Quiz: Electrical Engineering 29 Oct 2020

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

Each question carries 1 mark.
Negative marking: 1/3 mark
Time: 10 Minute

Q1. The relative permeability of a medium is equal to (with M=magnetization of medium and H=magnetic field strength)
(a) 1+M/H
(b) 1-M/H
(c) 1+√(M/H)
(d) 1-√(M/H)

Q2. For the circuit given in figure below the power delivered by the 2-volt source is given by

(a) 4 W
(b) -4 W
(c) 2 W
(d) -2 W

Q3. Elements R, L and C are connected in parallel, the impedance of the parallel combination can be expressed as
Z(s)=10s/(s^2+s+400)
The value of the individual elements R, L and C are
(a) 10 Ω, 40 H and 0.1 F
(b) 10 Ω, 1/40 H and 0.1 F
(c) 4 Ω, 1 H and 0.1 F
(d) 1 Ω, 40 H and 10 F

Q4. Which of the following represents the operating region slips of a 1-phase induction motor?
(a) 0<s<2
(b) -2<s<2
(c) 0<s<1
(d) -1<s<1

Q5. If the torque/weight ratio of an instrument is low, then it can be concluded that
(a) The meter will have a uniform scale
(b) The meter will have the non-uniform scale
(c) The sensitivity of the meter will be high
(d) The sensitivity of the meter will be low

Q6. The Ybus matrix of a 100-bus interconnected system is 80% sparse. Then, the number of transmission lines in the system must be
(a) 1500
(b) 1900
(c) 3000
(d) 950

Q7. Consider the following statements in connection with frequency domain specifications of a control system:
1.Resonant peak and peak overshoot are both functions of the damping ratio ξ only.
2.The resonant frequency ω_r=ω_n for ξ > 0.707.
3.Higher the resonant peak, higher is the maximum overshoot of the step response.
Which of the statements given above are correct?
(a) 1 and 2
(b) 2 and 3
(c) 1 and 3
(d) 1, 2 and 3

Q8. An ideal transformer has an input voltage of 480V. The output current and voltage are 10A and 120V respectively. Determine the value of input current-
(a) 25 A
(b) 2.5 A
(c) 0.25 A
(d) 0.025 A

Q9. The no-load current drawn by transformer is usually what percentage of the full-load current?
(a) 0.2 to 0.5 %
(b) 12 to 15 %
(c) 20 to 30 %
(d) 2 to 5 %

Q10. The secondary winding of which of the following transformers is always kept closed?
(a) Current transformer
(b) Voltage transformer
(c) Power transformer
(d) Step-down transformer

SOLUTIONS
S1. Ans.(a)
Sol. Magnetic susceptibility(X)=M/H=µ_r-1
Where, µ_r=relative permeability
∴µ_r=1+X=1+M/H

S2. Ans.(c)
Sol.

Here, i=2/1=2 A
On applying KCL at node a: I=i-1=2-1=1 A
i.e. the current delivered by the voltage source=1 A
∴power delivered by the voltage source=2×1=2 W

S3. Ans.(b)
Sol. Given that, Z(s)=10s/(s^2+s+400)
∴Y(s)=1/Z(s) =(s^2+s+400)/10s=s/10+1/10+40/s=sC+1/R+1/SL
On comparing: C=1/10 F,R=10 Ω and L=1/40 H

S4. Ans.(a)
Sol. In a 1-phase induction motor, due to both forward and backward torque slips, the operating region lies between 0 and 2. In motoring mode slip is 0 to 1 and breaking mode 1 to 2.

S5. Ans.(d)
Sol. The low torque/weight ratio signifies heavy weight of the moving system due to which the instrument incurs frictional losses hence the sensitivity will be low.

S6. Ans.(d)
Sol. Number of buses, N=100
Sparsity, s=0.8
Then, the number of transmission lines is given as: N_T=(N^2 (1-s)-N)/2=(100^2 (1-0.8)-100)/2
⇒N_T=(100×19)/2=950

S7. Ans. (c)
Sol. → Resonant Peak (M_r) = 1/(2ξ √(1-ξ²))
Peak overshoot = e (-ξπ)/√(1-ξ²)
Resonant frequency (wr) = Wn √(1-2ξ²)
For ξ = 0
wr = Wn.

S8. Ans.(b)
Sol. V₁I₁ = V₂I₂
⇒ I₁ = (V₂I₂)/(V₁)
=(120×10)/480=2.5 A

S9. Ans.(d)
Sol. Current required to set up magnetic flux in a magnetic material is rather low. Hence no-load primary current is a very small fraction of the rated full load current

S10. Ans.(a)
Sol. Because if CT secondary is not shorted then it can develop a very high voltage across secondary which may damage transformer insulation.

