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

Leave a comment

Your email address will not be published. Required fields are marked *