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# Multiple choice question for engineering

## Set 1

1. In case of a single-pulse width modulation with the pulse width = 2d, to eliminate the 5th harmonic from the output voltage waveform, the value of the pulse width (2d) must be equal to
a) 72°
b) 86°
c) 91°
d) 5°

Answer: c [Reason:] To eliminate the nth harmonic, nd = π. Therefore, d = π/n = π/5 = 36° Hence, 2d = 72°.

2. Several equidistant pulses per half cycle are used in ___________ type of modulation technique.
a) single-pulse
b) multiple-pulse
c) sine-pulse
d) equidistant-pulse

Answer: b [Reason:] In MPM, several equidistant pulses per half cycle are used.

3. In the multiple-pulse width modulation method, the Fourier coefficient an is
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 [Reason:] As the positive and the negative half cycles are identical the coefficient an = 0.

4. In the multiple-pulse width modulation method with two pulses per half cycle of width = d each and ɣ as the distance between the first pulse and ωt=0, has the Fourier coefficient bn =
a) (8Vs/π) [ cos(ɣπ/2) cos(nd/2) ].
b) 0
c) (4Vs/nπ) [sin(nɣ) sin(nd/2)].
d) (8Vs/nπ) [sin(nɣ) sin(nd/2)].

Answer: d [Reason:] The Fourier analysis is as under: bn = (4/π) ∫ Vs sin⁡ nωt .d(ωt) , Where the integration would run from (ɣ + d/2) to (ɣ – d/2) bn = (8Vs/nπ) [sin(nɣ) sin(nd/2)].

5. The amplitude of the nth harmonic of the two-pulse MPM waveform is given by __________
Let d be the width of a single pulse and ɣ be the distance from 0 to the centre of the first pulse.
a) (8Vs/nπ) sin(nɣ) sin(nd/2)
b) (4Vs/nπ) sin(nɣ) sin(nd/2)
c) (8Vs/nπ) sin(nπ) sin(nd/2)
d) (8Vs/nπ) sin(nɣ/2) sin(nd/2)

Answer: a [Reason:] The Fourier series representation of such a wave will be given by bn = (8Vs/nπ) [sin(nɣ) sin(nd/2)] an = 0 Therefore, Hence, the amplitude is the term excluding the sin nωt factor.

6. In case of MPM with two pulses per half cycle of width = d each and ɣ as the distance between the first pulse and ωt=0, for eliminating the nth harmonic from the output voltage, which of the following condition must be satisfied?
a) d = 2π
b) d = π
c) d = n°
d) d = 2π/n

Answer: d [Reason:] If d = 2π/n, sin (π) = 0 and the whole term is eliminated.

7. In case of MPM with two pulses per half cycle of width = d each and ɣ as the distance between the first pulse and ωt=0, for eliminating the nth harmonic from the output voltage, the value of gamma (ɣ) must be equal to
a) 0
b) π
c) π/n
d) d/n

Answer: c [Reason:] If gamma = π/n , the whole nth output voltage comes becomes zero.

8. Find the peak value of the fundamental component of voltage with MPM with two pulses having pulse width = 36° and ɣ = 54°. The Fourier representation of the waveform is as follows. a) 0.7484 x Vs
b) 1.414 x Vs
c) 0.637 x Vs
d) 2.54 x Vs

Answer: c [Reason:] For the fundamental component put n = 1. n = 36° and ɣ = 54° . . . (given) Vo1 = 8Vs/π x sin54 x sin18 = 0.637 Vs.

9. In the MPM method, the comparator is given _______ and _______ types of waveform at its input.
a) square, sine
b) square, quasi-square
c) sine, triangular
d) square, triangular

Answer: d [Reason:] To generate the modulated waves of equal width, square wave which is the reference signal is compared with the triangular wave which is the carrier signal wave.

