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# RVUNL’21 EE: Weekly Practices Quiz 08-Aug-2021

RVUNL’21 EE: Weekly Practices Quiz 08-Aug-2021

Each question carries 1 mark.
Negative marking: 1/4 mark
Total Questions: 12
Time: 12 min.

Q1. Three conductance G₁ = 0.5 S, G₂ = 0.3 S and G₃ = 0.2 S are in parallel. If the total circuit current is 4 A, current in G₁ is
(a) 1.2 A
(b) 2 A
(c) 0.8 A
(d) None of the above

Q2. The rotor power output of a 3-phase induction motor is 15 kW and the corresponding slip is 4 %. The rotor copper losses will be,
(a) 500 W
(b) 525 W
(c) 625 W
(d) 600 W

Q3. Corona loss does not depend on
(a) Atmosphere
(b) Conductor size
(c) Line voltage
(d) Height of the conductor

Q4. Which one of the following relays has the capability of anticipating the possible major fault in a transformer?
(a) Differential relay
(b) Overcurrent relay
(c) Buchholz relay
(d) Over-fluxing relay

Q5. Slip of a 3-phase induction motor may be expressed as
(a) Rotor power input/rotor copper loss
(b) Rotor copper input/rotor core loss
(c) Rotor copper loss/rotor power input
(d) Rotor copper loss/ input power

Q6. A dc series motor drawing an armature current of I_a is operating under saturated magnetic conditions, torque developed in the motor is proportional to
(a) 1/I_a
(b) 1/I_a^2
(c) I_a^2
(d) I_a

Q7. Fusing factor of fuse is always
(a) More than 1
(b) Less than 1
(c) Infinity
(d) Zero

Q8. In a three-phase induction motor, the ratio of starting current to full load current is 12.56 and the ratio of starting torque to full load torque is 1.6. percentage slip at full load is………
(a) 10 %
(b) 1%
(c) 12.56 %
(d) None of the above

Q9. The output signal of a common-collector amplifier always:
(a) Larger than the input signal
(b) In phase with the input signal
(c) Out of phase with the input signal
(d) Exactly equal to the input signal

Q10. An amplifier has a gain of 20 without feedback. If 10% of the output voltage is fed back by means of resistance negative feedback circuit, the overall gain would be
(a) 19.80
(b) 16.55
(c) 10.85
(d) 6.67

Q11. Phase reversed of 180⁰ as compared to input occurs in the output of
(a) CE amplifier
(b) CB amplifier
(c) CC amplifier
(d) All of the above

Q12. Which of the following is/are represents lag compensator?
(a) ((s+2))/((s+1))
(b) ((s+2))/((s+5))
(c) Both a & b
(d) None of the above

SOLUTIONS
S1. Ans.(b)
Sol. Total conductance, G_T = G₁ + G₂ + G₃ = 0.5 + 0.3 + 0.2 = 1 S
∴ Current in G₁ = I_total×G_1/G_T =4×0.5/1=2 A

S2. Ans.(c)
Sol. Rotor power output (P_m )=15 kW
∴P_m=P_g (1-s)
∴P_g=P_m/(1-s)=15/(1-0.04)=15.625 kW
Rotor copper loss=sP_g=0.04×15.625=0.625 kW=625 W

S3. Ans.(d)
Sol. corona loss(P_c )=242.2 ×10^(-5)×((f+25))/δ √(r/d) (v_0-v_c )^2 kw/km/phase
Where, 𝛿=relative density of air
So, corona loss does not depend on height of conductor.

S4. Ans.(c)
Sol. Buchholz relay is suitable to anticipate major fault in a transformer.

S5. Ans.(c)
Sol. rotor power input= air gap power=(3I^2 R_2)/s
Rotor copper loss=3I^2 R_2
∴SLIP(s)=(3I^2 R_2)/((3I^2 R_2)/s)=(Rotor copper loss)/(rotor power input)

S6. Ans.(d)
Sol. We know Torque Tα ΦIa, As the motor is operating under saturation condition so torque will be independent of flux i.e., T α Ia.

S7. Ans.(a)
Sol. Fusing factor=(minimum fusing current)/(current rating of fuse)
Fusing factor of fuse is always >1.

S8. Ans.(b)
Sol. In a three-phase induction motor, T_st/T_fl =(I_st/I_fl )^2 s_fl
∴1.6=(12.56)^2 s_fl
∴s_fl=0.01 or 1%

S9. Ans.(b)
Sol. The output signal of a common-collector amplifier always in phase with input signal.

 Parameter Common Base Common Emitter Common Collector Voltage Gain High, Same as CE High Less than Unity Current Gain Less than Unity High High Power Gain Moderate High Moderate Phase inversion No Yes No

S10. Ans.(d)
Sol. Given that, A=20 and β=10%=0.1
∵feedback is Negative.
∴A_v=A/(1+Aβ)=20/(1+20×0.1 )=20/3=6.67

S11. Ans.(a)
Sol. Phase reversed of 180⁰ as compared to input occurs in the output of CE amplifier.

S12. Ans.(a)
Sol. transfer function for phase lag compensator will be
G(s)=((1+Ts))/((1+βTs))
Where, β>1.
For phase lag compensator, pole will be near to origin.
Hence, option (a) is right.

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