# Power Electronics Questions and Answers – Single Phase FW AC-DC-3

This set of Power Electronics Questions and Answers for experienced people focuses on “Single-Phase FW AC-DC-3”.

1. A motor load is connected to a single-phase full converter B-2 type controlled rectifier. The net energy is transferred from ac source to the motor (dc load) when
a) π+α > 90
b) π-α > α
c) π+α > α
d) π-α > 90

Explanation: Converter will work as a line commuted inverter when π-α > α.

2. The below shown rectifier configuration has continues load current, find the expression for the average value of output voltage when the supply Vs = Vm sinωt is connected. a) (Vm/π)cosα
b) (2Vm/π)cosα
c) (Vm/π) (1+cosα)
d) (2Vm/π) (1+cosα)

Explanation: Vo = 1/π x [∫ Vm sinωt d(ωt)] Where the integration runs from α to π+α.

3. Find the expression of the rms value of output voltage for a single-phase M-2 type rectifier with RL load and continues load current. Transformer ratio is 1:1 with supply voltage Vs = Vm sinωt
a) Vm/√2
b) Vs
c) 2Vs
d) Vs/√2

Explanation: Vor2 = 1/π x [∫ Vm2 sin2ωt d(ωt)], Where the integration runs from α to π+α
Vor = Vm2/2 = Vs.

4. A single-phase full controlled converted with RLE load will act like a line-commutated inverter when the firing angle α
a) α > 180°
b) α > 90°
c) α < 90°
d) α = 90°

Explanation: It will act like a inverter i.e. most of the current would flow from the battery or back emf E to the source. For α=90° it will not act like a converter nor an inverter.

5. In converter operation, with output voltage = Vo and RLE load.
a) Vo < E
b) Vo = E
c) Vo > E
d) None of the mentioned

Explanation: The output voltage obtained by a converter is always greater than the counter or back emf E.
Vo = IR + E.

6. In inverter operation, with output voltage = Vo and a RLE load connected
a) Vo < E
b) Vo = E
c) Vo > E
d) None of the mentioned

Explanation: In inverter operation Vo < E.

7. A motor load is connected to a single-phase full converter B-2 type controlled rectifier, the net energy is transferred from ac source to the motor (dc load) when
a) π+α > 90
b) π-α < α
c) π+α < α
d) π-α > α

Explanation: It works as a converter when π-α < α.

8. The below shown rectifier configuration has continues load current, find the expression of the RMS value of output voltage when the supply Vs = Vm sinωt
a) Vs2/2
b) 2Vs/π
c) Vs
d) √Vs/2

Explanation: Vor2 = 1/π x [∫Vm2 sin2ωt d(ωt)] Where the integration runs from α to π+α
Vor = Vm2/2 = Vs.

9. Choose the correct statement
a) M-2 type connection requires SCRs with higher PIV as compared to those in a B-2 type
b) M-2 type connection requires SCRs with lower PIV as compared to those in a B-2 type
c) The average output voltage in M-2 type is more than that obtained from a B-2 type configuration of the same rating
d) The average output voltage in M-2 type is less than that obtained from a B-2 type configuration of the same rating

Explanation: The PIV of diodes in M-2 is 2Vm, whereas that in B-2 type of connection is Vm.

10. An SCR has the peak forward voltage = 1000 V. Find the maximum voltage that the SCR can handle if employed in a M-2 type full controlled converter circuit. Use factor of safety (FOS) = 2.5
a) 500 V
b) 400 V
c) 200 V
d) 1000 V

Explanation: In M-2 type configuration maximum voltage handled is 2Vm
Therefore, 1000/2×2.5 = 200 V.

Sanfoundry Global Education & Learning Series – Power Electronics.

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