds in a parallel magnetic
circuit.

S9. Ans.(b)
Sol. A good ammeter has very low resistance. So, a moving coil galvanometer is converted into an ammeter by connecting a very low shunt resistance in parallel to the galvanometer resistance.
A good voltmeter has very high resistance. So, a moving coil galvanometer is converted into a voltmeter by connecting a high resistance in series with the galvanometer resistance.

S10. Ans.(c)
Sol. Megger is used for the measurement of insulation resistance.

Quiz: Electrical Engineering 23 Sep 2020

Quiz: Electrical Engineering
Exam: SSC JE
Topic: Miscellaneous

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

Q1. A DC motor starter is
(a) Fixed resistance
(b) Variable resistance
(c) Variable inductance
(d) Variable capacitor

Q2. Which type of winding is generally preferred for generating large currents on DC generators?
(a) Progressive wave winding
(b) Retrogressive wave winding
(c) Lap winding
(d) Current depends on design

Q3. The range of efficiency for shaded pole motors is
(a) 95% to 99%
(b) 80% to 90%
(c) 50% to 75%
(d) 5% to 35%

Q4. Demand factor is
(a) (maximum demand)/(connected load)
(b) (maximum demand)/(average load)
(c) (connected load)/(maximum demand)
(d) (average load)/(maximum demand)

Q5. High tension cables are generally used up to
(a) 11 KV
(b) 33 KV
(c) 66 KV
(d) 132 KV

Q6. What does the bedding on the cable consists of?
(a) Jute strands
(b) Hessian tape
(c) Paper tape compounded with a fibrous material
(d) Any of these

Q7. A 6 pole Lap wound dc generator has 300 conductors. Emf induced per conductor is 5 volts. This generator will generate emf of:
(a) 60 V
(b) 250 V
(c) 300 V
(d) 1800 V

Q8. …………. are integrating instruments?
(a) Ammeters
(b) Voltmeters
(c) Wattmeter
(d) Ampere-hour and Watt-hour meters

Q9. Resistances can be measured with the help of a …………
(a) Wattmeter
(b) voltmeter
(c) ammeter
(d) ohmmeter and resistance bridge

Q10. ………….. instruments indicate the instantaneous value of the electrical quantity being measured at the time at which it is being measured?
(a) Absolute
(b) Indicating
(c) Recording
(d) Integrating

SOLUTIONS
S1. Ans.(b)
Sol. Sol. At the time of starting of motor it is at rest and no back e.m.f. is generated. On application of full voltage, armature winding draws a heavy current due to small armature resistance. This high armature current may damage the armature windings, commutator and brushes. To prevent high armature current during the starting of motors, variable resistance is connected in series with the armature winding. The starting resistance is reduced as the motor speeds up. The resistance is cut off fully when the motor attains full speed. This arrangement is known as starter. For very small D.C. motor (e.g. 6v, 12v, motor) starter is not required and it can be started directly.

S2. Ans.(c)
Sol. LAP winding: large current & low voltage applications
WAVE winding: high voltage & low current applications

S3. Ans.(d)
Sol. The range of efficiency for shaded pole motors is 5% to35%.
Shaded pole motors are used where low torque is acceptable (such as fans) and are usually less than ¼ HP. Due to their very low efficiency, shaded pole motors should only be used in applications where the motor is very small or operates for very short period of time (e.g. shower fan motor).

S4. Ans.(a)
Sol. Demand factor is the ratio of the sum of the maximum demand of a system (or part of a system) to the total connected load on the system (or part of the system) under consideration. Demand factor is always less than one.

S5. Ans.(a)
Sol. The classification of cables on the basis of voltage is more common, according to which they can be divided into the following categories:
1. Low-tension cables — up to 1000 V or 1 KV
2. High-tension cables — the operating voltage of high-tension cables is up to 11000 V or 11 KV
3. Super-tension cables — the operating voltage of super tension cable is from 22 kV to 33 kV
4. Extra high-tension cables — from 33 kV to 66 kV
5. Extra super voltage cables — beyond 132 kV

S6. Ans.(d)
Sol. Bedding:
Over the metallic sheath is applied a layer of bedding which consists of a fibrous material like jute or hessian tape. The purpose of bedding is to protect the metallic sheath against corrosion and from mechanical injury due to armouring.