10. In MPM, the square wave is the ________ signal whereas the triangular wave is the ________ signal.
a) reference, carrier
b) base, reference
c) carrier, reference
d) none of the mentioned

Answer: a [Reason:] To generate the modulated waves of equal width, square wave which is the reference signal is compared with the triangular wave which is the carrier signal wave.

## Set 2

1. In the multiple pulse width modulation method, the firing pulses are generate during the interval when the
a) triangular wave exceeds the square modulating wave
b) square modulating wave exceeds the triangular wave
c) square wave amplitude is same as the triangular wave’s amplitude
d) none of the mentioned

Answer: a [Reason:] The firing pulses to turn on the SCRs (or any other equivalent device) are generated when the triangular carrier signal exceeds the square reference signal.

2. In MPM, ____________ order harmonics can be eliminated by a proper choice of __________ and _________
a) higher, d, ɣ
b) lower, d, ɣ
c) higher and lower, d, ɣ
d) none of the mentioned

Answer: d [Reason:] In multiple pulse width modulation, the lower order harmonics can be eliminated by proper choice of 2d and ɣ.

3. In ___________ type of modulation method, the pulse width is not equal for all the pulses.
a) multiple pulse width modulation
b) single pulse width modulation
c) sinusoidal pulse width modulation
d) none of the mentioned

Answer: c [Reason:] In SPWM, the pulse width is a sinusoidal function of the angular position of the pulse in a cycle.

4. In sinusoidal pulse width modulation, __________ wave is compared with a ___________ type of wave.
a) square, sinusoidal
b) sinusoidal, triangular
c) sinusoidal, quasi-square
d) none of the mentioned

Answer: b [Reason:] In SPWM, a high-frequency triangular wave is compared with a sinusoidal reference wave of the desired frequency.

5. In the sinusoidal pulse width modulation, __________ is the carrier wave signal.
a) square wave
b) triangular wave
c) sinusoidal wave
d) quasi-square wave

Answer: b [Reason:] In SPWM, a high-frequency triangular wave is compared with a sinusoidal reference wave of the desired frequency.

6. In the sinusoidal pulse width modulation, ____________ is the reference wave signal.
a) square wave
b) triangular wave
c) sinusoidal wave
d) quasi-square wave

Answer: c [Reason:] In SPWM, a high-frequency triangular wave is compared with a sinusoidal reference wave of the desired frequency.

7. In sinusoidal pulse width modulation, the comparator output is high when the
a) triangular wave has magnitude higher than the sinusoidal wave
b) sinusoidal wave has magnitude higher than the triangular wave
c) triangular wave has magnitude equal to the sinusoidal wave
d) none of the mentioned

Answer: b [Reason:] The comparator output is high when the sinusoidal wave (reference signal) has magnitude higher than the triangular wave (carrier signal).

8. In PWM, the comparator output is further given to a ____________
a) integrator
b) scr devices
c) trigger pulse generator
d) snubber circuit

Answer: c [Reason:] The comparator output is processed in a trigger pulse generator in such a manner that the output voltage wave of the inverter has a pulse width in agreement with the comparator output pulse width.

9. The modulation index (MI) is given by
Vr = peak value of the reference wave.
Vc = peak value of the carrier wave.
a) Vr/Vc
b) Vc/Vr
c) (1 + Vc/Vr)
d) 1/(Vc Vr)

Answer: a [Reason:] MI = Vr/Vc.

10. By controlling the modulation index (MI), __________ can be controlled.
a) gain
b) output frequency
c) harmonic content of the output voltage
d) cosine component of the output voltage

Answer: c [Reason:] MI controls the output voltage waveform.

## Set 3

1. In pulse width modulated inverters, the output voltage is controlled by controlling the
a) input frequency
b) modulating index
c) amplification factor
d) none of the mentioned

Answer: b [Reason:] MI = Vr/Vc. MI controls the output voltage waveform, as it is proportional to the fundamental component of the output voltage.

