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Quiz: Electrical Engineering 26 June 2021

Quiz: Electrical Engineering
Exam: AE/JE
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/4 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. 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

Q3. The starting current of a three-phase induction motor is 5 times the rated current, while the rated slip is 4%. The ratio of starting torque to full-load torque is…….
(a) 0.6
(b) 0.8
(c) 1.0
(d) 1.2

Q4. Fig. shows part of a closed circuit. What is the value of V_A–V_B?

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

Q5. 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

Q6. 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

Q7. 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

Q8. A 3-phase synchronous motor draws 100 A from the line at unity power factor at rated load. Considering the same line voltage and load, the line current at a power factor of 0.5 lagging is:
(a) 100 A
(b) 50 A
(c) 200 A
(d) 150 A

Q9. Match List-I (instrument Type) with List-II (Used for) the following:

CODES:
A B C
(a) 1 2 1
(b) 1 3 1
(c) 1 2 3
(d) 3 1 2

Q10. Given a unity feedback system with G(s)=K/(s(s+4))
The value of K for damping ratio of 0.5 is
(a) 1
(b) 2
(c) 4
(d) 16

SOLUTIONS

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

S3. Ans.(c)
Sol. Tst/Tfl =(Ist/Ifl )^2×sfl=((5Ifl)/Ifl )^2×0.04=25×0.04=1

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

S6. Ans.(c)
Sol. Reserve capacity = Plant capacity – Max. demand

S7. Ans.(c)
Sol. Line current,` I_L=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.

S8. Ans.(c)
Sol. P=VI cos⁡ϕ
For same line voltage and load, P=constant
i.e., I cos⁡ϕ=constant
`I1 cos⁡ϕ1=I2 cos⁡ϕ2`
∴100×1=I_2×0.5
⇒I_2=200 A

S9. Ans.(c)
Sol. PMMC: for DC measurement
Moving iron: measures rms value
Moving iron connected through CT: for only AC measurement
Electrodynamometer: AC and DC measurement

S10. Ans.(d)
Sol. Characteristic equation is s² + 4s +k = 0
Compared to second ordered standard characteristic equation:
Natural frequency ωn = √k and 2ξωn = 4
Given: ξ = 0.5,
∴ωn = 4 = √k Therefore, k = 16

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