S7. Ans.(b)
Sol. Emf induced per conductor=5 V
For lap wound, number of parallel paths(A)=P=6
No. of conductor per parallel path=300/6=50
∴total emf generated by generator=50×5=250 V

S8. Ans.(d)
Sol. Integrating instruments: These instruments record the consumption of the total quantity of electricity, energy etc., during a particular period of time. That is, these instruments totalize events over a specified period of time. No indication of the rate or variation or the amount at a particular instant are available from them. Some widely used integrating instruments are: Ampere-hour meter: kilowatt-hour (kWh) meter.

S9. Ans.(d)
Sol. ohmmeter and resistance bridge

S10. Ans.(b)
Sol. Indicating instruments: Indicating instruments indicate, generally the quantity to be measured by means of a pointer which moves on a scale. Examples are ammeter, voltmeter, wattmeter etc.

Quiz: Electrical Engineering 16 Sep 2020

Quiz: Electrical Engineering
Exam: SSC JE
Topic: Miscellaneous

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

Q1. In power transformers, it is found that the iron or core losses practically varies very less, this is because
(a) Constant leakage flux
(b) Constant core flux
(c) Constant load current
(d) Constant leakage and core flux

Q2. Which of the conditions is must be fulfilled for a satisfactory parallel operation of transformers?
(a) Correct polarity
(b) Same pu impedance
(c) Same voltage ratio
(d) Leakage impedance should be inversely proportional to KVA of the transformer

Q3. The qualities aspired to obtain a good permanent magnet is/are
(a) High coercivity
(b) High residual flux
(c) Low coercivity
(d) High residual flux and high coercivity

Q4. The pitch factor, in rotating electrical machinery, is defined as the ratio of resultant EMF of a
(a) Full pitch coil to the phase emf
(b) Chorded coil to that of full pitched coil
(c) Full pitched coil to that of a chorded coil
(d) Chorded coil to the phase emf

Q5. What is the maximum load that can be connected in a circuit containing only lighting points?
(a) 500 watts
(b) 750 watts
(c) 800 watts
(d) 1000 watts

Q6. The iron core in a transformer provides a _____path to the main flux
(a) Low reluctance
(b) High reluctance
(c) Low resistance
(d) High conductivity

Q7. For the network shown in the figure, the value of current in 8Ω resistor is

(a) 408A
(b) 2.4A
(c) 1.5A
(d) 1.2A

Q8. 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

Q9. 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

Q10. A diesel station supplies the following loads to various consumers:
Industrial consumer = 1500 kW
Commercial load = 750 kW
Domestic power = 100 kW
Domestic light = 450 kW
If the maximum demand on the station is 2500 kW, then diversity factor is
(a) 1.5
(b) 1.92
(c) 2
(d) 1.12

SOLUTIONS
S1. Ans.(b)
Sol. Due to the constant core flux, the iron losses also remain constant.

S2. Ans.(a)
Sol. out of all, polarity must be correct for the connected transformers.

S3. Ans.(d)
Sol. both high residual flux as well as coercivity is desired for a good permanent magnet.

S4. Ans.(b)
Sol. Kp= Resultant EMF of a chorded coil/Resultant EMF of a full pitched coil = cos⁡(ε/2)

S5. Ans.(c)
Sol. The maximum load that can be connected in a circuit containing only lighting points is 800 watts.

S6. Ans.(a)
Sol. The iron core in a transformer provides a low reluctance path to the main flux.

S7. Ans.(b)
Sol.

Here 12ohm and 8-ohm resistors are in series and both are parallel connected to 20ohm resistor.
∴R_eqv=20‖20=10Ω
∴I=48/10=4.8 A
And I8Ω= 4.8×20/(20+20)=2.4 A

S8. 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

S9. 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.

