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UPRVUNL’21 EE: Daily Practices Quiz 09-Aug-2021

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

Q1. For a series RLC circuit, R = 10Ω, inductive reactance X_L = 20 Ω and capacitive reactance X_C = 20 Ω. If the applied voltage is 100 V AC, then current in the circuit is:
(a) 5A
(b) 10A
(c) 20A
(d) 0A

Q2. To measure the current in a 1-𝟇 transmission line, the primary winding of current transformer is connected in:
(a) Series with the line carrying current
(b) Parallel with the line carrying current
(c) Parallel to the load
(d) Parallel to the source

Q3. In a 4-pole dc machine, a coil span of 120 electrical degrees is equal to:
(a) 60 mechanical degrees
(b) 120 mechanical degrees
(c) 180 mechanical degrees
(d) 240 mechanical degrees

Q4. 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) 625 W
(b) 525 W
(c) 500W
(d) 600W

Q5. For induction motors at low values of slip:
(a) The torque slip is approximately a hyperbola
(b) The torque slip is approximately a straight line
(c) The torque slip curve is at its maximum
(d) The torque slip is approximately a parabola

Q6. A generating station supplies the following loads 15000 kW, 8500 kW, 6000 kW and 450 kW. The station has maximum demand of 22000 kW. Calculate the diversity factor.
(a) 0.68
(b) 1.34
(c) 1.91
(d) 0.52

SOLUTIONS
S1. Ans.(b)
Sol. according to Question, X_L=X_C = 20 Ω.
So, it is a case of series Resonance.
For RLC circuit at Resonance Z = R = 10 Ω
∴ I = V/R=100/10
= 10 A

S2. Ans.(a)
Sol. The primary winding of the current transformer is connected in series with the transmission line whose current is to be measured whereas the potential transformer is connected in parallel with the line

S3. Ans.(a)
Sol. Ө_electrical=P/2 Ө_mechanical
Where, P= Numbers of pole
∴Ө_mechanical=2/4×120=60

S4. Ans.(a)
Sol. p_cu/P_md =S/((1-s))
P_cu=s/(1-S)×P_md
= 0.04/(1-0.04)×15×10³
= 625 watts

S5. Ans.(b)
Sol. For induction motor –
T=(KS E₂²R₂)/(R₂²+(S×_2)²)
at low value of slip, S×_2≪R₂
⇒ T=(KSE₂²R₂)/(R₂²)
⇒ T∝S

S6. Ans.(c)
Sol. Diversity factor = (sum of individual max.demand)/(coincident maximum demand of whole system)
=(15000+12000+8500+6000+450)/22000
=41950/22000
=1.91
Note: If the value of the diversity factor is greater than 1, then it is a good diversity factor, and 1.0 represents a poor diversity factor. A high diversity factor has the effect of reducing the maximum demand.

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