Quiz: Electrical Engineering 2 April 2020

Quiz: Electrical Engineering
Exam: UPSSSC JE
Topic: Miscellaneous

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

Q1. Which dc motor is generally preferred for cranes and hoists?
(a) Series motor
(b) Shunt motor
(c) Cumulatively compound motor
(d) Differentially compound motor
L1Difficulty 3
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QCreator Balwant Kumar

Q2. Two heaters rated at 1000 W, 250 V each are connected in series across a 250 V, 50 Hz
AC mains. The total power drawn from the supply would be
(a) 1000 watt
(b) 500 watt
(c) 250 watt
(d) 2000 watt
L1Difficulty 3
QTags Alternators
QCreator Balwant Kumar

Q3. A synchronous phase modifier as compared to a synchronous motor used for mechanical
load has
(a) larger shaft and higher speed.
(b) smaller shaft and higher speed.
(c) larger shaft and smaller speed.
(d) smaller shaft and smaller speed.
L1Difficulty 3
QTags Alternators
QCreator Balwant Kumar

Q4. Which motor has widest variety of methods for speed control?
(a) DC shunt motor
(b) Synchronous motor
(c) Slip-ring induction motor
(d) Schrage motor
L1Difficulty 3
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QCreator Balwant Kumar

Q5. The efficiency for maximum power transfer to the load is
(a) 25%
(b) 50%
(c) 75%
(d) 100%
L1Difficulty 3
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QCreator Balwant Kumar

Q6. The transformer used for AC welding sets is
(a) booster type
(b) step up type
(c) step down type
(d) equal turn type
L1Difficulty 3
QTags Alternators
QCreator Balwant Kumar

Q7. A transformer has maximum efficiency at full load, when iron losses are 800 watts,
copper losses at half load will be
(a) 1600 W
(b) 800 W
(c) 400 W
(d) 200 W
L1Difficulty 3
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QCreator Balwant Kumar

Q8. Two d.c. series motors are connected in series to produce a torque T. Now if the motors
are connected in parallel, the torque produced will be
(a) T/4
(b) T/2
(c) 2 T
(d) 4 T
L1Difficulty 3
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QCreator Balwant Kumar

Q9.Inter-poles are used in __________.
(a) lap wound machines
(b) wave wound machines
(c) both lap and wave wound machines
(d) None of these
L1Difficulty 3
QTags Alternators
QCreator Balwant Kumar

Q10.In motor circuit static frequency changers are used for __________.
(a) improved cooling
(b) Power factor improvement
(c) reversal of direction
(d) speed regulation
L1Difficulty 3
QTags Alternators
QCreator Balwant Kumar

Solution

S1. Ans.(a)
Sol. Cranes and hoists require high staring torque. For the higher starting torque requirements dc series motors or 3-ϕ induction motors are used Particularly for traction (Railway) application dc series motor is used. Here, the option (a) is correct.

S2. (b)
The heaters of 1000 W, 250 V will have the resistance of
R=V²/P=((250)²)/1000=62.5 Ω

∴ R_eq=125 Ω
∴ P=V²/R_eq =((250)²)/125=500 W

S3. Ans.(b)
Sol. Synchronous phase modifier is nothing but over excited synchronous motor operating at leading power factor and used for the voltage control. This differs from a normal synchronous motor as it has no external load on shaft due to which it has smaller shaft. Also for no load the motor can operate at highest possible speed. Hence, for synchronous phase modifier the design consideration parameters are smaller shaft and higher speed.
S4. Ans.(a)
Sol. Dc shunt motor has the best speed regulation i.e. for the variation in load the speed changes minimum. Hence, we can have a wide ranged speed in the shunt motor applications.
Speed regulation = (N(NL)-N(FL))/N(FL) This speed regulation is low in case of shunt motor.
S5. (b) At maximum power transfer condition, the load resistance = Source resistance
i.e., R_L=R_S (in dc circuit)

From above circuit,
I=V_s/(R_s+R_L )
∴Load power=I^2 R_L = Output power
Loss = I^2 R_s
∵ η =output/input=output/(output+loss)
∵ η=(I^2 R_L)/(I^2 R_L+I^2 R_s )
∵ R_L=R_s
∴ η=(I²R_L)/(2I^2 R_L )=1/2 or η = 50%

S6. Ans.(c)
Sol. In the electric welding transformer the secondary side has low voltage and high current (typically 200 to 600 amps) because for electric welding high current is required. Hence, the transformer required for welding is step down type.
S7. Ans.(d)
Sol. For maximum efficiency (ηmax):
Copper loss (I^2 R)=Iron loss (p_i) So, for ηmax at full load
I_ft^2.R=800 W
At half load, I=I_η/2
∴ Coper loss at half load
P_cu=(I_η/2)^2.R
=800/4=200 Watts

S8. Ans.(a)

Sol. For DC motors:
Torque, T= kϕ I_a
ϕ → flux and I_a → current
For series motor ϕ ∝ I_a
∴ T=k.I_a.I_a
∝ T∝I_a^2
For series connection:
Draw input power = V.I

T_A=T_B=kϕ.I=I
Or T=kI^2 …..(i)
For parallel connection:

Each motor will hve the half valu of current now (if the drawn power is constant).
∴ T_parallel∝ϕ.I/2 (for each motor)
Or T_parallel∝1/2 .1/2 (asϕ ∝ I)
Or, T_parallel=kI²/4
From equation (i) and (ii), T_parallel=T/4

S9. (c)
Both lap and wave wound machines.
S10. (d)
Static frequency changers are used for speed regulation.

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