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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

101.

Which one of the following load is suitable for leading power factor load in Single phase Half-bridge inverter?(a) C load(b) R-L-C overdamped(c) R-L-C underdamped(d) L-C loadThis question was posed to me in homework.This key question is from Solid-State Switching Circuits topic in portion Introduction to Solid-State Switching Circuits of Electric Drives

Answer»
Discussion
102.

Which one of the following load is suitable for lagging power factor load in Single phase Half-bridge inverter?(a) C load(b) R-L-C overdamped(c) R-L-C underdamped(d) L-C loadThe question was asked during an interview for a job.The question is from Solid-State Switching Circuits in division Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Right choice is (B) R-L-C overdamped

The BEST I can explain: R-L-C overdamped loads are GENERALLY lagging power factor loads. They require forced COMMUTATION. Anti-Parallel DIODES do not help in the commutation process.
Discussion
103.

Calculate the value of the fundamental displacement factor for 1-Φ Full wave semi-converter if the firing angle value is 95^o.(a) .60(b) .68(c) .62(d) .67This question was posed to me by my college director while I was bunking the class.Asked question is from Solid-State Switching Circuits topic in portion Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Right OPTION is (d) .67

To EXPLAIN: Fundamental DISPLACEMENT factor is the cosine of angle difference between the fundamental voltage and fundamental current. The fundamental displacement factor for 1-Φ FULL wave semi-converter is cos(∝÷2)=cos(47.5^o)=.67.
Discussion
104.

Calculate the value of the fundamental displacement factor for 1-Φ Full wave bridge rectifier if the firing α=17°.(a) .98(b) .19(c) .56(d) .95The question was asked during an interview.Enquiry is from Solid-State Switching Circuits in division Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Right answer is (d) .95

Best EXPLANATION: Fundamental displacement factor is the cosine of angle DIFFERENCE between the fundamental voltage and fundamental CURRENT. D.F=cos(∝)=cos(17°)=0.95.
Discussion
105.

Calculate the value of the conduction angle for R-L load if the value of β and α are 56° and 18°.(a) 48°(b) 38°(c) 57°(d) 15°I have been asked this question during an internship interview.This interesting question is from Solid-State Switching Circuits topic in portion Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

The correct answer is (B) 38°

For explanation I WOULD say: The conduction angle for R-L LOAD is β-α=56°-18°=38°. R-L load is a current stiff type of load. The current in the circuit only flows from α to β.
Discussion
106.

Calculate the value of the conduction angle for diode for R-L load with a freewheeling diode if the value of β is 226°. (Discontinuous conduction mode)(a) 40°(b) 45°(c) 50°(d) 46°This question was posed to me during an online exam.My doubt is from Solid-State Switching Circuits in division Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Right CHOICE is (b) 45°

The EXPLANATION: The CONDUCTION angle for DIODE for R-L load with a freewheeling diode is β-π=226°-180°=45°. R-L load is a current STIFF type of load.
Discussion
107.

Calculate the value of the conduction angle for SCR for R-L load with a freewheeling diode if the value of β and α are 65° and 12°.(a) 168°(b) 170°(c) 130°(d) 180°I had been asked this question in an internship interview.Question is from Solid-State Switching Circuits topic in portion Introduction to Solid-State Switching Circuits of Electric Drives

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Discussion
108.

Full form of ADC is ___________(a) Analog to Digital converter(b) Discrete to Analog converter(c) Analog to Digital converter(d) Discrete to Avalanche converterI have been asked this question in exam.This intriguing question originated from Solid-State Switching Circuits topic in division Introduction to Solid-State Switching Circuits of Electric Drives

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Discussion
109.

Full form of DAC is ___________(a) Digital to Analog converter(b) Discrete to Analog converter(c) Digital to AC converter(d) Discrete to Avalanche converterThis question was addressed to me in a national level competition.My enquiry is from Solid-State Switching Circuits topic in portion Introduction to Solid-State Switching Circuits of Electric Drives

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Discussion
110.