S10. Ans.(d)
Sol. Diversity Factor = Sum of Individual Maximum Demands / Maximum Demand of the System.
Diversity factor=(1500+750+100+450)/2500=1.12

Quiz: Electrical Engineering 9 Sep 2020

Quiz: Electrical Engineering
Exam: SSC JE
Topic: Miscellaneous

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

Q1. In a power plant if the maximum demand on the plant is equal to the plant capacity then
(a) Load factor will be nearly 60%
(b) Plant reserve capacity will be zero
(c) Diversity Factor will be unity
(d) Load factor will be unity

Q2. For the circuit shown in figure, the voltage across the capacitor during steady state condition is

(a) 0 V
(b) 4 V
(c) 6 V
(d) 12 V

Q3. Find R_AB for the circuit shown in figure.

(a) 18Ω
(b) 30Ω
(c) 45Ω
(d) 68Ω

Q4. A supply voltage of 230 V, 50 Hz is fed to a residential building. Write down its equation for instantaneous value.
(a) 163 sin 314·16 t
(b) 230 sin 314·16 t
(c) 325 sin 314·16 t
(d) 361sin314·16 t

Q5. The AC bridge used for measurement of dielectric loss of capacitor is
(a) Anderson bridge
(b) Schering bridge
(c) Wien bridge
(d) Hay’s bridge

Q6. In a transformer, Low voltage windings are placed nearer to the core in the case of concentric windings because
(a) it reduces hysteresis loss
(b) it reduces eddy current loss
(c) it reduces insulation requirement
(d) it reduces leakage fluxes

Q7. Emf induced in the dc generator armature winding is
(a) AC
(b) DC
(c) AC and DC
(d) None of the above

Q8. In a dc machine 72 number of coils are used. Find the number of commutator segments required?
(a) 72
(b) 36
(c) 37
(d) 74

Q9. Reactance relay is used for protection in:
(a) Long transmission lines
(b) Short transmission lines
(c) Medium transmission lines
(d) Both long transmission lines and short transmission lines

Q10. The earthing electrode should be situated at a place at least ………. meters away from the building whose installation system is being earthed?
(a) 4
(b) 2.5
(c) 1.5
(d) 0.5

SOLUTIONS
S1. Ans.(b)
Sol. Reserve capacity = Plant capacity – Max. Demand
If the plant capacity is equal to Max. Demand then the Reserve capacity will be zero.

S2. Ans.(d)
Sol. under steady state condition, capacitor behaves as open circuit. So, source voltage will appear across the capacitor.
So, voltage across capacitor during steady state= 12 V

S3. Ans.(a)
Sol. given circuit is a balanced bridge.
So, RAB=(20+10)‖ |(30+15)=(30×45)/75=18 Ω

S4. Ans.(c)
Sol. V=Vm sin⁡(2πft)=230×√2 sin⁡(2×3.14×50 t)=325 sin⁡(314.16 t)

S5. Ans.(b)
Sol. Schering bridge is used for measurement of dielectric loss of capacitor.

S6. Ans.(c)
Sol. In concentric arrangement of placing of HV and LV winding, less insulation is required when LV winding is placed near the core which is at ground potential. The HV winding is placed around the LV.
If high voltage winding is placed near the core, the more insulation will be required and the more insulation lead to increase in the size and cost of the transformer.

S7. Ans.(a)
Sol. The emf induced in the dc generator armature winding is AC, but we need DC current from DC generator, so to convert this AC current to DC current mechanical rectifier called as commutator is used.

S8. Ans.(a)
Sol. In DC machines, Number of coils = Number of commutator segments.

S9. Ans.(b)
Sol. Relay used for different configuration of transmission lines……
Short transmission line: reactance relay
Medium transmission line: impedance relay
Long transmission line: mho relay

S10. Ans.(c)
Sol. The earthing electrode should be situated at a place at least 1.5 meters away from the building whose installation system is being earthed
The earthing electrode should always be placed in a vertical position inside the earth or pit so that it may be in contact with all the different earth layers.

