88 + Interview Questions in Application Of Diodes in Electronic Devices & Circuits Page 2 InterviewSolution

51.	The ripple factor for an l section filter is_______(a) ϒ= 1/6√2ω^2LC(b) ϒ= 6√2ω^2LC(c) ϒ= 6√3ω^2LC(d) ϒ= 1/6√3ω^2LCThe question was asked in final exam.The origin of the question is L-Section Filter topic in chapter Application of Diodes of Electronic Devices & Circuits
Answer» The correct choice is (a) ϒ= 1/6√2ω^2LC Easiest explanation: The RIPPLE factor is the ratio of ROOT mean square (RMS) VALUE of ripple VOLTAGE to absolute value of dc component. It can also be expressed as the peak to peak value.

Discussion

52.	The scale of a voltmeter is uniform. Its type is _________(a) Moving Iron(b) Induction(c) Moving coil permanent magnet(d) Moving coil dynamometerThis question was addressed to me in an international level competition.The above asked question is from The Rectifier Voltmeter in chapter Application of Diodes of Electronic Devices & Circuits
Answer» Right answer is (c) Moving coil permanent magnet Easy explanation: Moving coil permanent magnet INSTRUMENTS have permanent magnets. It is suited for DC MEASUREMENT because here deflection is proportional to the voltage because resistance is constant for a MATERIAL of the meter and hence if voltage polarity is reversed, deflection of the POINTER will also be reversed so it is used only for DC measurement.

Discussion

53.	Transformer utilization factor of bridge full wave rectifier _________(a) 0.623(b) 0.812(c) 0.693(d) 0.825I have been asked this question during an online exam.This interesting question is from Bridge Rectifier topic in section Application of Diodes of Electronic Devices & Circuits
Answer» CORRECT OPTION is (b) 0.812 Easy EXPLANATION: Transformer utilization factor is the ratio of AC power delivered to LOAD to the DC power rating. This factor indicates effectiveness of transformer usage by rectifier. For bridge full wave rectifier it’s equal to 0.693.

Discussion

54.	In a centre tapped full wave rectifier, the input sine wave is 250sin100t. The output ripple frequency will be _________(a) 50Hz(b) 100Hz(c) 25Hz(d) 200HzI got this question in my homework.The origin of the question is Full-wave Rectifier in section Application of Diodes of Electronic Devices & Circuits
Answer» Right OPTION is (b) 100Hz Easy EXPLANATION: The equation of SINE wave is in the form Vmsinωt. So, by comparing we get ω=100. FREQUENCY, f =ω/2=50Hz. The output of centre tapped full wave rectifier has double the frequency of inpu. Hence, fout = 100Hz.

Discussion

55.	The value of inductance at which the current in a choke filter does not fall to zero is_________(a) peak inductance(b) cut-in inductance(c) critical inductance(d) damping inductanceThe question was asked in quiz.My question is from L-Section Filter topic in chapter Application of Diodes of Electronic Devices & Circuits
Answer» RIGHT answer is (c) critical inductance Explanation: When the value of inductance is INCREASED, a value is REACHED where the diodes supplies CURRENT CONTINUOUSLY. This value of inductance is called critical inductance.

Discussion

56.	Electronic voltmeters can be designed to measure ____________(a) Only very small voltages(b) Only very high voltages(c) Both very small and very high voltages(d) Neither high nor small voltagesThe question was posed to me during a job interview.The query is from The Rectifier Voltmeter in division Application of Diodes of Electronic Devices & Circuits
Answer» Right option is (C) Both very small and very high voltages The explanation is: A voltmeter is an instrument that measures the difference in electrical potential between TWO points in an electric circuit. An analog voltmeter moves a POINTER across a scale in proportion to the circuit’s voltage; a digital voltmeter provides a numerical DISPLAY. Thus it is suitable for measuring both small and high voltages.

