Quiz: Electrical Engineering
Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute
Q1. Which type of winding is generally preferred for generating large currents on DC generators?
(a) Progressive wave winding
(b) Retrogressive wave winding
(c) Lap winding
(d) Current depends on design
Q2. The range of efficiency for shaded pole motors is
(a) 95% to 99%
(b) 80% to 90%
(c) 50% to 75%
(d) 5% to 35%
Q3. A 6 pole Lap wound dc generator has 300 conductors. Emf induced per conductor is 5 volts. This generator will generate emf of:
(a) 60 V
(b) 250 V
(c) 300 V
(d) 1800 V
Q4. If the fault current is 3000 A, for an IDMT relay with a plug setting of 50% and CT ratio of 400/5, the plug setting multiplier (PSM) would be:
Q5. Transposition of conductors in large transmission lines results in
(a) Same average capacitance for various phase over the entire length of the conductor
(b) Same average inductance for various phase over the entire length of the conductor
(c) Same average capacitance and inductance for various phase over the entire length of the conductor
(d) Averaging is valid only for inductance and not capacitance
Q6. If the maximum current of a half-wave rectified wave is 10 A, its RMS value is given by
(a) 7.07 A
(b) 5.0 A
(c) 6.37 A
(d) 10 A
Q7. In a rectifier the ripple voltage is 150 mV and the dc value is 15 V, then ripple factor is given by:
Q8. What are the average and RMS values of half wave rectified output if peak voltage is 5 V?
(a) 1.59 V, 2.5 V
(b) 3.18 V, 2.5 V
(c) 1.59 V, 5 V
(d) 3.18 V, 5 V
Q9. In a common emitter configuration if β=100 and base current is 50 µA. Then the value of α is…………
Q10. Which of the following is/are represents lag compensator?
(c) Both a & b
(d) None of the above
Sol. LAP winding: large current & low voltage applications
WAVE winding: high voltage & low current applications
Sol. The range of efficiency for shaded pole motors is 5% to35%.
Shaded pole motors are used where low torque is acceptable (such as fans) and are usually less than ¼ HP. Due to their very low efficiency, shaded pole motors should only be used in applications where the motor is very small or operates for very short period of time (e.g., shower fan motor).
Sol. Emf induced per conductor=5 V
For Lap wound, number of parallel paths(A)=P=6
No. of conductor per parallel path=300/6=50
∴total emf generated by generator=50×5=250 V
Sol. Pickup current=% current setting× rated secondary current of CT=0.5×5=2.5
PSM=I_f/(Relay current setting×CT ratio)=3000/((0.5×5)×(400/5) )=15.
Sol. The transposition of the conductor equalises the mutual inductance and capacitance between the lines.
Sol. For half wave rectifier: I_rms=I_m/2=10/2=5 A
Sol. Ripple factor=(ripple voltage)/(dc value)=(150×10^(-3))/15=1/100=0.01
Sol. For half-wave rectifier: V_avg=V_m/π=5/3.14=1.59 V
And V_rms=V_m/2=5/2=2.5 V
Sol. transfer function for phase lag compensator will be
For phase lag compensator, pole will be near to origin.
Hence, option (a) is right.