250+ TOP MCQs on PWM Inverters-1 and Answers

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Power Electronics Multiple Choice Questions on “PWM Inverters-1”.

1. In the single-pulse width modulation method, the output voltage waveform is symmetrical about __________
A. π
B. 2π
C. π/2
D. π/4
Answer: C
Clarification: The waveform is a positive in the first half cycle and symmetrical about π/2 in the first half.

2. In the single-pulse width modulation method, the output voltage waveform is symmetrical about ____________ in the negative half cycle.
A. 2π
B. 3π/2
C. π/2
D. 3π/4
Answer: B
Clarification: In the negative half the wave is symmetrical about 3π/2.

3. The shape of the output voltage waveform in a single PWM is
A. square wave
B. triangular wave
C. quasi-square wave
D. sine wave
Answer: C
Clarification: Positive and the negative half cycles of the output voltage are symmetrical about π/2 and 3π/2 respectively. The shape of the waveform obtained is called as quasi-square wave.

4. In the single-pulse width modulation method, the Fourier coefficient bn is given by
A. (Vs/π) [ sin(nπ/2) sin(nD. ].
B. 0
C. (4Vs/nπ) [sin(nπ/2) sin(nD.].
D. (2Vs/nπ) [sin(nπ/2) sin(nD.].
Answer: C
Clarification: The Fourier analysis is as under:
bn = (2/π) ∫ Vs sin⁡ nωt .d(ωt) , Where the integration would run from (π/2 + D. to (π/2 – D.
2d is the width of the pulse.

5. In the single-pulse width modulation method, the Fourier coefficient an is given by
A. (Vs/π) [ cos(nπ/2) cos(nD. ].
B. 0
C. (4Vs/nπ) [sin(nπ/2) sin(nD.].
D. (2Vs/nπ) [sin(nπ/2) sin(nD.].
Answer: B
Clarification: As the positive and the negative half cycles are identical the coefficient an = 0.

6. In the single-pulse width modulation method, when the pulse width of 2d is equal to its maximum value of π radians, then the fundamental component of output voltage is given by
A. Vs
B. 4Vs/π
C. 0
D. 2Vs/π
Answer: B
Clarification: The Fourier representation of the output voltage is given by
power-electronics-questions-answers-pwm-inverters-1-q6
Put 2d = π & n = 1.

7. In case of a single-pulse width modulation with the pulse width = 2d, the peak value of the fundamental component of voltage is given by the expression
A. 4Vs/π
B. Vs
C. (4Vs/π) sin 2d
D. (4Vs/π) sin d
Answer: D
Clarification: For the fundamental component put n = 1.
power-electronics-questions-answers-pwm-inverters-1-q6
Vo = (4Vs/π) sin (D. sin (ωt)
Hence the peak value is (4Vs/π) sin d.

8. In case of a single-pulse width modulation with the pulse width = 2d, to eliminate the nth harmonic from the output voltage
A. d = π
B. 2d = π
C. nd = π
D. nd = 2π
Answer: C
Clarification: To eliminate, the nth harmonic, nd is made equal to π radians, or d = π/n.
From the below expression,
power-electronics-questions-answers-pwm-inverters-1-q6
when nd = π. sin nd = 0 hence, that output voltage harmonic is eliminated.