Discussion

57.	The resistance of an ideal voltmeter is __________(a) low(b) infinite(c) zero(d) highI have been asked this question during an interview for a job.My doubt is from The Rectifier Voltmeter topic in chapter Application of Diodes of Electronic Devices & Circuits
Answer» The correct answer is (B) infinite The explanation is: The internal resistance of an ideal voltmeter is infinity and the internal resistance of an ideal ammeter is ZERO. Ammeter is CONNECTED in series and voltmeter is connected in parallel with the electric appliance. The resistance of an idea voltmeter is infinite so that it draws no current from the CIRCUIT under test.

Discussion

58.	If input frequency is 50Hz then ripple frequency of bridge full wave rectifier will be equal to_________(a) 200Hz(b) 50Hz(c) 45Hz(d) 100HzI have been asked this question in unit test.My enquiry is from Bridge Rectifier topic in portion Application of Diodes of Electronic Devices & Circuits
Answer» The correct choice is (d) 100Hz The best EXPLANATION: SINCE in the output of bridge rectifier one HALF cycle is repeated, the FREQUENCY will betwice as that of input frequency. So, f=100Hz.

Discussion

59.	Transformer utilisation factor of a half wave rectifier is _________(a) 0.234(b) 0.279(c) 0.287(d) 0.453This question was addressed to me in an interview for job.Origin of the question is Half-Wave Rectifier in chapter Application of Diodes of Electronic Devices & Circuits
Answer» Correct answer is (C) 0.287 Easy explanation: TRANSFORMER utilisation factor is the ratio of AC power delivered to load to the DC power rating. This factor indicates effectiveness of transformer usage by RECTIFIER. For a half wave rectifier, it’s low and EQUAL to 0.287.

Discussion

60.	For a given CLC filter, the operating frequency is 50Hz and 10µF capacitors used. The load resistance is 3460Ω with an inductance of 10henry. Calculate the ripple factor.(a) 0.165%(b) 0.142%(c) 0.178%(d) 0.321%The question was asked during an online exam.This key question is from CLC Filter in division Application of Diodes of Electronic Devices & Circuits
Answer» RIGHT answer is (a) 0.165% EASIEST EXPLANATION: We have, ϒ=5700 / (LC1C2RL) =5700 / (1010010^-12*3460) =0.165%.

Discussion

61.	The output dc voltage of an LC filter is_______(a) VDC=2Vm/π + IDCR(b) VDC=Vm/π – IDCR(c) VDC=2Vm/π – 2IDCR(d) VDC=2Vm/π – IDCRThis question was posed to me in an international level competition.I'm obligated to ask this question of L-Section Filter in section Application of Diodes of Electronic Devices & Circuits
Answer» Right option is (d) VDC=2Vm/π – IDCR Best explanation: The VALUE for VDC is same as that of inductor filter. If inductor has no dc resistance, then VDC=2Vm/π. If R is the SERIES resistance of TRANSFORMER, then VDC=2Vm/π – IDCR.

Discussion

62.	The output of a rectifier is pulsating because_________(a) It has a pulse variations(b) It gives a dc output(c) It contains both dc and ac components(d) It gives only ac componentsI have been asked this question in an interview for internship.I need to ask this question from Inductor Filters in chapter Application of Diodes of Electronic Devices & Circuits
Answer» Right OPTION is (C) It contains both dc and ac components The EXPLANATION is: For any electronic devices, a steady dc output is required.The filter is USED for this purpose. The ac components are REMOVED by using a filter.

Discussion

63.	Transformer utilization factor of a centre tapped full wave rectifier is_________(a) 0.623(b) 0.678(c) 0.693(d) 0.625I have been asked this question in a national level competition.This interesting question is from Full-wave Rectifier topic in section Application of Diodes of Electronic Devices & Circuits
Answer» Right option is (C) 0.693 The best explanation: Transformer utilisation factor is the RATIO of AC power delivered to load to the DC power rating. This factor INDICATES EFFECTIVENESS of transformer usage by rectifier. For a half wave rectifier, it’s low and EQUAL to 0.693.

