This set of Power Electronics Inteview Questions and Answers for freshers focuses on “Single-Phase HW AC-DC-4”.
1. A single-phase half-wave rectifier with a FD is supplied by Vs = 240 V, 50 Hz, AC source with a load R = 10 Ω, L = 0.8 mH. The firing angle is so adjusted such that the output voltage obtained is 100 V. Find the firing angle.
Explanation: Vo = Vm/2π cos(firing angle).
2. For the below shown circuit, a motor load (RLE) is connected to the SCR through supply Vs. The minimum value of the firing angle to turn on the SCR would be
Explanation: The firing angle should be such that the value of V exceeds E. So SCR turns on at excatlly Vm sinωt = E
Therefore ωt = min. angle = Sin-1(E/Vm).
3. Choose the correct statement with respect to the below given circuit.
a) The load current can never be zero
b) The source current can never be zero
c) The load voltage can never be zero
d) All the given statements are false
Explanation: Even if no current is flowing through the load when the SCR is off, voltage = E will exists at the load terminals at all times.
4. By using a freewheeling diode (FD) in a rectifier with RL load, the power consumed by the load
c) is not affected
d) decreases to zero
Explanation: The FD feeds inductor current again to the load.
5. A 230 V, 50 Hz, one-pulse SCR controlled converter has extinction angle β = 210°. Find the circuit turn-off time
a) 10 m-sec
c) 8.3 m-sec
d) 5.4 m-sec
t = 2π-β/ω
Where, π = 90°
ω = 2xπx50
β = 210°.
6. A 230V, 50Hz, single-pulse SCR is feeding a RL load with α = 40° and β = 210°. Find the value of average output voltage
a) 54 V
b) 106 V
c) 84 V
d) 32 V
Vo = Vm/2π x (cosα-cosβ)
Where Vm = √2Vs.
7. A single-pulse transformer with secondary voltage of 230 V, 50 Hz, delivers power to bulb of R = 10 Ω through a half-wave controlled rectifier circuit. For α = 60°, find the average current in the bulb
a) 2.5 A
b) 7.7 A
c) 9.6 A
d) 3 A
Explaantion: First find the average voltage, than Io = Vo/R
Vo = (Vm/2π) x (1+cosα).
8. A single-pulse transformer with secondary voltage of 230 V, 50 Hz, delivers power to bulb of R = 10 Ω through a half-wave controlled rectifier circuit. For α = 60° and output AC power of 2127 Watts, find the rectification efficiency
a) 98.6 %
b) 42 %
c) 28 %
d) 19 %
Vo = (Vm/2π) x (1+cosα) = 77.64 V
Pdc = Vo2xR = 602.8 W
Rectification efficiency = Pdc/Pac = 28.32 %.
Explanation: The maximum negative voltage faced by the SCR is –Vm which is nothing but √2Vs.
10. An SCR rectifier circuit is designed such that the average output voltage is 77.64 V & RMS value of output voltage is 145.873 V. Find the voltage ripple factor.
Explanation: FF = 145.873/77.64 = 1.879
VRF = √(FF2-1) = 1.5908.
Sanfoundry Global Education & Learning Series – Power Electronics.
To practice all areas of Power Electronics for freshers attending interviews, here is complete set of 1000+ Multiple Choice Questions and Answers.