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Quiz: Electrical Engineering 05 June 2021

Quiz: Electrical Engineering
Exam: DFCCIL
Topic: Miscellaneous

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

Q1. A smaller air gap in induction motor helps to
(a) Reduce the chance of cogging
(b) Reduce the magnetizing current
(c) Reduce the chance of crawling
(d) Increase the starting torque

Q2. A full-pitched coil in a 6-pole machine has a mechanical span of
(a) 30⁰
(b) 90⁰
(c) 60⁰
(d) 180⁰

Q3. If the no load voltage of a certain generator is 210 V and the rated voltage is 200 V, then the voltage regulation is
(a)2.5%
(b) 5%
(c) 10 %
(d) 15 %

Q4. The inductance of a high Q inductor can be measured using a
(a) Schering bridge
(b) Wein bridge
(c) Maxwell bridge
(d) Hay’s bridge

Q5. Illumination of one lumen per sq. meter is called . . .
(a)lumen meter
(b)lux
(c)foot candle
(d)candela

Q6. Voltage regulation due to the Ferranti effect may be:
(a) Zero
(b) Positive
(c) Negative
(d) All of the above

Q7. The time taken by a fuse to interrupt the circuit in the event of a fault is known as
(a) Current rating
(b) Fusing factor
(c) Fusing current
(d) Cut-off factor

Q8. Find final value of the following function:
F(s) = 12(s+1)/s(s+2) (s-3)
(a) 0
(b) -2
(c) 2
(d) can’t be determined

Q9. Kirchhoff’s Current law is based on the law of conservation of _
(a) energy
(b) momentum
(c) mass
(d) charge

Q10. A series R-C circuit having R = 5 Ω and C = 100 μF has a time constant of
(a) 20 μsec.
(b) 5 × 10^(-4) sec.
(c) 0.005 sec.
(d) 500 sec.

SOLUTIONS
S1. Ans.(b)
Sol. Due to smaller air-gap in an Induction motor:
Reluctance is minimum
Inductance is high
Magnetizing current is of smaller value as I_m α 1/L

S2. Ans.(c)
Sol. The coil span of a full pitch coil is always 180° electrical.
Ө_e=P/2 Ө_m
∴180⁰=6/2×Ө_m
⇒Ө_m=60⁰

S3. Ans.(b)
V.R.% = (No load voltage-Full load voltage )/(Full load voltage)×100
=(210-200)/200×100=5%

S4. Ans.(d)
Sol. Bridges used for measurement of inductance:
Anderson bridge (Q<1). Maxwell bridge (1<Q<10).

Hay bridge Q>10

S5. Ans.(b)
Sol. The amount of light that is cast on a surface is called illuminance, which is measured
in lux. This can be thought of as light intensity within a specific area.
LUX is an abbreviation for Lumens per square meter.

S6. Ans.(c)
Sol. Voltage regulation: (VS-VR)/VR
Due to Ferranti effect: VS<VR
Hence, voltage regulation is negative in case of Ferranti effect.

S7. Ans.(d)
Sol. The time taken by a fuse to interrupt the circuit in the event of a fault is known as cut-off factor.
Current rating: minimum value of current at which the fuse element or fuse wire melts.
Fusing current: current at which fuse melts.
Fusing factor: ratio of minimum fusing current to current rating of fuse element

S8. Ans.(d)
Sol.
final value=lim(s⟶0)⁡sF(s) is not applicable as system is not stable due to one pole on RHS.

S9. Ans.(d)
Sol. Kirchhoff’s current law is based on the law of conservation of charge i.e., charge that flows in = charge that flows out.

S10. Ans.(b)
Sol. Time constant is
τ = RC = 5 × 100 × 10^(-6)= 5 × 10^(-4) sec.

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