Quiz: Electrical Engineering 03 Sep 2020

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

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

Q1. If an ammeter is connected, like a voltmeter, across the load circuit,
(a)The reading will be low
(b)Almost no current will flow through the meter
(c)The loading effect will be low
(d)An inadmissibly high current will flow through the meter and meter may burn out

Q2. Moving iron and PMMC instruments can be distinguished from each other by looking at
(a)Pointer
(b)Terminal size
(c)Scale
(d)Scale range

Q3. If a voltmeter is connected, like an ammeter, in series with the load,
(a)Almost no current will flow in the circuit
(b)The measurement reading will be too high
(c)The meter will burn out
(d)An inadmissibly high current will flow

Q4. The internal resistance of a milli-ammeter must be very low for
(a)High sensitivity
(b)High accuracy
(c)High precision
(d)minimum effect on the current in the circuit.

Q5. The inductance of a certain moving iron ammeter is expressed as L = 2+5θ-θ2 µH. The control spring torque is 2 × 10-6 Nm/radian. Find the deflection in radians for a current of 2A
(a)1
(b)2
(c)3
(d)4

Q6. If the instrument is to have a wide range, the instrument should have
(a) Linear scale
(b) Square-law scale
(c) Exponential scale
(d) Logarithmic scale

Q7. Which of the following meters has a linear scale?
(a) Thermocouple meter
(b) Moving iron meter
(c) Hot wore meter
(d) Moving coil meter

Q8. No eddy current and hysteresis losses occur in
(a) Electrostatic instruments
(b) PMMC instruments
(c) Moving iron instruments
(d) Electro-dynamo meter instruments

Q9. The deflection expression θ ∝ V^2 dc/dθ, applies to the:
(a)Moving iron type instrument
(b)Electrodynamic type instrument
(c)Electrostatic type instrument
(d)Induction type instrument

Q10. A, 0-150V voltmeter has a guaranteed accuracy of 1 per cent full-scale reading. The voltage measured by this instrument is 83V. The limiting error in per cent will be—
(a)2·81%
(b)1·81%
(c)3·2%
(d)None of these

SOLUTIONS

S1. Ans.(d)
Sol. current coil has very small resistance. When it is connected in place of the voltage coil, then the whole current will flow through it and will cause damage to the current coil.

S2. Ans.(c)
Sol. A PMMC instruments has linear scale whereas moving iron instrument has non-linear /non-uniform scale.

S3. Ans.(a)
Sol. Voltmeter has very high resistance to ensure that its connection do not alter flow of current in the circuit. Now if it is connected in series then no current will be there in the circuit due to its high resistance. Hence it is connected in parallel to the load across which potential difference is to be measured.

S4. Ans.(d)
Sol. A high value of internal resistance will vary the current considerably in a milli-ammeter as the current is in milli range. So, the internal resistance of a milli-ammeter must be very low.

S5. Ans.(b)
Sol. The deflecting torque of a moving iron instrument (T) = 1/2 I^2 dL/dӨ
I = 2A; T = 2 × 10-6 Nm/radian; dL/dθ = 5-2θ µH;
∴2×10^(-6)=1/2×4×(5-2Ө)×10^(-6)
So, 5-2θ = 1; θ = 2.

S6. Ans.(d)
Sol.
If the scale is directly proportional to the quantity being measured, then it follows the uniform law
If the scale is directly proportional to the square of the quantity being measured, then it follows the square law
If an instrument has a cramped scale for larger values, then it follows logarithmic law; In this case, the instrument instruments will have a wide range

S7. Ans.(d)
Sol. Moving coil meter

S8. Ans.(a)
Sol. No eddy current and hysteresis losses occur in Electrostatic instruments.

S9. Ans.(c)
Sol. The deflection expression θ ∝ v2 dc/dθ, applies to the electrostatic instruments.