In single phase RLE load, calculate the Peak inverse voltage using the data: (Vs)r.m.s=89 V, f=50 Hz,E=440 V.(a) 521.2 V(b) 527.8 V(c) 597 V(d) 529 VThe question was posed to me during an interview.I'd like to ask this question from Solid-State Switching Circuits topic in division Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Right choice is (c) 597 V

Easy explanation: In a SINGLE PHASE, RLE load the peak INVERSE VOLTAGE across the thyristor VT=Vm+E=529 V.
Discussion
111.

Calculate the PIV for the Full-wave bridge rectifier if the peak value of the supply voltage is 230.(a) 324.4 V(b) 325.2 V(c) 524.4 V(d) 626.8 VThis question was addressed to me during a job interview.I would like to ask this question from Solid-State Switching Circuits topic in division Introduction to Solid-State Switching Circuits of Electric Drives

Answer» RIGHT choice is (a) 324.4 V

To elaborate: The PEAK inverse voltage for the Full-wave BRIDGE rectifier is Vm=√2×230=325.2 V. The peak inverse is the maximum NEGATIVE voltage across the thyristor.
Discussion
112.

Full form of PIV.(a) Peak inverse voltage(b) Peak insert voltage(c) Paas Inverse volatile(d) Peak insert volatileThis question was addressed to me in an online quiz.My doubt is from Solid-State Switching Circuits in division Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Right OPTION is (a) Peak inverse voltage

Explanation: PIV stands for Peak inverse voltage. It is the MAXIMUM NEGATIVE voltage after which thyristor will BREAKDOWN.
Discussion
113.

In single phase RLE load, calculate the angle at which conduction starts using the data: (Vs)=14sin(Ωt), f=50 Hz, E=10 V.(a) 45.58°(b) 46.26°(c) 47.26°(d) 49.56°I had been asked this question during an interview for a job.The question is from Solid-State Switching Circuits in division Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Correct ANSWER is (a) 45.58°

Explanation: The ANGLE at which conduction STARTS when Vmsin(θ)=E. The conduction will remain from ϴ to π-θ. Θ=SIN-(E÷Vm)=sin-(.71)=45.58°.
Discussion
114.

In single phase RLE load, calculate the voltage across the thyristor when current decays to zero using the data: (Vs)r.m.s=340 V, f=50 Hz, R=2 Ω, E=150 V, β=160°.(a) 45.89 V(b) 74.45 V(c) 54.85 V(d) 84.48 VI had been asked this question in an online quiz.This intriguing question originated from Solid-State Switching Circuits in section Introduction to Solid-State Switching Circuits of Electric Drives

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The correct answer is (b) 74.45 V

Easiest EXPLANATION: In SINGLE PHASE RLE load the VOLTAGE across the thyristor when current DECAYS to zero VT=Vmsin(β)-E=340×√2sin(160°)-90=74.45 V.
Discussion
115.

Displacement factor depends upon the shape of the waveform.(a) True(b) FalseI got this question in class test.This question is from Solid-State Switching Circuits in portion Introduction to Solid-State Switching Circuits of Electric Drives

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Correct answer is (a) True

The best I can explain: Displacement factor depends upon the SHAPE of the WAVEFORM. Displacement factor is the cosine of the ANGLE between the fundamental voltage and fundamental current. It depends upon the TYPE of load, converter type.
Discussion
116.

Calculate the value of the Input power factor using the data: Fundamental displacement factor=.96, Distortion factor=.97.(a) .93(b) .84(c) .48(d) .89This question was posed to me during an interview.The query is from Solid-State Switching Circuits topic in division Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Right choice is (a) .93

To elaborate: The VALUE of the input POWER FACTOR is a product of the (Fundamental DISPLACEMENT factor)×(Distortion factor)=g×F.D.F=.97×.96=.93.
Discussion
117.