Quiz: Electrical Engineering 24 Oct 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 synchronous machine, if the filed flux is ahead of the armature field axis, in the direction of rotation, the machine works as-
(a) asynchronous motor
(b) asynchronous generator
(c) synchronous motor
(d) synchronous generator

Q2. Synchronous motor can be made self-starting by providing
(a) Damper winding on rotor poles
(b) Damper winding on stator
(c) Damper winding on stator as well as rotor poles
(d) None of these

Q3. The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240×10-6 N-m on full scale. If magnetic flux density is 1Wb/m2, range of meter is
(a) 1 mA
(b) 2 mA
(c) 3 mA
(d) 4 mA

Q4. If the cathode of the thyristor is made positive with respect to the anode & no gate current is applied then: –
(a) Only the middle junction is forward biased
(b) Only the middle junction is reversed biased
(c) All the junctions are forward biased
(d) All the junctions are reversed biased

Q5. In a dynamometer type wattmeter, the pressure coil connected across the load terminal is
(a) Highly inductive
(b) Highly capacitive
(c) Highly resistive
(d) Non inductive

Q6. Due to the inductance in the pressure coil of dynamometer type wattmeter, the reading will be
(a) High for both leading and lagging power factors
(b) Low for both leading and lagging power factors
(c) High for lagging power factor and low for leading power factor
(d) Low for lagging power factor and high for leading power factor

Q7. If the number of turns of a coil is increased, its inductance
(a) Is decreased
(b) Remains the same
(c) Is increased
(d) None of the above

Q8. Mho 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

Q9. Which of the following type of instrument is used only for D.C?
(a) Moving iron-attraction type
(b) Moving iron-repulsion type
(c) Permanent magnet type
(d) Hot wire type

Q10. At a frequency less than the resonant frequency_____:
(a) series circuit is capacitive and parallel circuit is inductive
(b) series circuit is inductive and parallel circuit is capacitive
(c) both circuits are inductive
(d) both circuits are capacitive

SOLUTIONS
S1. Ans.(d)
Sol.

In and alternator, field flux is a head of armature field flux in the direction of rotation.

S2. Ans.(a)
Sol. Synchronous motor is not self-starting. It can be made self-starting by providing “Damper winding on ROTOR pole faces and around the pole shoes.
Following is the primary function of damper winding: –
(i) Provides starting torque
(ii) Reduce hunting
(iii) Used for salient pole rotors
(iv) Under normal running condition, damper winding in synchronous machine does not carry and current because the relative velocity between the rotating magnetic field (RMF) of stator and the rotor is zero.

S3. Ans.(b)
Sol. In case of moving coil meter, Controlling torque Tc = NBIA.
Where, N = no. of turns of the coil, B – magnetic flux density, = A = area of coil.
So, I = 240×10-6 / (100 × 0.04 × 0.03) = 2mA.

S4. Ans.(a)
Sol. Anode must be positive with respect to the cathode so that SCR will be forward biased. To turn on an SCR, a current is to be passed through the gate terminal. Hence it is a current operated device.

With the gate open, if the cathode is made positive with respect to the anode, the junctions J1 and J3 become reverse biased and J2 forward biased. Still the current flowing is leakage current, which can be neglected as it is very small. The SCR is said to be in reverse blocking state and voltage applied is called reverse voltage.

S5. Ans.(c)
Sol. In a dynamometer type wattmeter, the pressure coil connected across the load terminal is highly Resistive.

S6. Ans.(c)
Sol. Because of the inductance, the current of the pressure coils lags behind the voltage. Thus, the power factor of the wattmeter becomes lagging, and the meter reads high reading and vice-versa.

S7. Ans.(c)
Sol. inductance of a coil is given by: L=(N^2µA)/l
∴L α N^2
Hence, L increases if number of turns(N) increases.

S8. Ans.(a)
Sol. Relay used for different configuration of transmission lines……
Short transmission line: reactance relay
Medium transmission line: impedance relay
Long transmission line: mho relay

S9. Ans.(c)
Sol. PMMC
→ only for DC
→ Scale is uniform
→ Eddy current damping is used
→ F = BIL
→ Torque/Weight ratio is high to reduce frictional errors.

