UPPSC-LECTURER'21 EE: Daily Practices Quiz 09-NOV-2021

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UPPSC polytechnic lecturer QUIZ: EE

UPPSC-LECTURER’21 EE: Daily Practices Quiz 09-NOV-2021

Each question carries 3 marks.
Negative marking: 1 mark each or 1/3rd
Total Questions: 06
Time: 08 min.

Q1. Which among the following is a current controlled device?
(a) FET
(b) BJT
(c) IGBT
(d) MOSFET

Q2. When the firing angle α of a single phase fully controlled rectifier feeding constant dc current into the load is 30°, the displacement power factor of the rectifier is
(a) 2
(b) 5
(c) √3/2
(d) √2/3

Q3. The primary and secondary voltages of an autotransformer are 250 V and 200 V, respectively. The secondary current is 100 A. What will be the economy of copper in the auto-transformer?
(a) 50%
(b) 80%
(c) 60%
(d) 40%

Q4. Due to the inductance in the pressure coil of dynamometer type wattmeter, the reading will be
(a) High for both leading and lagging power factors
(b) Low for both leading and lagging power factors
(c) High for lagging power factor and low for leading power factor
(d) Low for lagging power factor and high for leading power factor

Q5. If the number of turns of a coil is increased, its inductance
(a) Is decreased
(b) Remains the same
(c) Is increased
(d) None of the above

Q6. A parallel RLC circuit resonates at a frequency of 2 MHz. At a frequency 2.2 MHz, the nature of circuit impedance will be
(a) Resistive
(b) Inductive
(c) Capacitive
(d) None of the above

SOLUTIONS

S1. Ans.(b)
Sol.
Current Controlled Device: BJT
Voltage-controlled devices: MOSFET, IGBT, FET.

S2. Ans.(c)
Sol. Displacement factor=cos⁡α=cos⁡30^0= √3/2
NOTE:
Displacement factor for half controlled bridge converter=cos⁡α/2

S3. Ans.(b)
Sol. Transformation ratio(K)= LV side/HV side=200/250=0.8
∴ Economy of copper = K = 0.8 = 80%.

S4. Ans.(c)
Sol. Because of the inductance, the current of the pressure coils lags behind the voltage. Thus, the power factor of the wattmeter becomes lagging, and the meter reads high reading and vice-versa.

S5. Ans.(c)
Sol. the inductance of a coil is given by: L=(N^2µA)/l
∴L α N^2
Hence, L increases if the number of turns(N) increases.

S6. Ans.(c)
Sol. As given frequency is greater than the resonant frequency. So, nature will be capacitive for a parallel resonant circuits.