For highly inductive load current remains continuous.(a) True(b) FalseThe question was posed to me in an internship interview.I would like to ask this question from Solid-State Switching Circuits topic in section Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Correct option is (a) True

Easy explanation: For highly INDUCTIVE load current remains continuous. INDUCTOR opposes the change in the current. When current is about to EXTINCT the next CONDUCTING cycle starts due to which currently remains CONSTANT.
Discussion
118.

Calculate the value of the fundamental displacement factor for 1-φ Full wave semi-converter if the firing angle value is 0^o.(a) 1(b) .8(c) .4(d) .2The question was posed to me in an online interview.My query is from Solid-State Switching Circuits topic in chapter Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Correct answer is (a) 1

Explanation: Fundamental displacement factor is the COSINE of angle difference between the fundamental voltage and fundamental current. The fundamental displacement factor for 1-φ Full WAVE semi-converter is cos(∝÷2)=cos(0^o)=1.
Discussion
119.

Calculate the I0 r.m.s for Single-phase Half Wave rectifier for R-L load (Highly inductive) using the data: Vm=110 V, ∝=16°, β=31°, R=1 Ω.(a) 4.56 A(b) 1.82 A(c) 4.81 A(d) 9.15 AI have been asked this question during an interview for a job.Question is taken from Solid-State Switching Circuits in chapter Introduction to Solid-State Switching Circuits of Electric Drives

Answer» RIGHT option is (B) 1.82 A

The explanation: In Half-wave CONTROLLED rectifier, the r.m.s value of the current is Vm(cos(∝) – cos(β))÷2πR=110(cos(16°) – cos(31°))÷6.28=1.82 A. The thyristor will CONDUCT from ∝ to β.
Discussion
120.

Calculate the I0 avg for Single-phase Half Wave rectifier for R-L load using the data: Vm=56 V, ∝=15°, β=30°, R=2 Ω.(a) .56 A(b) .44 A(c) .26 A(d) .89 AThis question was posed to me in class test.The origin of the question is Solid-State Switching Circuits topic in section Introduction to Solid-State Switching Circuits of Electric Drives

Answer» CORRECT answer is (a) .56 A

The explanation is: In Half-wave controlled rectifier, the AVERAGE value of the CURRENT is Vm(cos(∝)-cos(β))÷2πR=56(cos(15°)- cos(30°))÷12.56=.44 A. The thyristor will conduct from ∝ to β.
Discussion
121.

Calculate the V0 avg for Single-phase Half Wave rectifier for R-L load using the data: Vm=24 V, ∝=30°, β=60°.(a) 1.39 V(b) 8.45 V(c) 4.55 V(d) 1.48 VThe question was asked during an internship interview.I'd like to ask this question from Solid-State Switching Circuits topic in chapter Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Correct answer is (a) 1.39 V

For explanation I would say: In Half-wave CONTROLLED rectifier, the average VALUE of the VOLTAGE is VM(cos(∝)-cos(β))÷2π=24(cos(30°)- cos(60°))÷6.28=1.39 V. The thyristor will conduct from ∝ to β.
Discussion
122.

Calculate the value of the conduction angle for R-L load if the value of β and α are 19° and 29°.(a) 10°(b) 70°(c) 30°(d) 80°The question was posed to me during an interview for a job.My question is from Solid-State Switching Circuits topic in section Introduction to Solid-State Switching Circuits of Electric Drives

Answer» CORRECT choice is (a) 10°

Best explanation: The conduction angle for R-L load is β-α=29°-19°=10°. R-L load is a current stiff TYPE of load. The current in the CIRCUIT only FLOWS from α to β.
Discussion
123.

The problem of short-circuiting in 1-Φ Full wave semi-converter is very common.(a) True(b) FalseThe question was asked during an interview.The doubt is from Solid-State Switching Circuits topic in division Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Correct option is (a) True

For explanation: The problem of short-circuiting in 1-Φ Full wave semi-converter is very COMMON. Diodes protect the THYRISTOR from short-circuiting. They provide the gap from (α, π+α) to avoid the conduction of ONE LEG THYRISTORS.
Discussion
124.