S10. Ans.(a)
Sol. For series Resonant RLC circuit

Parallel RLC circuit

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 21 Oct 2020

Quiz: Electrical Engineering
Exam: RSMSSB-JEn
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/3 mark
Time: 10 Minute

Q1. In a three-phase system, the relation V_L=V_ph is applicable to a ________.
(a) star-connected without neutral point
(b) star-connected load
(c) single-phase system also
(d) delta-connected load

Q2. In a star connected three-phase balanced system. If the line current is 125A then the phase current is?
(a) 216.5 A
(b) 72.16 A
(c) 250 A
(d) 125 A

Q3. A 8-pole lap-wound generator has 400 conductors; the e.m.f. induced per conductor
being 5 V. The generated voltage of the generator is ………
(a) 60 V
(b) 1500 V
(c) 360 V
(d) 250 V

Q4. The curve B in Fig. below is the external characteristic of a ……… generator.

(a) shunt
(b) series
(c) flat-compounded
(d) under-compounded

Q5. Out of the following, ……… wave is the peakiest.
(a) sinusoidal
(b) square
(c) rectangular
(d) triangular

Q6. When a 15-V square wave is connected across a 50-V a.c. voltmeter, it will read ………
(a) 15 V
(b) 15×√2 V
(c) 15/√2 V
(d) none of the above

Q7. An alternating voltage υ=200 √2 sin 100 t is connected to a 1 μ F capacitor
through an a.c. ammeter. The reading of the ammeter shall be
(a) 40 mA
(b) 80 mA
(c) 100 mA
(d) 20 mA

Q8. A choke is preferred to a resistance for limiting current in an a.c. circuit because
(a) choke is cheap
(b) there is no wastage of energy
(c) current becomes wattless
(d) current strength increases

Q9. The value of current ‘I’ is flowing in the 1Ω resistor in the circuit shown in the figure
below will be

(a) 10 A
(b) 6 A
(c) 5 A
(d) zero

Q10. The no load speed of DC series motor is
(a) very small
(b) medium
(c) very high
(d) small

SOLUTIONS
S1. Ans.(d)
Sol. for delta connection –
Line voltage (VL )= Phase voltage (Vph)
Line current (IL) = √3× Phase current (Iph )

S2. Ans.(d)
Sol. for star (Y) – connection: –
Line current (IL) = Phase current (Iph )
Line voltage (VL )=√3×Phase voltage (Vph )
Or, phase voltage = (Line voltage)/√3
⇒ ▭(Vph=VL/√3) – For star connection
S3. Ans.(d)
Sol. No of conductors per parallel path
=400/8=50
∴ Eg=50×5=250 V

S4. Ans.(c)
Sol. Flat-compounded. (voltage regulation=zero)

S5. Ans.(d)
Sol. That wave is peakiest which has the greatest form factor (triangular wave).
The Form Factor for the various sinusoidal waveforms are as follows:
For a sine wave, it is π/2√2 = 1.11072073
For a half-wave rectified sine wave, it is π/2 = 1.5707963
For a full-wave rectified sine wave, it is π/2√2 = 1.11072073
For a square wave, it is equal to 1
For triangle waveform, it is 2/√3 = 1.15470054
For sawtooth waveform, it is 2/√3 = 1.15470054

S6. Ans.(a)
Sol. A voltmeter records the r.m.s. value and the r.m.s. value of 15-V square wave is 15 V.

S7. Ans.(d)
Sol. V(r.m.s.)=Vm/√2=200 √2 /√2=200V; ω=100 rad/s; X_C=1/ωC=1/(100×1×10^(-6) )=10^4 Ω
∴ I(r.m.s.)=V(r.m.s.)/X_C =200/10^4 =20×10^(-3) A=20 mA

S8. Ans.(b)
Sol. In an AC circuit, the coil of high inductance and negligible resistance used to control current, is called the choke coil. The power factor of such a coil is given by
cos⁡φ=R/√(R^2+ω^2 L^2 )≈ R/ωL (as R <<<ωL).
cos⁡φ is very small. Thus, the power absorbed by the coil is very small. The only loss of energy is due to hysteresis in the iron core, which is much less than the loss of energy in the resistance that can also reduce the current if placed instead of the choke coil.

S9. Ans.(c)
Sol.
Method-1: From ohm’s law, if we know the voltage across a resistor, then we can determine the current through resistor R as I=V/R.
Here, V = 5 V (constant)
So, I=5/1 = 5A

Note: The voltage across any current source is purely arbitrarily. The voltage across it depends purely upon the voltage source connected in parallel across it. Hence, in present case voltage across the current source = 5 V.
Method2:
Applying superposition theorem:
Current due to 5 V source: current source open circuit & I=5/1=5 A
Current due to 5 A source: voltage source short circuit and current flows through short-circuit. So, current through 1-ohm resistor=0 A
∴I_TOTAL=5+0=5 A

S10. Ans.(c)
Sol. The load-speed curve of DC series motor is shown below.

It is clear that the speed of the DC series motor at no load is very high. So, it is never stared at no load.