S10. Ans.(b)
Sol. Given voltmeter range 0–150V
measured value = 83V
The magnitude of the limiting error
= 0·01 × 150V = 1·5V
The percentage error at a meter indication of 83V will be
= 1.5/83 × 100 = 1.81 %

Quiz: Electrical Engineering 29 Aug 2020

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

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

Q1. In a transmission system the feeder supplies power to
(a)Generating stations
(b) Service mains
(c)Distributors
(d)All of the above

Q2. Feeder is designed mainly from the point of view of
(a)Its current carrying capacity
(b)Voltage drop in it
(c)Operating voltage
(d) Operating frequency

Q3. Distributors are designed from the point of view of
(a)Its current carrying capacity
(b)Operating voltage
(c)Voltage drop in it
(d)Operating frequency

Q4. Which of the following is/are active element?
(a)Current source
(b)Voltage source
(c)Both voltage source and current source
(d)None of the above

Q5. Superposition theorem is not applicable for
(a)Current calculations
(b)Voltage calculations
(c)Power calculations
(d)None of the above

Q6. The speed-torque regimes in a DC motor and the control methods suitable for the same are given, respectively, in Group-II and Group-I
Group-I                           Group-II
P. Field control                 1. Below base speed
Q. Armature control         2. Above base speed
3. Above base torque
4. below base torque
The match between the control method and the speed-torque regime is as follows:
Code:
P      Q
(a) 1       3
(b) 2       1
(c) 2        3
(d) 1       4

Q7. Two capacitance, C1 = 150 ± 2.4µF and C2 = 120 ± 1.5 µF are connected in parallel. What is the limiting error of the resultant capacitance C (in µF)?
(a)0.9 µF
(b)2.4 µF
(c)1.5 µF
(d) 3.9 µF

Q8. A 0 – 200 V voltmeter has a guaranteed accuracy of 1% of full-scale reading. The voltage measured by this instrument is 50 V. What is the limiting error?
(a)1 %
(b)2 %
(c)3 %
(d) 4 %

Q9. Which of the following is done in speed control of DC motor by field flux control?
(a)Inserting an additional resistance in the armature circuit
(b)Inserting an additional resistance in the field circuit
(c)Inserting resistance to armature circuit and field circuit
(d)Any of the above

Q10. The acceptable value of grounding resistance to domestic application is
(a)0.1 Ω
(b)1 Ω
(c)10 Ω
(d)100 Ω

SOLUTIONS
S1. Ans.(c)
Sol. In a transmission system the feeder supplies power to distributors.

S2. Ans.(a)
Sol. Feeders are the conductors which connect the substations or generating stations to the areas to be fed by these stations. Generally, from feeders no tapping is taken to the consumer, so current loading in the feeder remains the same throughout the conductors. so, feeders are designed based on the current carrying capacity.

S3. Ans.(c)
Sol. Distributors are the conductors from which numerous tapping are taken for providing power supply to the consumers. Therefore, the current loading on the distributors varies along the length. So, distributors are designed from the point of voltage drop in it.

S4. Ans.(c)
Sol. Active elements are capable of delivering energy independently for long or infinite time.
Both voltage source and current source are active elements and they can change energy level of a circuit.

S5. Ans.(c)
Sol. Superposition theorem is only applicable for linear quantities. Whereas power is a non-linear quantity. i.e. P=I^2 R
So, superposition theorem is not applicable for power calculations.

S6. Ans.(b)
Sol. For below base speed we use armature control method and for above base speed we use field control method.

S7. Ans.(d)
Sol. Equivalent capacitance of the given parallel combination is C = C1 + C2. So, limiting error of C is (2.4 + 1.5) = 3.9 µF

S8. Ans.(d)
Sol. limiting error=(full scale reading)/(measured value)×accuracy at full scale
∴ limiting error=200/50×1=4 %

S9. Ans.(b)
Sol. By inserting a resistance cause to reduce field current and hence the flux is also reduced. The reduction in flux will result in an increase in speed. Because of saturation of iron, the flux cannot increase beyond its normal value. So, the motor runs at a speed higher than normal speed.