Diodes in 1-Φ Full wave semi-converter protects the thyristor from short-circuiting.(a) True(b) FalseI had been asked this question in a job interview.My question is taken from Solid-State Switching Circuits topic in section Introduction to Solid-State Switching Circuits of Electric Drives

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The correct answer is (a) True

For explanation: Diodes in 1-Φ Full WAVE semi-converter PROTECTS the thyristor from short-circuiting. They PROVIDE the gap from (α, π+α) to AVOID the conduction of one leg thyristors.
Discussion
125.

Calculate the r.m.s value of diode current in 1-Φ Full wave semi-converter for the load (Highly inductive) current=.2 A and α=74°. (F.D configuration)(a) .154 A(b) .248 A(c) .128 A(d) .587 AThis question was posed to me during an internship interview.This intriguing question comes from Solid-State Switching Circuits topic in division Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

The CORRECT CHOICE is (c) .128 A

Explanation: The r.m.s VALUE of diode current in 1-Φ Full wave semi-converter is Io√(α÷π). It is the r.m.s value of the diode current. Ir.m.s = Io√(α÷π)=.2√(.41)=.128 A.
Discussion
126.

Calculate the average value of diode current in 1-Φ Full wave semi-converter for the load (Highly inductive) current=5.2 A and α=11°. (F.D configuration)(a) .32 A(b) .31 A(c) .25 A(d) .27 AThis question was posed to me during an interview.My query is from Solid-State Switching Circuits topic in chapter Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

The CORRECT option is (b) .31 A

Best explanation: The average value of diode current in 1-Φ FULL wave semi-converter is Io(α÷π). It is the average value of the diode current. Iavg= Io(α÷π)=5.2(.061)=.31 A.
Discussion
127.

Calculate the average value of diode current in 1-Φ Full wave semi-converter for the load (Highly inductive) current=75.2 A and α=41°. (Asymmetrical configuration)(a) 46.16 A(b) 42.15 A(c) 41.78 A(d) 41.18 AThis question was posed to me in my homework.Origin of the question is Solid-State Switching Circuits topic in division Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

The correct OPTION is (a) 46.16 A

Easiest explanation: The average VALUE of DIODE current in 1-Φ FULL wave semi-converter is Io(π+α÷2π). It is the average value of the diode current. Iavg = Io(π+α÷2π)=75.2(.61)=46.16 A.
Discussion
128.

Calculate the average value of thyristor current in 1-Φ Full wave semi-converter for the load (Highly inductive) current=25.65 A and α=18°. (Asymmetrical configuration)(a) 11.54 A(b) 12.15 A(c) 15.48 A(d) 14.52 AThe question was asked during a job interview.This intriguing question comes from Solid-State Switching Circuits in portion Introduction to Solid-State Switching Circuits of Electric Drives

Answer» RIGHT option is (a) 11.54 A

For explanation: The average VALUE of thyristor current in 1-Φ FULL wave semi-converter is Io(π-α÷2π). It is the average value of the thyristor current. Iavg = Io(π-α÷2π)=25.65(.45)=11.54 A.
Discussion
129.

Calculate the r.m.s value of diode current in 1-Φ Full wave semi-converter for the load (Highly inductive) current=5.1 A and α=115°. (Asymmetrical configuration)(a) 4.21 A(b) 4.61 A(c) 4.71 A(d) 4.52 AThe question was asked in homework.Query is from Solid-State Switching Circuits in portion Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Correct answer is (B) 4.61 A

For EXPLANATION I would say: The r.m.s value of DIODE CURRENT in 1-Φ FULL wave semi-converter is Io√π+α÷2π. It is the r.m.s value of the diode current. I(r.m.s) = Io√π+α÷π=5.1(√.819)=4.61 A.
Discussion
130.

