# Power Electronics Questions and Answers – Single Phase VSI-4

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This set of Power Electronics Multiple Choice Questions & Answers (MCQs) focuses on “Single Phase VSI-4”.

1. A single phase full bridge inverter has a dc voltage source Vs = 230 V. Find the rms value of the fundamental component of output voltage.
a) 90 V
b) 207 V
c) 350 V
d) 196 V

Explanation: The fundamental component of voltage = (4Vs/π) sinωt.
Peak value Vm = 4Vs/π
Rms value = 4Vs/π√2.

2. A single phase full bridge inverter has load R = 2 Ω, and dc voltage source Vs = 230 V. Find the rms value of the fundamental load current.
a) 96 A
b) 0 A
c) 103 A
d) none of the mentioned

Explanation: The fundamental component of voltage = (4Vs/π) sinωt.
Peak value Vm = 4Vs/π
Rms voltage = 4Vs/π√2 = 207 V
Current = 207/2 = 103.5 A.

3. A certain full bridge type inverter circuit has its rms value of fundamental load current component given by W. The fundamental frequency component of the load current would be given by
a) W sin ωt
b) (W/√2) sin ωt
c) √2 W sin ωt
d) sin ωt

Explanation: The fundamental frequency component of the load current is given by Im sin ωt
As W = Irms . . . (Given)
Im = √2 Irms = √2 W.
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4. In a half wave bridge inverter circuit, the power delivered to the load by each source is given by
a) Vs x Is
b) (Vs x Is)/2
c) 2(Vs x Is)
d) None of the mentioned

Explanation: Power delivered by each source (Vs/2) each is (Vs/2) x Is.

5. In a half wave circuit, forced commutation is essential when the
c) source voltage is below 150 V
d) none of the mentioned

Explanation: When the load is resistive (R load) , the diodes do not conduct, hence they cannot help stop the conduction of the SCRs. Hence, forced commutation in such cases becomes essential.

6. A single phase full bridge inverter circuit, has load R = 2 Ω and dc source Vs = 230 V. Find the value of power delivered to the load in watts only due to the fundamental component of the load current.
a) 5361.5 W
b) 2142.5 W
c) 21424.5 W
d) 214.2 W

Explanation: The fundamental component of voltage = (4Vs/π) sinωt
Peak value Vm = 4Vs/π
Rms voltage = 4Vs/π√2 = 207 V
RMS Current (Irms) = 207/2 = 103.5 A
P = (Irms)2 x R = 21424.5 W.

7. A single phase full bridge inverter is fed from a dc source such that the fundamental component of output voltage = 230 V. Find the rms value of SCR and diode current respectively, for a R load of 2 Ω.
a) 115 A, 80 A
b) 81.33 A, 36.2 A
c) 36.2 A, 0 A
d) 81.33 A, 0 A

Explanation: Fundamental component of load current = V/R = 230/2 = 115 A.
SCR current = 115/2 = 81.33 A
Diode current = 0 as the diodes do not come into picture for R loads.

8. For a full bridge inverter with the following load: R = 2 Ω, XL = 8 Ω and XC = 6 Ω.
a) The output voltage lags the current by 45°
b) The output current lags the voltage by 45°
c) The output current lags the voltage by 90°
d) The output current lags the voltage by more than 90°

Explanation: As the inductive effect is more than the capacitive effect, of course the current will lag the voltage by an angle P.
P = tan-1 (XL – XC)/R = tan-1 (1) = 45°.

9. A single phase full bridge inverter has RLC load. The dc input voltage is 230 V and the output frequency is 50 Hz. Find the expression for the load voltage up to the fifth harmonic.
a) 292 sin 314t + 97.62 sin 314t + 58.57 sin 318t + 28.31 sin 318t + 3.686 sin 318t
b) 292 sin 314t + 97.62 sin (3 x 314t) + 58.57 sin (5 x 318t)
c) 292 sin 314t + 97.62 sin (2 x 314t) + 58.57 sin (3 x 318t) + 28.31 sin (4 x 318t) + 3.686 sin (5 x 318t)
d) 292 sin 512t + 25.62 sin 249t + 6.74 sin 508t

Explanation: In a single phase full bridge inverter only odd harmonics are present. i.e. 1,3,5 etc.
Vo = (4Vs/π) sin ωt + (4Vs/3π) sin 3ωt + (4Vs/5π) sin 5ωt
(4Vs/π) = 292 V
ωt = 2 x f x π x t = 2 x 3.14 x 50 x t = 314t.

10. A single phase full bridge inverter has RLC load with R = 4 Ω, L = 35 mH and C = 155 μF. The dc input voltage is 230 V and the output frequency is 50 Hz. Find the rms value of the fundamental load current.
a) 28.31 A
b) 20.02 A
c) 16.69 A
d) 26.90 A

Explanation: Average value of fundamental output voltage = 4Vs/π = 292.85 V
XL = 2 x 3.14 x 50 x 0.035 = 10.99 Ω
XC = 1/(2 x 3.14 x 50 x 155 x 10-6) = 20.54 Ω
Z = 10.345 Ω
I = V/Z = 28.31
I(rms) = V/Z√2 = (292.85)/(1.414 x 10.345) = 20.02 A.

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