Discussion

64.	Efficiency of a half wave rectifier is(a) 50%(b) 60%(c) 40.6%(d) 46%The question was posed to me in an online interview.The above asked question is from Half-Wave Rectifier in section Application of Diodes of Electronic Devices & Circuits
Answer» Correct answer is (c) 40.6% Best explanation: EFFICIENCY of a rectifier is the EFFECTIVENESS to CONVERT AC to DC. For half wave it’s 40.6%. It’s given by, Vout/Vin*100.

Discussion

65.	The below figure arrives to a conclusion that _________(a) for Vi > 0, V0=-(R2/R1)Vi(b) for Vi > 0, V0=0(c) Vi < 0, V0=-(R2/R1)Vi(d) Vi < 0, V0=0The question was posed to me during an interview for a job.This intriguing question comes from Half-Wave Rectifier topic in division Application of Diodes of Electronic Devices & Circuits
Answer» Correct choice is (B) for Vi > 0, V0=0 For EXPLANATION: The given op-amp is in inverting mode and this makes the output voltage to have a phase shift of 180°. The output voltage is now NEGATIVE. So, the diode 1 is reverse BIASED and diode 2 is forward biased. Then output is clearly ZERO.

Discussion

66.	A half wave rectifier, operated from a 50Hz supply uses a 1000µF capacitance connected in parallel to the load of rectifier. What will be the minimum value of load resistance that can beconnected across the capacitor if the ripple% not exceeds 5?(a) 114.87Ω(b) 167.98Ω(c) 115.47Ω(d) 451.35ΩI had been asked this question in an interview for internship.This interesting question is from Capacitor Filters in portion Application of Diodes of Electronic Devices & Circuits
Answer» Correct OPTION is (C) 115.47Ω Easy explanation: For a HALF WAVE filter, ϒ=1/2√3 fCRL=1/2√35010^-3*RL RL=10^3/5√3=115.47Ω.

Discussion

67.	The inductor is placed in the L section filter because_________(a) It offers zero resistance to DC component(b) It offers infinite resistance to DC component(c) It bypasses the DC component(d) It bypasses the AC componentThis question was posed to me in an interview for internship.The doubt is from CLC Filter in section Application of Diodes of Electronic Devices & Circuits
Answer» RIGHT option is (a) It offers zero resistance to DC component To EXPLAIN: The INDUCTOR offers high reactance to ac component and zero resistance to dc component. So, it blocks the ac component which cannot be bypassed by the capacitors.

Discussion

68.	In a shunt capacitor filter, the mechanism that helps the removal of ripples is_________(a) The current passing through the capacitor(b) The property of capacitor to store electrical energy(c) The voltage variations produced by shunting the capacitor(d) Uniform charge flow through the rectifierThe question was asked during an interview.My doubt stems from Capacitor Filters topic in portion Application of Diodes of Electronic Devices & Circuits
Answer» RIGHT option is (b) The PROPERTY of capacitor to store electrical energy Best explanation: Filtering is FREQUENTLY done by shunting the load with capacitor. It depends on the fact that a capacitor stores energy when conducting and delivers energy during non conduction. Throughout this PROCESS, the ripples are eliminated.

Discussion

69.	Which of the instruments is most accurate?(a) PMMC(b) Moving iron(c) Thermo couple(d) Induction typeThis question was posed to me in an internship interview.I'd like to ask this question from The Rectifier Voltmeter topic in section Application of Diodes of Electronic Devices & Circuits
Answer» Correct answer is (a) PMMC For EXPLANATION I WOULD SAY: This torque in PMMC ensures the pointer comes to an equilibrium position i.e. at rest in the scale without oscillating to give an accurate reading. In PMMC as the coil moves in the magnetic field, eddy current sets up in a METAL former or core on which the coil is wound or in the circuit of the coil itself which opposes the motion of the coil resulting in the slow swing of a pointer and then come to rest quickly with very little oscillation. It has great accuracy.

