# Power Electronics Questions and Answers – Three Phase Converters-2

This set of Power Electronics Multiple Choice Questions & Answers (MCQs) focuses on “Three Phase Converters-2”.

1. A three phase full converter will require __________ number of SCRs.
a) 3
b) 6
c) 9
d) 2

Explanation: Three legs having two SCRs each, six in total.

2. A three phase six pulse full converter works as a ac to dc converter for firing angles in the range
a) α > 90
b) 90 < α < 180
c) 0 < α < 90
d) 0 < α < 360

Explanation: When α is less than 90°, the SCRs conduct for 120° and the current and voltage are positive on an average hence, the power flows from AC source to DC load.

3. For the below given circuit, α = 60°. T2 will start conduction at ωt = __________ Assume the inductor L value to be negligible.

a) 60°
b) 120°
c) 90°
d) 150°

Explanation: Assuming the phase sequence is R-Y-B. T1 would start conducting at 30+60 = 90°, T2 at 90+210/2 = 150°. This is because after T1, T3 would conduct from the upper group, as T2 belongs to the lower group it will start to conduct exactly between T1 and T3 i.e. between 90 and 210(90+120) which is 150°.

4. For a three phase full controlled converter, with 3 thyristors in the upper or positive group and 3 thyristors in the lower or negative group, at any given time
a) two thyristors are conducting from each group
b) one thyristor is conducting from each group
c) one thyristor is conducting from either of the groups
d) all 6 thyristors are conducting at a time

Explanation: Let’s say T1, T3 and T5 belong to the positive group and T2, T4 and T6 to the negative group. At any given time one SCR from each group conducts. e.g. T1 and T6 or T1 and T2.

5. In case of a three phase full controlled converter with 6 SCRs, commutation occurs every
a) 120°
b) 60°
c) 180°
d) 30°

Explanation: Every SCR conducts for 120°. This means that the SCRs from the positive group are fired 120° among themselves, same is true for SCRs from negative group. For example, if T1 starts conducting at 90° it will conduct till 90+120 = 210°. But while T1 is conducting, half of the time i.e. from 90 to 150, T6 is conducting and another half of the time T2 is conducting. Hence, commutation (change in the SCR which is conducting) takes place every 60 degrees irrespective of the firing angle. Construct the firing sequence table for better understanding.
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6. For the below given circuit, the conduction sequence for the negative group of SCRs is

a) T4-T6-T2
b) T1-T2-T3
c) T2-T6-1
d) T2-T4-T6

Explanation: The negative group of SCRs has T2, T4 and T6. The conduct as T2-T4-T6, as T2 is connected to the B phase, T4 to the R phase and like-wise.

7. For a three-phase full controlled converter with R load, the average value of output voltage is zero for
a) α = 0°
b) α = 90°
c) α = 180°
d) It can never be zero

Explanation: For α = 90 degrees, the voltage waveform is equally symmetrical about the ωt axis, hence the average value is zero. This can also be found by using the formula for average output voltage,
Vo = (3Vml/π) cos α,
For α = 90°, cosα = 0, Vo = zero.

8. A three-phase full converter charges a battery from a three-phase supply of 230 V. The battery emf is 200 V and the internal resistance of the battery is 0.5 Ω. Find the value of the continuous current which is flowing through the battery if its terminal voltage is 210 V
a) 10 A
b) 20 A
c) 0.5 A
d) 25 A

Explanation: Vo = 210 V
Vo = E + Io x R
210 = 200 + 0.5 x Io
Io (Current through the battery) = 20 A.

9. A three-phase full converter charges a battery from a three-phase supply of 230 V. The battery emf is 200 V. Find the value of the firing angle if the battery terminal voltage is 210 V.
a) 36.54°
b) 56.7°
c) 89.3°
d) 47.45°

Explanation: Vo = (3Vml/π) cos α
α = cos-1(210π/3√2×230) = 47.453°.

10. A three-phase full converter charges a battery from a three-phase supply of 230 V. Find the value of the power delivered to the load if a continues current of 20A is flowing through the battery of emf 200 V and internal resistance of 0.5 Ω.
a) 0 W
b) 5600 W
c) 4200 W
d) 1040 W

Explanation: Iavg = Irms = 20 A
P = E x Iavg + Irms2 x R = 4200 W.

Sanfoundry Global Education & Learning Series – Power Electronics.

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