S10. Ans.(b)
Sol. The acceptable value of grounding resistance to domestic application is nearly equal to 1 Ω.

Quiz: Electrical Engineering 22 Aug 2020

Quiz: Electrical Engineering
Exam: SSC JE
Topic: Miscellaneous

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

Q1. A moving-iron ammeter produces a full-scale torque of 240 µNm with a deflection of 120⁰ at a current of 10 A. The rate of change of self-inductance (µH/radian) of the instrument at full scale is
(a)2.0 µH/radian
(b)4.8 µH/radian
(c)12.0 µH/radian
(d)114.6 µH/radian

Q2. In wiring and winding an abbreviation for S.W.G is:
(a)Synthetic winding guard
(b)Simple winding gauge
(c)Standard wire gauge
(d)Single wire guard

Q3. Overlapping of conduction band and valence band occurs in
(a)Conductors only
(b)Semiconductors only
(c)Insulators only
(d)Conductors and semiconductors both

Q4. The pressure coil of an energy meter is…….
(a)Purely resistive
(b) Purely inductive
(c)Highly resistive
(d)Highly inductive

Q5. In DC compound Motor the shunt field and series field winding are made up of _________ & ____________ respectively
(a)High Resistance, Low Resistance
(b)Low Resistance, High Resistance
(c)Equal Resistance
(d)None of the above

Q6. The energy in joules stored in the magnetic field of 0.15 H inductance with a
180 mAcurrent will be
(a) 2.43 J
(b) 2.43×10^(-3) J
(c) 2.43×10^(-6) J
(d) 2.43×10^(-9) J

Q7. A 4-pole, 3-phase induction motor is running at 4% slip at full load. If the speed of the motor is 750 rpm, the supply frequency is
(a) 162/3 Hz
(b) 26 Hz
(c) 50 Hz
(d) 60 Hz

Q8. Two coils have inductances L1 = 1200 mH and L2 = 800 mH. They are connected in such a way that flux in the two coils aid each other and inductance is measured to be 2500 mH, then Mutual inductance between the coils is ___________ mH.
(a)500
(b)250
(c)225
(d)150

Q9. The magnetic field required to applied in opposite direction to reduce the residual magnetism to zero is called
(a)Hysteresis
(b)Retentivity
(c)Coercivity
(d)none of these

Q10. An alternating current varying sinusoidally has a peak value of 50 A and a frequency of 10 Hz. Mathematical expression for the current is:
(a) i=50*√2 sin⁡(2*π*10)t
(b) i=50/√2 sin⁡(2*π*10)t
(c) i=50*√2 sin⁡(π*10)t
(d) i=50 sin⁡(2*π*10)t

SOLUTIONS

S1. Ans.(b)
Sol. moving-iron ammeter full torque is given as: T_c=1/2 I^2 dL/dӨ
∴dL/dӨ=(2×240×10^(-6))/10^2 =4.8 µH/radian

S2. Ans.(c)
Sol. S.W. G= Standard wire gauge

S3. Ans.(a)
Sol. overlapping of conduction and valence band occurs in conductors only.

S4. Ans.(d)
Sol. The pressure coil of an energy meter is highly inductive while for wattmeter it is highly resistive.

S5. Ans.(a)
Sol. In DC compound Motor:
The shunt-winding is made up of wires of small cross-sections and has high resistance and the series winding will carry a large armature current and therefore it is made of wires of large cross-section and has a few turns only.

S6. Ans.(b)
Sol. Energy stored in magnetic field
=1/2 LI²
=1/2×0.15×(0.18)^2
=2.43×10^(-3) J

S7. Ans.(b)
Sol. S= 4% Nr =750 rpm
∴ Ns=Nr/((1-S))
=750/0.96
=781.25 rpm
But Ns = (120 f)/P
∴ (781.25×4)/120=f
⇒ f = 26 Hz

S8. Ans.(b)
Sol. Flux in two coils aid each other inductor of the combination


So, the mutual inductance is 250 mH.