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 14 Oct 2020

Quiz: Electrical Engineering
Exam: RSMSSB-JEn
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/3 mark
Time: 10 Minute

Q1. The energy stored in the magnetic field of a solenoid carrying a current of 5A is 0.5 J. What will be the stored energy if the number of turns is doubled and the current is halved?
(a) 0.25 J
(b) 0.5 J
(c) 1 J
(d) 2J

Q2. The reciprocal of susceptance is known as
(a) Reactance
(b) Admittance
(c) Susceptance
(d) Conductance

Q3. In a series circuit with R=10 Ω, Xc=12 Ω and XL=15 Ω, the current
(a) Leads the applied voltage
(b) Is in phase with the applied voltage
(c) Lags the applied voltage
(d) None of the above

Q4. The ratio of RMS and the average value of a particular waveform is called……
(a) Peak factor
(b) Crest factor
(c) Mean value
(d) Form factor

Q5. Loss of charge method can be used to measure which of the following?
(a) Insulation resistance of cable
(b) Resistance of shunt winding
(c) Resistance of transformer winding
(d) None of the above

Q6. Wien’s bridge is used to measure……….
(a) Energy
(b) Power
(c) Frequency
(d) Phase angle

Q7. In which of the type of transformer primary and secondary winding placed on the same central limb?
(a) Core type transformer
(b) Shell type transformer
(c) Cylindrical type transformer
(d) None of the above

Q8. In synchronous motor damper winding is placed at
(a) Yoke
(b) Stator
(c) Rotor
(d) Both stator and rotor

Q9. Load factor of a power station is generally
(a) Less than unity
(b) Equal to unity
(c) More than unity
(d) Equal to zero

Q10. What is the maximum load that is permitted in a power circuit?
(a) 2000 watts
(b) 3000 watts
(c) 5000 watts
(d) 10000 watts

SOLUTIONS
S1. Ans.(b)
Sol. As we know, energy stored in the inductor is E=1/2 L1 I1^2=0.5 J
For solenoid, L ⍺ N^2⇒L2=4L1 (Number of turns doubled)
And I2=I1/2
∴E^’=1/2 (4L1 ) (I1/2)^2=1/2 L1 I1^2=0.5 J

S2. Ans.(a)
Sol. susceptance ⍺ 1/ reactance
Admittance ⍺ 1/ impedance
Conductance ⍺ 1/ resistance

S3. Ans.(c)
Sol. Here, XL>XC, so the circuit is inductive in nature and the current lags the applied voltage.

S4. Ans.(d)
Sol. form factor= RMS/ avg. value
Crest or peak factor= max. value/ rms value

S5. Ans.(a)
Sol. Loss of charge method is used to measure high values of resistance (insulation resistance).
methods for measurement of high resistance:
Megger
Loss of charge method
Direct deflection method
Megaohm bridge

S6. Ans.(c)
Sol. Wien’s bridge is used to measure frequency.

S7. Ans.(b)
Sol. In shell type of transformer primary and secondary winding placed on the same central limb.

S8. Ans.(c)
Sol. synchronous motors are not self-starting machines. These are made self-starting by providing damper winding in the rotor poles.

S9. Ans.(a)
Sol. load factor=(average load)/(maximum demand )
Load factor is always less than 1.

S10. Ans.(b)
Sol. maximum load for lighting points circuit= 800 watts
Maximum load for power circuit= 3000 watts

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 10 Oct 2020

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

S10. Ans.(c)
Sol.

Req =(10×15)/25+12
= 6 + 12
= 18 Ω

Quiz: Electrical Engineering 08 Oct 2020

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

Each question carries 1 mark.
Negative marking: 1/3 mark
Time: 10 Minute

Q1. If the modulation index of an amplitude modulated wave is changed from 0 to 1, then the transmitted power will be
(a) Zero
(b) Doubled
(c) Unchanged
(d) Increase by 50 %

Q2. Three resistances P, Q, R each of 2Ω and an unknown resistance S form the four arms of a Wheatstone bridge circuit. When a resistance of 6Ω is connected in parallel to S the bridge gets balanced. What is the value of S?

