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DFCCIL’21 EE: Daily Practice Quiz. 23-Sep-2021

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DFCCIL’21 EE: Daily Practices Quiz

DFCCIL’21 EE: Daily Practices Quiz 23-Sep-2021

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

Q1. In a split-phase motor, the running winding should have:
(a) Low resistance and low inductance
(b) Low resistance and high inductance
(c) High resistance and low inductance
(d) High resistance and high inductance

Q2. The cause of surge voltage in power systems are
(a) Faults
(b) Lightning
(c) Switching operation
(d) Any of the above

Q3. By increasing spacing between phase conductors, the line capacitance will
(a) Increase
(b) Decrease
(c) Remain same
(d) None of the above

Q4. A shunt generator has a critical field resistance of 200 Ω at a speed of 800 rpm. If the speed of the generator is increased to 1,000 rpm, what is the change in the critical field resistance of the generator?
(a) Remains the same at 200 Ω
(b) Decreases to 160 Ω
(c) Increases to 250 Ω
(d) Increases to 312.50 Ω

Q6. A 6-pole, 50 Hz, 1-phase induction motor runs at a speed of 900 r.p.m. The frequency of currents in the cage rotor will be
(a) 5 Hz, 50 Hz
(b) 5 Hz, 55 Hz
(c) 5 Hz, 95 Hz
(d) 55 Hz, 95 Hz


S1. Ans.(b)
Sol. The Split Phase Motor is also known as a Resistance Start Motor. It has a single cage rotor, and its stator has two windings known as main winding and starting winding. Both the windings are displaced 90 degrees in space.
The main /running winding has very low resistance and a high inductive reactance.
the starting winding has high resistance and low inductive reactance.

S2. Ans.(d)
Sol. Overvoltage (surges) on power systems are due to various causes…
switching operation

S3. Ans.(b)
Sol. C=(2πϵ0)/ln⁡(d/r)
Where; d=spacing between conductors
And r=radius of conductors
Therefore, if we increase the spacing(d) between the phase conductors, the line capacitance will decrease.

S4. Ans.(c)
Sol. For generator, Eg ∝ N. So, the critical field resistance also increases slightly with speed.
Critical field resistance(Rf ) α N
∴(Rf2 )/Rf1 =N2/N1
⇒R_f2=1000/800×200=250 Ω

S6. Ans.(c)
Sol. Synchronous speed for forward rotating flux, Ns = 120 f/P = 120 × 50/6 = 1000 r.p.m.
Therefore, slip for forward rotating flux is
sf=s=(Ns-N)/Ns =(1000-900)/1000=0.1
∴ Rotor current frequency for sf is f’ = sf × f = 0.1 × 50 = 5 Hz.
Slip for backward rotating flux sb =(2 – s) =(2 – 0.1) = 1.9. Therefore, rotor current frequency, f” = (2 – s) × f = 1.9 × 50 = 95 Hz.

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