S9. Ans.(c)
Sol. from the B-H curve, the value of magnetising field requires to reduce residual magnetism to zero is called coercivity of the material.
RETENTIVITY: The value of the intensity of magnetisation of a material when the magnetising field is reduced to zero is called retentivity.

S10. Ans.(d)
Sol. i=im sin⁡(2πf)t

Quiz: Electrical Engineering 12 Aug 2020

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

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

Q1. In the circuit given below, determine Vab.

(a)2.5 V
(b)3.5 V
(c)7 V
(d)5V

Q2. The rotor power output of a three-phase induction motor is 15 KW and the corresponding slip is 4%. The rotor ohmic losses are
(a)600 W
(b)625 W
(c)650 W
(d)700 W

Q3. A current i=5+14.14 sin⁡(ωt+45^0) is passed through a center-zero PMMC and a moving-iron instrument respectively, the respective readings are:
(a)-5 and 15
(b)5 and √125
(c)-5 and 19.14
(d)5 and 10

Q4. The type of single-phase induction motor having the highest power factor at full load is:
(a)Shaded pole type
(b)Split-phase type
(c)Capacitor-start type
(d)Capacitor-run type

Q5. In house wiring, Black and Green wires indicate
(a)Earth and neutral respectively
(b)Phase and neutral respectively
(c)Phase and earth respectively
(d)Neutral and earth respectively

Q6. The commutator in a d.c. machine acts as
(a) a mechanical inverter
(b) a mechanical rectifier
(c) current controller
(d) either (a) or (b)

Q7. Speed control is possible for _______ and not possible for __________
(a) induction motor, synchronous motor
(b) induction motor, differential motor
(c) synchronous motor, synchronous-induction motor
(d) dc motor, induction motor

Q8. Which of the following can be done using a synchronous motor but not by induction motor?
(a) Power factor improvement
(b) Supplying mechanical load
(c) Power factor improvement and supply mechanical load
(d) None of the mentioned

Q9. The torque in an induction motor varies as _______________
(a) square of voltage
(b) proportion to voltage
(c) inversely proportion to voltage
(d) none of the above

Q10. For satisfactory performance of 3-phase 480V, 60 Hz induction motor, the supply voltage at 50 Hz should be equal to
(a)480 V
(b)350 V
(c)400 V
(d)420 V

SOLUTIONS

S1. Ans.(b)
Sol. Nodal analysis at point a: (Va-1)/2+Va/2-3=0
∴2Va=7
⇒Va=3.5=V_ab

S2. Ans.(b)
Sol. rotor power output of 3 phase I/M=P_g-P_cu
Where, Pg=air gap power and Pcu=rotor cu loss=sPg
According to question: Pg-Pcu=15
∴Pg-sPg=15
⇒Pg=15/(1-s)=15/(1-0.04)=15.625 KW
∴Rotor ohmic loss= sPg=0.04×15.625=0.625 KW=625 W

S3. Ans.(b)
Sol. Reading of Centre-zero PMMC =IDC=5
Reading of moving iron=Irms=√(5^2+〖(14.14/1.414)〗^2 )=√125

S4. Ans.(d)
Sol. capacitor-run type motor will have high power factor in which capacitor will be connected in running condition.

S5. Ans.(d)
Sol. In house wiring, Black and Green wires indicate neutral and earth respectively.

S6. Ans.(d)
Sol. Commutator acts as a reversing switch.

S7. Ans.(a)
Sol. Speed can be adjusted for an induction motor while it cannot be altered for a synchronous motor operating at normal speed. Synchronous motor rotates at synchronous speed.

S8. Ans.(a)
Sol. An over-excited synchronous motor has a leading power factor. This makes it useful for power factor correction of industrial loads.
Induction motor is a singly excited machine and it always needs exciting current to set up the flux in the machine.

S9. Ans.(a)
Sol. Torque in induction motor varies as square of the voltage.

S10. Ans.(c)
Sol. For satisfactory performance, the V / f ratio must be constant.