Discussion

70.	In a bridge full wave rectifier, the input sine wave is 250sin100t. The output ripple frequency will be_________(a) 50Hz(b) 200Hz(c) 100Hz(d) 25HzThis question was addressed to me in an interview for internship.My question is based upon Bridge Rectifier topic in section Application of Diodes of Electronic Devices & Circuits
Answer» Right OPTION is (c) 100Hz For EXPLANATION I would say: The equation of sine wave is in the form of Emsinωt. So, ω=100 and frequency (f)=ω/2=50Hz. SINCE OUTPUT of bridge RECTIFIER have double the frequency of input, f=100Hz.

Discussion

71.	Determine the maximum and minimum values of load current for which the Zener diode shunt regulator will maintain regulation when VIN=24V and R=500Ω. The Zener diode has a VZ=12V and (IZ)MAX=90mA.(a) 40mA, 0mA respectively(b) 36mA, 5mA respectively(c) 10mA, 6mA respectively(d) 21mA, 0mA respectivelyI got this question in an interview for internship.My question comes from Voltage Regulation Using Zener Diode topic in chapter Application of Diodes of Electronic Devices & Circuits
Answer» The correct answer is (d) 21mA, 0MA respectively The best I can explain: The current through the resistance R is given by, I=(VIN-VZ)/R= (24-12)/500=24mA. (IL)MAX=I-(IZ)MIN=24-3=21mA .This current is LESS than (IZ)MAX. So, we assume that all the input current flows through the Zener diode. Under this condition,(IL)MIN is 0mA.

Discussion

72.	The condition for the regulation curve in a choke filter is_________(a) LC≥RL/3ω(b) LC≤RL/3ω(c) L≥RL/3ω(d) LC≥RL3ωThis question was posed to me during an online interview.Query is from L-Section Filter topic in portion Application of Diodes of Electronic Devices & Circuits
Answer» CORRECT CHOICE is (a) LC≥RL/3ω The best I can explain: IDC should not exceed the NEGATIVE peak of ac component. So, the regulation curve in direct output voltage against load current for a filter is GIVEN the relationLC≥RL/3ω.

Discussion

73.	A full wave rectifier with a load resistance of 5KΩ uses an inductor filter of 15henry. The peak value of applied voltage is 250V and the frequency is 50 cycles per second. Calculate the dc load current.(a) 0.7mA(b) 17mA(c) 10.6mA(d) 20mAThis question was posed to me in examination.I want to ask this question from Inductor Filters topic in portion Application of Diodes of Electronic Devices & Circuits
Answer» RIGHT ANSWER is (c) 10.6mA Easy explanation: For a RECTIFIER with an inductor filter, VDC=2Vm/π, Idc=VDC/RL=2Vm/RLπ IDC=2250/(3.1415*103)=10.6mA.

Discussion

74.	What is the effect of an inductor filter on a multi frequency signal?(a) Dampens the AC signal(b) Dampens the DC signal(c) To reduce ripples(d) To change the currentI got this question in an interview for internship.This intriguing question originated from Inductor Filters topic in division Application of Diodes of Electronic Devices & Circuits
Answer» Correct ANSWER is (a) DAMPENS the AC signal To explain: Presence of inductor usually dampens the AC signal. Due to self INDUCTION induces opposing EMF or changes in the CURRENT.

Discussion

75.	In reverse bias, rectifier behaves as a___________(a) Resistor(b) Capacitor(c) Inductor(d) AmplifierThe question was asked during a job interview.This is a very interesting question from The Rectifier Voltmeter topic in section Application of Diodes of Electronic Devices & Circuits
Answer» The correct choice is (b) Capacitor The explanation is: The rectifier exhibits capacitance PROPERTIES when reverse biased, and tends to BYPASS higher frequencies. The METER reading may be in ERROR by as MUCH as 0.5% decrease for every 1 kHz rise in frequency.