(a) 1 Ω
(b) 2 Ω
(c) 3 Ω
(d) 6 Ω

Q3. To avoid cogging in induction motor, the number of stator slots should not be ______________ the number of rotor slots.
(a) greater than
(b) dependent on
(c) equal to
(d) less than

Q4. Which of the following device is commonly used for lightening protection?
(a) isolator
(b) Over voltage relay
(c) Circuit breaker
(d) ZnO Varistor

Q5. A SCHOTTKY diode is a:
(a) Both minority and majority carrier diode
(b) Minority carrier device
(c) Fast recovery diode
(d) Majority carrier device

Q6. Multi-meters cannot be used to measure:
(a) resistance
(b) current
(c) voltage
(d) frequency

Q7. A second-order servo has unity feed and an open-loop transfer function: G(S) = 500/s(s+15). The settling time is:
(a) 2 sec
(b) 0.53 sec
(c) 500 sec
(d)0.5 sec

Q8. A single phase, full-wave, fully controlled bridge converter for supplying high inductive load is fed by a voltage V_m sin⁡ωt. The average and rms value of output voltages are
(a) (2Vm)/π cos⁡α and Vm/√2
(b) Vm/π cos⁡α and Vm/√2
(c) (2Vm)/π cos⁡α and Vm
(d) Vm/π cos⁡α and √2 Vm

Q9. Which testing method is performed to determine No Load losses in DC motor?
(a) Running down test
(b) Field test
(c) Swinburne’s test
(d) Brake test

Q10. The pointer returns to its zero position on removing the source producing the deflecting torque. This happens due to:
(a) Controlling torque
(b) Balancing torque
(c) Damping torque
(d) Mass of pointer

SOLUTIONS
S1. Ans.(d)
Sol. the transmitted power in AM wave is: Pt=Pc (1+µ^2/2)
Where, µ= modulation index
Pc= power in carrier wave
When, µ=0, Pt=Pc
When, µ=1, Pt=Pc (1+1/2)=1.5 Pc
∴% increase in transmitted power=(1.5Pc-Pc)/Pc *100%=50%

S2. Ans.(c)
Sol. let X be the equivalent resistance between S and 6 Ω.
∴1/X=1/6 +1/S…………… (1)

For a balanced Wheatstone bridge, we get: P/Q=R/X
⇒2/2=2/X ⇒X=2 Ω
Putting this value in eq. (1), 1/2=1/6+1/S
⇒S=3 Ω

S3. Ans.(c)
Sol. To avoid cogging in induction motor, the number of stator slots should not be equal to the number of rotor slots.

S4. Ans.(d)
Sol. In modern era, gap less ZnO or zinc oxide surge arresters are mainly used for surge protection.

S5. Ans.(d)
Sol. Schottky diode is formed by joining a semiconductor region (usually N-type) with a metal such as gold, silver or platinum. So, it is a metal to semiconductor junction having only the majority carriers.

S6. Ans.(d)
Sol. A typical multi-meter can measure voltage, current, and resistance.

S7. Ans.(b)
Sol. characteristics eq.: 1+ G(s)H(s)=0
⇒1+500/(s^2+15s)=0
⇒s^2+15s+500=0
From above eq. 2ξωn=15
∴ξωn=15/2
Settling time=4/ ξωn=8/15=0.53 s

S8. Ans.(a)
Sol. A single phase, full-wave, fully controlled bridge converter for supplying high inductive load: Vavg=(2Vm)/π cos⁡α
And Vrms=Vm/√2

S9. Ans.(c)
Sol. Swinburne’s test is the most commonly used and simplest method of testing of shunt and compound wound DC machines which have constant flux.
In this method of testing no load losses are measured separately and eventually we can determine the efficiency.

S10. Ans.(a)
Sol. Controlling torque brings the pointer back to zero when the deflecting torque is removed. If it were not provided, the pointer once deflected would not return to zero position on removing the deflecting torque.

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 01 Oct 2020

Quiz: Electrical Engineering
Exam: RSMSSB-JEn
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/3 mark
Time: 10 Minute

Q1. The rotor of which motor does not have winding on it?
(a) Repulsion motor
(b) Universal motor
(c) Reluctance motor
(d) Hysteresis motor

Q2. The ratio of starting torque to running torque in a synchronous motor is:
(a) 0
(b) 1
(c) 2
(d) ꝏ

Q3. The torque-slip characteristics of a polyphase induction motor becomes almost linear at small value of slip, because in this region of slips:
(a) The effective rotor resistance is very large compared to the rotor reactance.
(b) The effective rotor resistance is very small compared to the rotor reactance.
(c) The effective rotor resistance is equal to the rotor reactance.
(d) The effective rotor resistance is equal to the stator reactance.

