DFCCIL’21 EE: Daily Practice Quiz. 17-July-2021

Each question carries 1 mark.
Negative marking: 1/4 mark
Total Questions: 06
Time: 08 min.

Q1. D.C. shunt motors are used in those applications where ……… is required.
(a) high starting torque
(b) practically constant speed
(c) high no-load speed
(d) variable speed

Q2. Cumulatively compounded motors are used where we require ………
(a) variable speed
(b) poor speed regulation
(c) sudden heavy loads for short duration
(d) none of the above

Q3. For 20% increase in current, the motor that will give the greatest increase in torque
is ……. motor.
(a) shunt
(b) series
(c) cumulatively compounded
(d) differentially compounded

Q4. 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

Q5. In a split phase induction motor, the running winding should have
(a) Low resistance and high inductance
(b) High resistance and low inductance
(c) Low resistance as well as low inductance
(d) High resistance as well as high inductance

Q6. Which one of the following statements is correct?
Corona loss increase with
(a) decreases in conductor size and increase in supply frequency
(b) increase in both conductor size and supply frequency
(c) decrease in both conductor size and supply frequency
(d) increase in conductor size and decrease in supply frequency

SOLUTIONS
S1. Ans.(b)
Sol. Practically constant speed

S2. Ans.(c)
Sol. Cumulatively compounded motors are used where we require sudden heavy loads
for short duration.

S3. Ans.(b)
Sol. Series motor because
T∝I_a^2 (Before saturation)

S4. 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

S5. Ans.(a)
Sol. In a split phase induction motor:
The main or running winding has very low resistance and a high inductive reactance
The starting winding has high resistance and low inductive reactance.
Split phase induction motor is also known as resistance start motor.

S6. Ans.(a)
Sol. corona loss = 242.2/δ×(f+25) √(r/D) (V-V_c )^2×10^(-5) kw\/ km\/phase
Where;
f = System frequency
r = radius of conductor
D = distance between conductor
V = phase voltage (kv,rms)
V_c= Critical disruptive phase voltage
Therefore,
Corona loss ∝ supply frequency and
Corona loss ∝ (Vph – Vc)²
And Vc ∝ r
Therefore, corona loss increases with the decrease in conductor size and increase in supply frequency.

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