Engineering Jobs   »   Quiz: Electrical Engineering 30 July 2020

Quiz: Electrical Engineering 30 July 2020

Quiz: Electrical Engineering
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. The amount of feedback applied to an amplifier reduces the gain by a factor of 10. The bandwidth
(a)Decreases by factor of 10
(b)Increases by factor of 10
(c)Remains the same
(d)None of the above

Q2. Transformer utilization factor of TUF = K signifies:
(a) That transformer for rectifier should be 1/K times larger than that for ac source
(b) That transformer for rectifier should be K-1 times larger than that for ac source
(c) That transformer for rectifier should be 1-K times larger than that for ac source
(d) That transformer for rectifier should be K time larger than that for ac source

Q3. A synchronous generator is feeding power to infinite bus bars at unity power factor. Its excitation is now increased. It will feed
(a)The same power but at leading power factor
(b)The same power but at lagging power factor
(c)More power at unity power factor
(d)Less power at unity power factor

Q4. A 400/200 volts transformer has pu impedance of 0.05. the HV side voltage required to circulate full load current during short circuit test is
(a)20 V
(b)40 V
(c)10 V
(d)5 V

Q5. In a D.C. machine, fractional pitch winding is used
(a)to increase the generated voltage
(b)to reduce sparking
(c)to save the copper because of shorter end connections
(d)due to (b) and (c) above

Q6. Why are shunt reactors connected at the receiving end of long transmission line system?
(a)To increase the terminal voltage
(b)To compensate voltage rise caused by capacitive charging at light load
(c)To improve power factor
(d)None of these

Q7. Which of the following transmission line have more initial cost?
(a)Overhead Transmission
(b)Underground transmission
(c)Both have almost the same initial cost
(d)None of the above

Q8. Which one of the following is the CORRECT relation between the peak and average value of an alternating current?
(a) Iavg=0.637IPeak
(b) Iavg=1.414IPeak
(c) Iavg=0.7071IPeak
(d) Iavg=0.8241IPeak

Q9. With regard to filtering property, lead compensator is
(a)Low pass filter
(b)High pass filter
(c)Band pass filter
(d)Band reject filter

Q10. The maximum phase occurs at the ………. of the two corner frequencies?
(a) arithmetic mean
(b) geometric mean
(c)Both a& b
(d)None of the above


S1. Ans.(b)
Sol. for an amplifier, the product of gain-bandwidth is always constant. So, if gain reduces by a factor of 10 then the bandwidth increases by a factor of 10.

S2. Ans.(a)
Sol. The transformer utilization factor (TUF) of a rectifier circuit is defined as the ratio of the DC power available at the load resistor to the AC rating of the secondary coil of a transformer.
TUF = (DC POWER output (pdc))/(Effective VA rating of transformer)
TUF indicate how much is the utilization of the transformer in a circuit.
▭(TUF=Pdc/Pac(rated) )

▭(■(TUF for full wave rectifier=0.812@TUF for half wave rectifier=0.287))

For ideal condition, TUF=1 and apart from this, TUF is always less than 1.

S3. Ans.(b)
Sol. from the V-curve of an alternator, it is clear that when excitation is increased, the alternator feeds a lagging power factor.

S4. Ans.(a)
Sol. pu quantity will be same on both side.
Given Z_pu=0.05
And we know 〖Vsc〗pu=〖Isc〗pu×Zpu and 〖Isc)pu=I_rated/(I_base)
As rated current on HV side is same as base current on HV side. Then, 〖I_sc〗_pu=1
And 〖Vsc〗pu=〖Isc〗pu×Zpu=1×0.05=0.05
∴Vsc (HV)=〖Vsc〗pu×Vbase=0.05×400=20 V

S5. Ans.(d)
Sol. Some advantages of short pitch winding are:
Due to shortening span, the copper required is less.
Low copper losses
Improve waveform due to reduction in harmonic
Fractional Pitch winding reduces sparking in DC machines

S6. Ans.(b)
Sol. Shunt Reactor compensation at the receiving end help to reduce the effect of capacitance thus reducing the Ferranti effect.
Shunt Reactor absorbs the excess reactive power during no load or light load condition and thus helps in stabilizing the voltage of Transmission Line.

S7. Ans.(b)
Sol. As the voltage level increases the cost of insulation is increased therefore the underground cable is restricted to low and medium voltages.

S8. Ans.(a)
Sol. For AC, Iavg=(2Ip)/π=0.637 Ip

S9. Ans.(b)
Sol. In terms of filtering property, lead compensator acts as high pass filter.
Following are the properties of lead compensator….
For sinusoidal input, output is leading
Acts as high pass filter
Reduces rise time and peak overshoot
Increases bandwidth & stability of the system.

S10. Ans.(b)
Sol. The maximum phase occurs at the GM (geometric mean) of the two corners

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