Q4. A shunt reactor of 100 MVAR is operated at 98% of its rated voltage and at 96% of its rated frequency. What is the reactive power absorbed by the reactor?
(a) 98 MVAR
(b) 104.02 MVAR
(c) 96.04 MVAR
(d) 100.04 MVAR

Q5. A dielectric material is placed in vacuum in a uniform electric field of E = 4 V/m. what is the electric field inside the material if the relative permittivity of dielectric material is 2?
(a) 0 V/m
(b) 2 V/m
(c) 4 V/m
(d) 8 V/m

Q6. A spring controlled moving iron voltmeter draws a current of 1 mA for full scale value of 100 V. If it draws a current of 0.5 mA, what is the meter reading?
(a) 25 V
(b) 50 V
(c) 100 V
(d) 200 V

Q7. Candela is a unit of:
(a) Luminous intensity
(b) Frequency
(c) Lamp efficiency
(d) Power

Q8. The gas filled in vacuum filament lamps is:
(a) Nitrogen
(b) Chlorine
(c) Argon
(d) Air

Q9. Total core loss is also referred as
(a) Eddy current loss
(b) Hysteresis loss
(c) Copper loss
(d) Magnetic loss

Q10. The primary to secondary turn ratio of a transformer is 1:2. If the primary is connected to 50 Hz supply, then the frequency of secondary supply in Hz is:
(a) 25
(b) 50
(c) 100
(d) 150

SOLUTIONS
S1. Ans.(d)
Sol. A Hysteresis Motor is a synchronous motor with a uniform air gap and without DC excitation.
The rotor of the motor is made up of smooth chrome steel cylinder and it has no winding.

S2. Ans.(a)
Sol. when supply is given to stator starts rotating and stator poles too rotate at synchronous speed. these poles interchange their position due to alternating nature of current. the rotor cannot pick up the quick reversal of stator. Due to this clockwise and counter clockwise directions of torque will develop making average torque zero.

S3. Ans.(a)
Sol. The torque-slip characteristics of a polyphase induction motor becomes almost linear at small value of slip, i.e. X2/R2 ≪1.

S4. Ans.(d)
Sol. The reactive power is given as


∴Q2=100.04 MVAR

S5. Ans.(b)
Sol. D1n=D2n
⇒ϵr1 E1n=ϵr2 E_2n
Where, ϵr1=relative permitivity of vaccum=1
And ϵr2=relative permitivity of medium=2
∴E2n=(1×4)/2=2 V/m

S6. Ans.(a)
Sol. For moving iron meter
Reading ∝ I^2
∴V2/V1 =(I2^2)/(I1^2 )
⇒V2=100×0.25=25 V

S7. Ans.(a)
Sol. Candela is a unit of Luminous intensity.

S8. Ans.(c)
Sol. Argon gas is used in fluorescent and incandescent light bulbs to stop the oxygen in the light bulbs from corroding the hot tungsten filament. The use of argon in lightbulbs prevents the evaporation of the tungsten filaments, which results in increased light bulb life.

S9. Ans.(d)
Sol. Core loss consisting of eddy current loss and hysteresis loss.
i.e. core loss= eddy current loss+ hysteresis loss.
Core loss is mainly due to magnetic properties. So, core loss is also called magnetic loss.
Copper loss is due heat generated due to resistance of winding.

S10. Ans.(b)
Sol. A Transformer is a static electrical machine which transfers AC electrical power from one circuit to the other circuit at the constant frequency, but the voltage level can be altered that means voltage can be increased or decreased according to the requirement.

Quiz: Electrical Engineering 30 Sep 2020

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

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

Q1. In the shown figure

1. If both ammeter and voltmeter are ideal, then reading of ammeter is zero, reading of voltmeter is 10 V
2. If both ammeter and voltmeter are ideal, then reading of ammeter is 10 A, reading of voltmeter is 0 V
3. If ammeter is non ideal, voltmeter is ideal, then reading of ammeter is 0 A, reading of voltmeter is 10 V
4. If ammeter is ideal, voltmeter is non ideal, then reading of ammeter is less than 10 A, reading of voltmeter is less than 10 V
Which of these statements are correct?
(a) 1,2 and 3 only
(b) 2,3 and 4 only
(c) 1, 3 and 4 only
(d) All of the above

Q2. A galvanometer having a coil resistance of 100 Ω gives a full-scale deflection, when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full-scale deflection for a current of 10 A is:
(a) 2 Ω
(b) 0.1 Ω
(c) 3 Ω
(d) 0.01 Ω

Q3. The Charging reactance of 50 km length of line is 1500 ohm. What will be the charging reactance for 100 km length of the line?
(a) 600 Ω
(b) 3000 Ω
(c) 750 Ω
(d) 1500 Ω

Q4. Turn off time of an SCR is measured from the instant:
(a) Anode current becomes zero
(b) Anode voltage becomes zero
(c) Gate current becomes zero
(d) Anode voltage and current become zero at the same time

Q5. The number of address lines required to address a memory size of 16K is:
(a) 16
(b) 14
(c) 12
(d) 15

Q6. Skin effect occurs when a conductor carries current at………frequencies.
(a) Medium
(b) High
(c) Low
(d) Very low

