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# Quiz: Electrical Engineering 27 May 2020

Quiz: Electrical Engineering
Exam: UPSSSC JE
Topic: Network theorems
Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. Eight cells each of e.m.f. E and internal resistance r are connected in series. If due to oversight, two cells are connected wrongly, then equivalent e.m.f. and internal resistance of the combination are:
(a) 10 E and 6r
(b) 6 E and 10r
(c) E and 3r
(d) 4E and 8r

Q2. An ammeter has a resistance of 60 Ω and requires 10 mA for full-scale deflection. The shunt resistance S required to allow the meter to measure currents up to 100 mA is
(a) 5.56 Ω
(b) 2.5 Ω
(c) 6.67 Ω
(d) 8.75 Ω

Q3. Two resistors of resistances 6 Ω and 9 Ω are connected in parallel. The power dissipated in 6 Ω resistor is 15 W. The power dissipated is 9 Ω resistor is
(a) 10 W
(b) 24 W
(c) 6 W
(d) 16 W

Q4. Three capacitors of capacitance 12 μ F each are available. The minimum and maximum capacitances which may be obtained are
(a) 8 μ F, 24 μ F
(b) 4 μ F, 36 μ F
(c) 6 μ F, 33 μ F
(d) none of above

Q5. The desired voltage regulation for a single phase, 50 Hz transformer is zero. Its pu ohmic drop is 4% and pu reactive drop is 4%. The required power factor angle for the operation in degrees
a) 600
b) 900
c) 300
d) 450

Q6. For a separately excited dc machine of 25 kW, 250 V and armature resistance of 0.25 ohms, is running at 3000 rpm with supply of 255 V.
The electromagnetic power produced at the armature in kilo watts is
a) 0.5
b) 10
c) 5
d) 8

Q7. An alternator is to be used for generation purpose in an aircraft. The most suited operating frequency of the alternator should be
a) of 400 Hz to remove extra bulk
b) of 400 Hz to reduce losses
c) of 50 Hz to remove extra bulk
d) none of the mentioned

Q8. A 3-phase synchronous alternator is connected to the infinite bus is over excited, which corresponds to
a) inductor
b) capacitor
c) variable inductor
d) any of the mentioned

Q9. Separately excited dc generators are used in
a) Ward Leonard system of speed control
b) Hopkinson’s testing
c) Voltage control
d) None of the mentioned

Q10. Maximum torque in dc series motor is limited by
a) commutation
b) heating
c) field control
d) all of the mentioned

Solution

S1. (d)
Equivalent e.m.f. = E + E + E + E + E+ E – E – E=4E
Internal resistance = r+ r + r + r + r + r +r + r = 8r

S2. (c)
Full-scale voltage across meter = IR=10 mA×60 Ω=0.6 V. When circuit current is 100 mA, current through shunt = 100 – 10 = 90 mA.
∴ S=(0.6 V)/(90 mA)=6.67 Ω
S3. (a)
P=V^2/R or V=√(PR.) Since voltage is same in a parallel circuit, P_1 R_1=P_2 R_2 or 6×15 = P_2×9
∴P_2=10 W

S4. (b)
The capacitance is minimum when the three capacitors are connected in series and maximum when connected in parallel. i.e. 4 μ F and 36 μ F.

S5. (d)
Sol. The power factor angle is, tan (φ)=pu resistive drop/pu reactive drop
= 4/4= 1
tan^(-1)⁡1= 45°
S6. (c)
Sol. The current in armature, Ia = 255-250/0.25 = 20 A
Power = Eb*Ia = 250*20 = 5000 W
S7. (a)
Sol. The components to be used in aircrafts are made to operate at very high frequencies so that the size of the components to be used is reduced.
S8. (b)
Sol. In the mentioned condition the generator will correspondence as capacitor.
S9. (a)
Sol. Separately excited dc generators are used in Ward Leonard system of speed control.

S10. (a)
Sol. Commutation is the process which reduces the induced emf and so the torque.

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