Engineering Jobs   »   Quiz: Electrical Engineering 27 Mar 2021

Quiz: Electrical Engineering 27 Mar 2021

Quiz: Electrical Engineering
Topic: Miscellaneous

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

Q1. In an induction motor, if the air gap is increased
(a) Breakdown torque will reduce
(b) Efficiency will improve
(c) Power factor will be lower
(d) Speed will reduce

Q2. In a series R-C circuit, the AC current is:
(a) Lagging behind the applied voltage
(b) Leading the applied voltage
(c) Out of phase with applied voltage
(d) In phase with the applied voltage

Q3. A fuse 10 A current rating of fuse element and it can be blown out at minimum fusing current of 15 A. The fusing factor will be:
(a) 1
(b) 0.66
(c) 3
(d) 1.5

Q4. The advantage of PMMC instrument is that it:
(a) Has low torque to weight ratio
(b) Can be used on both AC and DC
(c) Has high torque to weight ratio of moving parts
(d) Is free from friction error

Q5. For measurement of which of the following is Low Power Factor (LPF) wattmeter is NOT suitable?
(a) Load test on induction motor
(b) Open circuit test on single phase transformer
(c) Measurement of power in resistive loads
(d) Measurement of power in inductive loads

Q6. Illumination is measured using which of the following instruments?
(a) Voltmeter
(b) Stroboscope
(c) pH meter
(d) Luxmeter

Q7. Which of the following has the largest wavelength?
(a) Violet
(b) Red
(c) Green
(d) Blue

Q8. The flicker effect of fluorescent lamps is more pronounced at:
(a) higher frequencies
(b) Lower voltages
(c) Lower frequencies
(d) Higher voltages

Q9. A 230 V,50 Hz, 4 pole single phase induction motor is rotating in the clockwise forward direction at the speed of 1425 rpm. If the rotor resistance at standstill is 7.8 ohm, then what will be the effective rotor resistance in the backward branch of the equivalent circuit?
(a) 2 Ω
(b) 4 Ω
(c) 78 Ω
(d) 156 Ω

Q10. Which of the following statement is true for no-load current of the transformer?
(a) has high magnitude and low power factor
(b) has high magnitude and high-power factor
(c) has small magnitude and high-power factor
(d) has small magnitude and low power factor

S1. Ans.(c)
Sol. If the air gap is increased, the stator will draw more magnetizing current from the supply to maintain the flux in the air gap to the rotor. This will reduce the power factor.

S2. Ans.(b)
In case of pure resistive circuit, the phase angle between voltage and current is zero i.e., both are in phase.
In pure inductive circuit, the current lags the voltage by 90°
In series RL circuit, the current lags the voltage by an angle in between 0 to 90°
In case of pure capacitive circuit, phase angle is 90° i.e., current leads the voltage by 90°.
In case of series RC circuit, AC current leads the voltage by an angle in between 0 to 90°.

S3. Ans.(d)
Sol. Fusing factor=(minimum fusing current)/(rated current)=15/10=1.5
Note: the value of fusing factor is always more then 1.

S4. Ans.(c)
Sol. PMMC instruments:
High torque to weight ratio
Sensitivity high
Less friction errors
PMMC instruments (i.e., D’Arsonval meters) are only used for measuring the Direct Current (DC) current.

S5. Ans.(c)
Sol. LPF wattmeter:
Works with low pf system
Resistive loads have unity, it will give large error in such system.

S6. Ans.(d)
Sol. Illumination is measured using the “Luxmeter”. It measures the brightness in the form of electrical current stream.

S7. Ans.(b)
Sol. increasing order of wavelength: –

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S8. Ans.(c)
Sol. Flicker effect of fluorescent lamp is more pronounced at “Lower frequencies” and can be due to loosen connection of bulb or cold condition or tube needs replacement (i.e., black spot at end).

S9. Ans.(b)
Sol. In single phase induction motor, the slip of the rotor w.r.t forward and backward rotating field will be s and (2-s) respectively.
And N_S=120f/p=(120×50)/4=1500 rpm
∴slip(s)= (1500-1425)/1500=0.05
∴backward branch resistance=R_2/(2-s)=7.8/(2-0.05)=4Ω

S10. Ans.(d)
Sol. The no load current is the current drawn by the transformer winding when no external load is connected to the secondary of the transformer that is the secondary circuit is open. Theoretically this current should be zero. But actually, it is not zero but a small factor of the full load current.
Since no-load current lags voltage by the angle of nearly 90⁰, power factor being equal to cosine of the angle between current and voltage, it will be equal to value which is near to 0. Thus, power factor will be low.

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