Q7. The distribution or winding factor is defined in an alternator in terms of
(a) Power factor
(b) Electromotive force
(c) Power
(d) Current

Q8. A 3 phase, 50 Hz induction motor draws 50 kW from the mains. If the stator losses are 2 kW and rotor emf oscillates at 100 oscillations per minute the rotor copper losses will be:
(a) 1.8 kW
(b) 1.2 kW
(c) 1.9 kW
(d) 1.6 kW

Q9. Slip of an induction machine is 0.04 and the stator supply frequency is 150 Hz. What will be the frequency of the rotor induced emf?
(a) 0.06 Hz
(b) 6 Hz
(c) 16 Hz
(d) 4 Hz

Q10. The crawling in an induction motor is caused by
(a) Improper design of the machine
(b) Low voltage supply
(c) High loads
(d) Harmonics developed in the motor

SOLUTIONS
S1. Ans.(c)
Sol. If both are ideal, voltmeter has infinite resistance and hence, no current flows through the circuit. Thus, ammeter will read zero and voltmeter will read 10V.
If ammeter is non ideal but voltmeter is ideal, current through circuit will be 0 and hence ammeter will read 0A and voltmeter 10V.
If voltmeter is non ideal, finite current will flow through the circuit and hence there will be some voltage drop across the resistor. Therefore, Voltmeter will read less than 10V. Due to finite resistance of voltmeter, current will be less than 10A and hence ammeter will read less than 10A.

S2. Ans.(d)
Sol. Galvanometer is a very sensitive instrument therefore it cannot measure heavy currents. In order to convert a Galvanometer into an Ammeter, a very low shunt resistance is connected in parallel to Galvanometer. Value of shunt is so adjusted that most of the current passes through the shunt.
If resistance of galvanometer is Rg and it gives full-scale deflection when current Ig is passed through it. Then,
V = Ig Rg
V = 1 mA × 100 Ω
Let a shunt of resistance (Rs) is connected in parallel to galvanometer. If total current through the circuit is I.
I = 10 A = Is + Ig
V = Is Rs = (I – Ig) Rs
(I – Ig) Rs = Ig Rg
(10 – 10-3) Rs = 100 × 10-3
Rs ≈ 0.01 Ω

S3. Ans.(c)
Sol. charging reactnce(X_C ) α 1/L
∴X_C2/X_C1 =L1/L2
⇒X_C2=1500×50/100= 750 Ω

S4. Ans.(a)
Sol. Turn off time of an SCR is measured from the instant when anode current becomes zero.

S5. Ans.(b)
Sol. memory size=16K=16×1024 bytes=2^4×2^10=2^14 bytes
So, the number of address lines required to address a memory size of 16K is 14.

S6. Ans.(b)
Sol. The non-uniform distribution of electric current over the surface or skin of the conductor carrying a.c is called the skin effect. In other words, the concentration of charge is more near the surface as compared to the core of the conductor.
Factors affecting skin effect
Frequency – Skin effect increases with the increase in frequency.
Diameter – It increases with the increase in diameter of the conductor.
The shape of the conductor – Skin effect is more in the solid conductor and less in the stranded conductor because the surface area of the solid conductor is more.
Type of material – Skin effect increase with the increase in the permeability of the material (Permeability is the ability of material to support the formation of the magnetic field).

S7. Ans.(b)
Sol. The distribution or winding factor is defined in an alternator in terms of electromotive force(emf).
Kd=(phasor sum of coil emfs)/(arithmetic sum of coil emf)

S8. Ans.(d)
Sol. Slip (s)= fr/fs =(100/60)/50
= 1/30
Rotor input = stator input – stator loss
= 50-2 = 48 kw
Rotor copper loss = S × Rotor input
= 1/30×48
= 1.6 kw

S9. Ans.(b)
Sol. Rotor frequency (fr)= Sf
= 0.04×150
= 6 HZ

S10. Ans.(d)
Sol. Crawling is associated with squirrel cage induction motor due to which motor has a tendency to run at very low speed compared to its synchronous speed. The fundamental runs at synchronous speed and harmonics 3^rd,5^th,7^th may rotate in backward or forward direction at Ns/3,Ns/5,Ns/7 speed respectively.
Hence harmonics torque developed along with fundamental.
5^th Harmonics = 120°×5= 600°= -120°, so will run in backward direction with speed (Ns/S) and 7th Harmonics = 120×7 = 840°= +120°, so will run in forward direction with speed Ns/7.