Calculate the r.m.s value of thyristor current in 1-Φ Full wave semi-converter for the load (Highly inductive) current=2.2 A and α=155°. (Asymmetrical configuration)(a) .58 A(b) .57 A(c) .51 A(d) .52 AThe question was asked in quiz.This interesting question is from Solid-State Switching Circuits topic in division Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Correct CHOICE is (b) .57 A

Best EXPLANATION: The r.m.s value of source current in 1-Φ FULL wave semi-converter is Io√π-α÷2π. It is the r.m.s value of the thyristor current. I(r.m.s) = Io√π-α÷2π=2.2(√.069)=.57 A.
Discussion
131.

Calculate the r.m.s value of source current in 1-Φ Full wave semi-converter for the load (Highly inductive) current=51.2 A and α=15°.(a) 10.53 A(b) 14.52 A(c) 44.92 A(d) 49.02 AThis question was addressed to me in examination.My enquiry is from Solid-State Switching Circuits in section Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Correct choice is (d) 49.02 A

Explanation: The r.m.s value of SOURCE CURRENT in 1-Φ FULL wave semi-converter is Io√π-α÷π. It is the r.m.s value of the source current. I(r.m.s) = Io√π-α÷π = 51.2(√.916) = 49.02 A.
Discussion
132.

Calculate the fundamental component of source current in 1-Φ Full wave bridge rectifier for the load(Highly inductive) current=78 A.(a) 78 A(b) 45 A(c) 69 A(d) 13 AI have been asked this question during an internship interview.My question is taken from Solid-State Switching Circuits topic in portion Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Right choice is (a) 78 A

To explain I WOULD say: The fundamental COMPONENT of source current in 1-Φ Full wave bridge rectifier is IO. It is the r.m.s value of the source current. Is(r.m.s)=Io=78 A.
Discussion
133.

Calculate the circuit turn-off time for 1-Φ Full wave bridge rectifier for α=145°. Assume the value of ω=5 rad/sec.(a) 84.9 msec(b) 94.5 msec(c) 101.2 msec(d) 87.2 msecThis question was posed to me in my homework.This question is from Solid-State Switching Circuits in division Introduction to Solid-State Switching Circuits of Electric Drives

Answer» RIGHT answer is (d) 87.2 MSEC

Best EXPLANATION: The circuit turn-off time for 1-Φ Full WAVE bridge rectifier is (π-α)÷ω. The VALUE of circuit turn-off time is (π-145°)÷5=87.2 msec.
Discussion
134.

Calculate the fundamental component of source current in 1-Φ Full wave bridge rectifier for load(Highly inductive) current=3.14 A.(a) 2.82 A(b) 1.45 A(c) 3.69 A(d) 4.78 AThe question was posed to me in final exam.This question is from Solid-State Switching Circuits in chapter Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Right choice is (a) 2.82 A

The explanation: The fundamental COMPONENT of SOURCE current in 1-Φ Full wave bridge RECTIFIER is 2√2Io÷π. It is the r.m.s VALUE of the fundamental component. IS1(r.m.s) = 2√2Io÷π=2√2=2.82 A.
Discussion
135.

Calculate the value of the fundamental displacement factor for 1-Φ Full wave semi-converter if the firing angle value is 69°.(a) .48(b) .24(c) .82(d) .88I have been asked this question in semester exam.The question is from Solid-State Switching Circuits topic in section Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

The correct OPTION is (c) .82

Easiest EXPLANATION: FUNDAMENTAL DISPLACEMENT factor is the cosine of angle DIFFERENCE between the fundamental voltage and fundamental current. The fundamental displacement factor for 1-Φ Full wave semi-converter is cos(∝÷2)=cos(34.5°)=.82.
Discussion
136.

Calculate the value of the fundamental displacement factor for 1-Φ Full wave bridge rectifier if the firing angle value is 38°.(a) .22(b) .78(c) .33(d) .44I got this question during an interview for a job.This question is from Solid-State Switching Circuits in division Introduction to Solid-State Switching Circuits of Electric Drives

Answer» RIGHT option is (b) .78

The explanation is: FUNDAMENTAL displacement FACTOR is the cosine of angle difference between the fundamental VOLTAGE and fundamental current. D.F=cos(∝)=cos(38°)=0.78.
Discussion
137.