2. In case of sinusoidal pulse width modulation with MI < 1, if the number of pulses per half cycle (N) = 5, then
a) harmonics of order 5 and 7 become significant
b) harmonics of order 5 and 7 are eliminated
c) harmonics of order 9 and 11 become significant
d) harmonics of order 9 and 11 are eliminated

Answer: c [Reason:] For MI less than one, largest harmonic amplitudes in the output voltage are associated with harmonics of order 2N (+/-) 1. Thus, by increasing the number of pulses per half cycle (N), the order of the dominate harmonic can be raised, which can be filtered out easily.

3. In case of sinusoidal pulse width modulation with MI < 1, the order of the dominate harmonic can be raised by
a) increasing the number of pulses
b) reducing the number of pulses
c) lowering the input voltage frequency
d) raising the input voltage frequency

Answer: a [Reason:] For MI less than one, largest harmonic amplitudes in the output voltage are associated with harmonics of order 2N (+/-) 1. Thus, by increasing the number of pulses per half cycle (N), the order of the dominate harmonic can be raised.

4. In case of sinusoidal pulse width modulation with MI < 1, if the number of pulses per half cycle (N) = 6, then
a) harmonics of order 7 and 9 become significant
b) harmonics of order 7 and 9 are eliminated
c) harmonics of order 11 and 13 become significant
d) harmonics of order 11 and 13 are eliminated

Answer: c [Reason:] For MI less than one, largest harmonic amplitudes in the output voltage are associated with harmonics of order 2N (+/-) 1. Thus, by increasing the number of pulses per half cycle (N), the order of the dominate harmonic can be raised, which can be filtered out easily.

5. Increasing the number of pulses (N), ____________
a) reduces the output voltage amplitude
b) reduces the inverter efficiency
c) improves the inverter efficiency
d) none of the mentioned

Answer: b [Reason:] Increasing N, increasing the switching frequency of the SCRs. This amounts to more switching losses and therefore efficiency lowers down.

6. In single-phase modulation of PWM inverters, the pulse width is 120°. For an input voltage of 220 V dc, the rms value of output voltage is
a) 185 V
b) 254 V
c) 127 V
d) 179 V

Answer: d [Reason:] Vo = (Vs) x (2d/π)1/2 Where, d = 2π/3 and Vs = 220 V.

7. In MPM the amplitudes of square wave and triangular wave are respectively 1 V and 2 V. For generating 5 pulses per half cycle, the pulse width should be ___________
a) 36°
b) 24°
c) 12°
d) 18°

Answer: d [Reason:] In multiple pulse width modulation, the pulse width is given by [ 1 – (Vr/Vc) ] x (π/N) (in degrees) Where, Vr = 1 Vc = 2 N = 5. Hence, pulse width = 180/10 = 18°.

8. In an inverter, if the fundamental output frequency is 50 Hz, then the frequency of the lowest order harmonic will be
a) 50 Hz
b) 150 Hz
c) 250 Hz
d) 350 Hz

Answer: b [Reason:] The 3rd harmonic is the lowest order harmonic, Hence, 50 x 3 = 150 Hz.

9. Calculate the pulse width in case of MPM, if the amplitudes of square wave and triangular wave are respectively 2 V and 3 V respectively. 16 pulses per cycle are generated.
a) 18°
b) 7.5°
c) 6.4°
d) 9°

Answer: b [Reason:] In multiple pulse width modulation, the pulse width is given by [ 1 – (Vr/Vc) ] x (π/N) (in degrees) Where, Vr = 2 V Vc = 3 V N = Pulses per half cycle = 16/2 = 8 Hence, pulse width = 180/(3×4) = 7.5°.

10. In a PWM inverter, if the frequency of the lowest harmonic is 180 Hz, then the frequency of the fundamental component would be ___________
a) 50 Hz
b) 60 Hz
c) 540 Hz
d) 90 Hz

Answer: b [Reason:] The 3rd harmonic is the lowest order harmonic, Hence, 180/3 = 60 Hz.