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 26 Sep 2020

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Unit =1/Ωm (mho/m)
Or
Siemen/meter

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

Unit =1/ohm (mho or siemen)

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

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

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

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

Quiz: Electrical Engineering 24 Sep 2020

Quiz: Electrical Engineering
Exam: RSMSSB-JEn
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/3 mark
Time: 10 Minute

Q1. The relation between μ and H for a specimen of iron is as follows-
μ = [0.4/H + 12 × 10^(-4)] Henry / meter
The value of H which produces flux density of 1 Tesla will be-
(a) 250 A/m
(b) 500 A/m
(c) 750 A/m
(d) 103 A/m

Q2. The SI unit of magnetic flux is……
(a) Weber
(b) Maxwell
(c) Tesla
(d) Gauss

Q3. Which of the following is a diamagnetic material?
(a) Aluminium
(b) Oxygen
(c) Lead
(d) Nickel

Q4. Which one of the following statements is NOT TRUE about the MI type instruments? (a) MI type Instruments are suitable for both AC and DC circuits.
(b) Frictional error in MI type instruments is very less.
(c) The torque weight ratio of MI type instruments is high.
(d) The instrument cost is much higher as compared to PMMC type instruments.

Q5. In a shaded pole single-phase motor, the revolving field is produced by the use of
(a) inductor
(b) capacitor
(c) Resistor
(d) shading coils

Q6. A pole pitch in an electrical machine is
(a) Equal to 180° Electrical
(b) Equal to 180° Mechanical
(c) Less than 180° electrical
(d) Greater than 180° electrical

Q7. 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)

Q8. In a power plant if the plant capacity is equal to the maximum demand then
(a) Plant reserve capacity will be 50%
(b) Plant reserve capacity will be zero
(c) Load factor will be unity
(d) Load factor will be nearly 70%

Q9. The maximum demand i.e. sum of connected load to the simultaneous maximum demand is expressed through
(a) Derivative factor
(b) Demand factor
(c) Diversity factor
(d) Dividend factor

Q10. If a resistance is inserted in the rotor circuit of a slip ring induction motor, then compared to direct line starting:
(a) Both the starting current and torque reduce
(b) Both the starting current and torque increase
(c) The starting current reduces but starting torque increases
(d) The starting current reduces but starting torque

SOLUTIONS
S1. Ans.(b)
Sol. B = μH
or B = [0.4/H + 12 × 10^(-4)] H
or 1 = 0.4 + 12×10^(-4) H
H = 500 A/m

S2. Ans.(a)
Sol. Φ = BA = Weber*m2/m2 = Weber

S3. Ans.(c)
Sol. Example of diamagnetic material are: – Silver, lead, copper, water, mercury, carbon.
Aluminium & oxygen: paramagnetic
Nickel: ferromagnetic
Note:
1.If a value of μ_r is slightly less than unity then it is a diamagnetic material (μ_r< 1).
2. If the value of μ_ris slightly greater than unity, then it is paramagnetic material (μ_r≥ 1).
3. If the value of μ_ris very large than unity 〖(μ〗_r>>1), then it is the ferromagnetic material.

S4. Ans.(d)
Sol. The following are the advantages of the moving iron instruments.
Universal use – The MI instrument is independent of the direction of current and hence used for both AC and DC.
Less Friction Error – The friction error is very less in the moving iron instrument because their torque weight ratio is high. The torque weight ratio is high because their current carrying part is stationary and the moving parts are lighter in weight.
Cheapness – The MI instruments require a smaller number of turns as compared to PMMC instrument. Thus, it is cheaper.

S5. Ans.(d)
Sol. In a shaded pole single-phase motor, the revolving field is produced by the use of
Shading coils.
The shaded pole induction motor is simply a self-starting single-phase induction motor whose one of the poles is shaded by the copper ring. The copper ring is also called the shaded ring. This copper ring act as a secondary winding for the motor.

S6. Ans.(a)
Sol. 1 pole pitch=180⁰ electrical

S7. Ans.(a)
Sol. Demand factor = (Maximum demand)/(connected load)
Demand factor is less than 1.

S8. Ans.(b)
Sol. Plant capacity factor = (Peak load)/(Plant capacity)×load factor
If maximum demand = Plant capacity then
Load factor = capacity factor
And reserve capacity = Plant capacity – max. demand
= 0

S9. Ans.(c)
Sol. Diversity factor = (Sum of indivdula max.demand)/(Max demadnd of plant)
It is always greater than 1.

S10. Ans.(c)
Sol. Starting torque is directly proportional to the rotor resistance of an induction motor.
〖i.e.Tst α R2
In slip ring induction motor, the ends of the rotor windings are externally connected by a variable rheostat. So more the resistance, more the torque. When we add resistance to the rotor, the torque is high, the slip is high and the current is reduced. Therefore, slip ring induction motor is used for providing high starting torque.

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.