Calculate the pulse number if the supply frequency is 2π and the output frequency is π÷6.(a) 4(b) 12(c) 16(d) 8I have been asked this question during an interview for a job.I want to ask this question from Solid-State Switching Circuits in chapter Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

The correct choice is (b) 12

The best explanation: The pulse NUMBER can be CALCULATED using the ratio of input FREQUENCY to the output frequency. The VALUE of pulse number (P) is 2π÷π÷6=12. It is a twelve-pulse converter.
Discussion
138.

Calculate the output frequency for the six-pulse converter if the supply frequency is 10 Hz.(a) 40 Hz(b) 30 Hz(c) 60 Hz(d) 80 HzI got this question in a national level competition.My doubt is from Solid-State Switching Circuits in portion Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

The CORRECT choice is (c) 60 HZ

The EXPLANATION is: The output of a SIX-pulse converter CONSISTS of six pulses in one cycle. The output frequency of the six pulse converter is 6×supply frequency=6×10=60 Hz.
Discussion
139.

Calculate the string efficiency if the de-rating factor is .429.(a) 48.1 %(b) 57.1 %(c) 47.8 %(d) 46.5 %The question was asked in semester exam.The above asked question is from Solid-State Switching Circuits in section Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

The correct OPTION is (B) 57.1 %

Easiest explanation: The string efficiency is CALCULATED for series and parallel CONNECTION of SCRs. The value of string efficiency is 1-(De-rating factor)=1-.429=57.1 %.
Discussion
140.

Calculate the circuit turn-off time for Half-wave controlled rectifier for a ω=5 rad/sec for resistive load.(a) .62 sec(b) .42 sec(c) .58 sec(d) .64 secI got this question in an international level competition.This question is from Solid-State Switching Circuits topic in division Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Correct answer is (a) .62 sec

Explanation: The VALUE of the CIRCUIT turn-off for Half-wave controlled RECTIFIER for a ω=5 rad/sec for the RESISTIVE load is a π÷Ω=π÷5=.62 sec.
Discussion
141.

In Half-wave controlled rectifier calculate the average value of the current for 2.5 Ω resistive load if the supply is sin(5.2t) and firing angle is 26°.(a) 0.8 V(b) 0.15 V(c) 0.12 V(d) 0.21 VThe question was posed to me in my homework.I would like to ask this question from Solid-State Switching Circuits in division Introduction to Solid-State Switching Circuits of Electric Drives

Answer» RIGHT option is (C) 0.12 V

Best explanation: In Half-wave controlled rectifier, the average VALUE of the current is Vm(1+cos(∝))÷2πR=(1+cos(26°))÷6.28×2.5=.12 V. The THYRISTOR will conduct from ∝ to π.
Discussion
142.

In Half-wave uncontrolled rectifier calculate the average value of the current for 3 Ω resistive load if the supply is 34sin(11t).(a) 3.6 A(b) 2.6 A(c) 2.5 A(d) 3.1 AI got this question in an interview for job.This intriguing question comes from Solid-State Switching Circuits topic in chapter Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Correct answer is (a) 3.6 A

For explanation I would say: In Half-wave UNCONTROLLED RECTIFIER, the average value of the current is Vm÷πR=34÷3π=3.6 A. The diode will conduct only for the positive half cycle. The CONDUCTION period of the diode is π.
Discussion
143.

The conduction period of diode in Half-wave uncontrolled rectifier for resistive load is ______________(a) π(b) 2π(c) 3π(d) 4πI got this question in unit test.Origin of the question is Solid-State Switching Circuits in section Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Correct answer is (a) π

For EXPLANATION I would say: The conduction period of the DIODE in Half-wave UNCONTROLLED rectifier for the resistive load is π. For the negative A.C supply diode will be reverse biased.
Discussion
144.