## Set 4

1. In case of multiple pulse width modulation method, if the amplitudes of the reference wave and the carrier wave are made equal then, the pulse width =
a) ∞
b) 0
c) 100 °
d) none of the mentioned

Answer: b [Reason:] In multiple pulse width modulation, the pulse width is given by [ 1 – (Vr/Vc) ] x (π/N) (in degrees) If Vr = Vc Pulse width = 0.

2. In an inverter, if the fundamental output frequency is 45 Hz, then the frequency of the second lowest order harmonic will be
a) 45 Hz
b) 135 Hz
c) 225 Hz
d) 9 Hz

Answer: c [Reason:] The 5th harmonic is the second lowest order harmonic, as only odd numbers of harmonics are present. Hence, 45 x 5 = 225 Hz.

3. In an inverter, if the fundamental output frequency is 45 Hz, then the frequency of the lowest order harmonic will be
a) 45 Hz
b) 225 Hz
c) 15 Hz
d) 135 Hz

Answer: d [Reason:] The 3rd harmonic is the lowest order harmonic, as only odd numbers of harmonics are present. Hence, 45 x 3 = 135 Hz.

4. A VSI will have a better performance if its
a) load inductance is small and source inductance is large
b) both load inductance and source inductance are small
c) both load inductance and source inductance are large
d) none of the mentioned

Answer: b [Reason:] Higher value of source inductance will increase the overlap angle and cause commutation issues. Hence, both load inductance and source inductance should be small.

5. A single-phase bridge inverter has a square wave output voltage waveform, with odd harmonics present. What is the percentage of the fifth harmonic component to the fundamental component?
a) 50 %
b) 25 %
c) 20 %
d) 5 %

Answer: c [Reason:] For the fundamental comoponent, n = 1 and for the fifth harmonic n = 5 1/5 = 0.2 = 20 %.

6. Find, the maximum rms value of the fundamental component of the output voltage for the below given single pulse width modulated circuit. Take Vs = 68 V. a) 86.6 sin(d)
b) 64 sin(d)
c) 244 sin(d)
d) none of the mentioned

Answer: a [Reason:] Vo1 = (4Vs/π) sin(d) = 86.6 sin(d).

7. Control of frequency and control of voltage in 3-phase inverters is
a) possible only through inverter control circuit
b) possible through the control circuit of inverter and converter
c) possible through inverter control of frequency and through converter control for voltage
d) none of the mentioned

Answer: c [Reason:] Control of frequency and control of voltage in 3-phase inverters is possible through inverter control of frequency and through converter control for voltage.

8. Output voltage of a single-phase bridge inverter, fed from a fixed dc source is varied by
a) varying the switching frequency
b) pulse-width modulation
c) pulse amplitude modulation
d) all of the mentioned

Answer: b [Reason:] The output voltage is controlled by PWM techniques.

9. A single-phase bridge inverter, fed from a 230 V dc is connected to the load R = 10 Ω and L = 0.03 H. Determine the fundamental component of rms output current. Fundamental output frequency of the square wave output = 50 Hz.
a) 30 A
b) 15 A
c) 2.3 A
d) 20.7 A

Answer: b [Reason:] Vo1 (rms fundamental voltage) = 4Vs/π√2 = 207.10 V. Now, the load impedance at the fundamental frequency i.e. at 50 Hz will be Z1 = [102 + (2π x 50 x 0.03)2]1/2 = 13.7414 Ω Hence, I = V1/Z1 = (207.10)/(13.7414) = 15.07 A.

10. A single-phase bridge inverter, fed from a 230 V dc is connected to the load R = 10 Ω and L = 0.03 H. The output is a quasi-square wave with an on period of 0.5 of a cycle. Determine the fundamental component of rms output voltage.
a) 207.10 V
b) 146.42 V
c) 265.4 V
d) 129 V

Answer: b [Reason:] For quai-square wave or single-pulse width modulated wave pulse width 2d = 0.5 x 180° = 90° or d = 45° Hence, Vo1 = (4Vs/π√2) sind = (4 x 230)/( π.√2) sin45° = 146.423 V.

