Quiz Electrical Engineering
Each Question carries 2 Mark
Negative Marking: 1/3
Time: 20 Minutes
Q3. In the circuit shown below, the current source of 4 A…….
(a) absorbs 400 W
(b) supplies 368 W
(c) supplies 400 W
(d) absorbs 368 W
Q5. Shunt capacitance in an EHV line is restored to
(a) improve the stability
(b) improve the voltage
(c) reduce fault level
(d) none of above
Q6. A power factor meter connected in a circuit indicates pf of 0.6 lagging. To improve the power factor, we have to insert the following component, in the circuit.
(b) Both inductor and resistors
Q7. In a 360 kV network, 280 kV is recorded at a 360 kV bus. The reactive power absorbed by a shunt rated for 50 MVAR, 360 kV connected at the bus is ___.
(a) 40.8 MVAR
(b) 30.24 MVAR
(c) 33.67 MVAR
(d) 41.55 MVAR
Q8. A source V_S = 200 cos ωt delivers power to a load at power factor 0.8 lag. The reactive power is 300 VAR. the active power is given by?
(a) 200 Watts
(b) 225 Watts
(c) 400 Watts
(d) 300 Watts
Q9. Consider a phase-controlled converter shown in the figure. The thyristor is fired at an angle α in every positive half of input voltage. If peak value of instantaneous output is 230, then the firing angle α is close to
Q10. A 10kW, 400V, 3 phase induction motor is started with direct-on-line starter. The motor current and power factor at starting will be __________respectively.
(a) 20A, 0.6
(b) 20A, 0.2
(c) 120A, 0.2
(d) 120A, 0.6
Sol. Reactive power plays important role in power System stability.
During light load condition shunt compensator is able to minimize the line overvoltage.
During heavy load demand, shunt compensator is able to maintain appropriate voltage regulation.
Sol. When circuit is lagging in nature, its power factor can be improved by adding capacitor:
Cos ϕ=R/Z;Z =√(R^2+(X_L-X_C )^2 )
When capacitor is added, (X_L-X_C ) reduces. So, Z decreases and cos ϕ increases.
Sol. Impedance = (KV^2)/MVAR=360^2/50=2592 Ω
Record kV = 280 kV
Reactive power absorbed = V^2/Z=280^2/2592
= 30.24 MVAR
Sol. cosϕ = 0.8
Reactive power (Q) = V_rms I_rms sinθ
300=200/√2 .I_rms sin36.86°
Active power (P) = V_rms I_rms cosϕ
Sol. V_i = 230 V
Frequency f = 50 Hz
Firing angle α = 180° – cos^(-1) (230/(230√2))
Sol. power (P) = 10 kW
Voltage (V) = 400 V
Current (I) = 10000/400=25A
Therefore, starting current which reaches to 4 to 6 times
I.e., 100 to 150 A
Hence 120 A, 0.2 p.f. is most appropriate.