In Half-wave uncontrolled rectifier calculate the power dissipation across the 8 Ω resistor if the supply is 29sin(22t).(a) 26.2 W(b) 24.2 W(c) 26.1 W(d) 29.1 WThis question was posed to me during an online interview.My doubt is from Solid-State Switching Circuits topic in section Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Correct OPTION is (a) 26.2 W

Easiest explanation: In HALF-wave uncontrolled rectifier, the r.m.s value of the VOLTAGE is Vm÷2=29÷2=14.5 V. The diode will conduct only for the positive half cycle. Power dissipation ACROSS the RESISTOR is V^2r.m.s÷R=14.5^2÷8=26.2 W.
Discussion
145.

In Half-wave uncontrolled rectifier calculate the r.m.s value of the voltage if the supply is 89sin(41t).(a) 91.5 V(b) 44.5 V(c) 25.1 V(d) 15.1 VThis question was addressed to me in an interview.My enquiry is from Solid-State Switching Circuits topic in division Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

The CORRECT choice is (b) 44.5 V

Best explanation: In Half-wave uncontrolled rectifier, the r.m.s value of the voltage is Vm÷2=89÷2=44.5 V. The DIODE will CONDUCT only for the positive half cycle. The conduction PERIOD of the diode is π.
Discussion
146.

R-L-C underdamped loads are generally lagging power factor loads.(a) True(b) FalseThis question was posed to me in homework.Asked question is from Solid-State Switching Circuits topic in chapter Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

The CORRECT option is (b) False

The best I can explain: R-L-C underdamped loads are GENERALLY leading power FACTOR loads. They do not REQUIRE forced commutation. Anti-Parallel diodes help in the commutation process.
Discussion
147.

In Half-wave uncontrolled rectifier calculate the average value of the voltage if the supply is 3sin(5t).(a) .95 V(b) .92 V(c) .93 V(d) .94 VThe question was posed to me in examination.This is a very interesting question from Solid-State Switching Circuits in section Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

The correct answer is (a) .95 V

To explain I would say: In HALF-wave uncontrolled rectifier, the average value of the voltage is Vm÷π=3÷π=.95 V. The DIODE will conduct only for the POSITIVE half cycle. The conduction period of the diode is π.
Discussion
148.

Calculate the extinction angle in purely inductive load if the firing angle is 13°.(a) 328°(b) 347°(c) 349°(d) 315°This question was addressed to me in homework.This interesting question is from Solid-State Switching Circuits topic in section Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Right answer is (b) 347°

Explanation: The EXTINCTION angle in the purely inductive LOAD is 2π-∝=360°-13°=347°. The extinction angle is the angle at which current in the circuit becomes ZERO. The average value of the VOLTAGE in a purely inductive load is zero.
Discussion
149.

Calculate the conduction angle in purely inductive load if the firing angle is 165°.(a) 78°(b) 55°(c) 30°(d) 19°The question was posed to me at a job interview.Query is from Solid-State Switching Circuits in section Introduction to Solid-State Switching Circuits of Electric Drives

Answer» RIGHT answer is (c) 30°

Explanation: The conduction ANGLE in the PURELY inductive load is β-α=2(π-∝)=2(180°-165°)=30°. The conduction angle is the angle for which the current exists in the CIRCUIT. The AVERAGE value of the voltage in a purely inductive load is zero.
Discussion
150.

In Half-wave controlled rectifier calculate the average value of the voltage if the supply is 13sin(25t) and firing angle is 13°.(a) 4.08 V(b) 4.15 V(c) 3.46 V(d) 5.48 VI had been asked this question by my school teacher while I was bunking the class.Enquiry is from Solid-State Switching Circuits in section Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

The correct choice is (a) 4.08 V

The best EXPLANATION: In Half-wave CONTROLLED RECTIFIER, the AVERAGE value of the voltage is VM(1+cos(∝))÷2π=13(1+cos(13°))÷6.28=4.08 V. The thyristor will conduct from ∝ to π.
Discussion