## Set 5

1. If Vr is the rms value of the inverter output voltage and V1 is the rms value of the fundamental component, then the total harmonic distortion (THD) is given by
a) Vr/V1
b) (Vr + V1r
c) [ (Vr/V1)2 – 1 ]1/2
d) [ (Vr/V1)1/2 + 1 ]2

Answer: c [Reason:] THD is the ratio of rms value of all the harmonic components to the rms value of the fundamental component. Rms value of all the harmonic components VH = (Vr2 – V12)1/2 THD = (Vr2 – V12)1/2/ V1.

2. The distortion factor (μ) is the ratio of
a) total rms output voltage to fundamental rms output voltage
b) fundamental rms output voltage to fundamental average output voltage
c) total rms output voltage to rms value of all the harmonic components
d) fundamental rms output voltage to total rms output voltage

Answer: d [Reason:] μ = V1/Vr.

3. A single-phase half bridge inverter is connected to a 230 V dc source which is feeding a R load of 10 Ω. Determine the fundamental power delivered to the load.
a) 1.07 W
b) 107.2 W
c) 1.07 kW
d) 107.2 kW

Answer: c [Reason:] The fundamental power is the power delivered to the load due to the fundamental components of voltage and current. V1 (rms) = 2Vs/√2π = 103.552 V I1 (rms) = V1/R = 10.3552 A P = (I1)2 x R = 1072.3 W = 1.0723 kW.

4. A single-phase half bridge inverter is connected to a 230 V dc source which is feeding a R load of 10 Ω. Determine the average current through each SCR inverter switch.
a) 11.5 A
b) 5.75 A
c) 23 A
d) none of the mentioned

Answer: b [Reason:] Peak current through each SCR = Vs/2R = 11.5 A As each SCR would conduct for 180° of the total 360° cycle, average current = peak current/2 = 11.5/2 = 5.75 A.

5. Find the distortion factor (μ), for a single phase half wave bridge inverter with dc source Vs = 1 kV.
a) 0.87
b) 1
c) 0.9
d) 0.7

Answer: c [Reason:] μ = V1/Vr V1 = rms value of fundamental component = 2Vs/π√2 Vr = Total rms output voltage = Vs/2 μ = (2Vs/π√2) x (2/Vs) = 2√2/π = 0.9.

6. A single phase inverter gives rms value of output voltage as 115 V and the fundamental output voltage of as 103.5 V. Find the THD (Total Harmonic Distortion).
a) 0.4 %
b) 40.8 %
c) 48.3 %
d) 4.83 %

Answer: c [Reason:] Vr = 115 V and V1 = 103.5 V THD =[ (Vr2 – V12)1/2/ V1 ] x 100 %.

7. What would be the harmonic factor of lowest order harmonic in case of a half wave bridge inverter?
a) 1/1
b) 1/3
c) 1/2
d) Insufficient data

Answer: b [Reason:] Let Vs be the input voltage. The 3rd harmonic is the lowest order harmonic. Vo (rms) = 2Vs/(n x π x √2) For fundamental component n = 1 and for the lowest order harmonic n = 3 ρ3 = V3/V1 = 1/3 or 33.33 %.

8. A single phase full bridge inverter using transistors and diodes is feeding a R load of 3 Ω with the dc input voltage of 60 V. Find the rms output voltage and the peak reverse blocking voltage of each transistor.
a) 30 V, 60 V
b) 30 V, 30 V
c) 60 V, 60 V
d) 60 V, 30 V

Answer: c [Reason:] Rms output voltage = PIV of each transistor = Vs. Vs = 60 V.

9. A single phase full bridge inverter using transistors and diodes is feeding a R load of 3 Ω with the dc input voltage of 60 V. Find the fundamental frequency output power.
a) 1200 W
b) 856 W
c) 972 W
d) 760 W