Discussion

76.	In a centre tapped full wave rectifier, RL=1KΩ and for diode Rf=10Ω. The primary voltage is 800sinωt with transformer turns ratio=2. The ripple factor will be _________(a) 54%(b) 48%(c) 26%(d) 81%This question was posed to me during an online interview.My question is taken from Full-wave Rectifier topic in division Application of Diodes of Electronic Devices & Circuits
Answer» CORRECT ANSWER is (b) 48% To elaborate: The ripple factor ϒ= [(IRMS/IAVG)^2 – 1]^1/2. IRMS =Im /√2=Vm/(Rf+RL)√2=200/1.01=198. (SECONDARY line to line VOLTAGE is 800/2=400. Due to centre tap Vm=400/2=200) IRMS=198/√2=140mA, IAVG=2*198/π=126mA. ϒ=[(140/126)^2-1]^1/2=0.48. So, ϒ=48%.

Discussion

77.	AC voltmeter consists of_________(a) half wave rectifier(b) full wave rectifier(c) center tap rectifier(d) bridge wave rectifierThis question was posed to me in my homework.The query is from The Rectifier Voltmeter in section Application of Diodes of Electronic Devices & Circuits
Answer» The correct choice is (d) bridge wave rectifier To elaborate: The bridge rectifier provides a FULL wave PULSATING dc. Due to the inertia of the movable coil, the meter indicates a steady deflection proportional to the average value of the current. The meter scale is usually calibrated to give the RMS value an alternating sine wave INPUT.

Discussion

78.	A transistor in a series voltage regulator acts like a variable resistor. The value of its resistance is determined by _______(a) emitter current(b) base current(c) collector current(d) it is not controlled by the transistor terminalsI had been asked this question at a job interview.Query is from Voltage Regulation Using Zener Diode in section Application of Diodes of Electronic Devices & Circuits
Answer» Right CHOICE is (b) base CURRENT For explanation I would say: The principle is based on the fact that a large fraction of the INCREASE in input voltage appears ACROSS the transistor so that the output voltage remains to be constant. When input voltage is increased, the output voltage ALSO increases which biases the transistor towards less current.

Discussion

79.	What is the number of capacitors and inductors used in a CLC filter?(a) 1, 2 respectively(b) 2, 1 respectively(c) 1, 1 respectively(d) 2, 2 respectivelyThis question was addressed to me in an online quiz.The query is from CLC Filter topic in division Application of Diodes of Electronic Devices & Circuits
Answer» RIGHT option is (b) 2, 1 respectively To explain: A very smooth output can be obtained by a filter consisting of ONE inductor and two capacitors connected across each other. They are arranged in the FORM of letter ‘pi’. So, these are also called as pi filters.

Discussion

80.	A 100µF capacitor when used as a filter has 15V ac across it with a load resistor of 2.5KΩ. If the filter is the full wave and supply frequency is 50Hz, what is the percentage of ripple frequency in the output?(a) 2.456%(b) 1.154%(c) 3.785%(d) 3.675%I have been asked this question in an online interview.This interesting question is from Capacitor Filters in chapter Application of Diodes of Electronic Devices & Circuits
Answer» Right answer is (B) 1.154% Easiest EXPLANATION: For a FULL WAVE rectifier, ϒ=1/4√3 fCRL =1/4√35010^-3*2.5 =0.01154. So, ripple is 1.154%.

Discussion

81.	The charge (q) lost by the capacitor during the discharge time for shunt capacitor filter.(a) IDCT(b) IDC/T(c) IDC2T(d) IDC/2TI had been asked this question in homework.This intriguing question originated from Capacitor Filters topic in chapter Application of Diodes of Electronic Devices & Circuits
Answer» The CORRECT ANSWER is (a) IDC*T For EXPLANATION: The ‘T’ is the total non conducting time of CAPACITOR. The charge per unit time will give the current flow.

Discussion

82.	The cut-in point of a capacitor filter is_________(a) The instant at which the conduction starts(b) The instant at which the conduction stops(c) The time after which the output is not filtered(d) The time during which the output is perfectly filteredI got this question in final exam.I'm obligated to ask this question of Capacitor Filters in section Application of Diodes of Electronic Devices & Circuits
Answer» Right option is (a) The INSTANT at which the conduction starts For EXPLANATION: The CAPACITOR charges when the diode is in ON state and discharges during the OFF state of the diode. The instant at which the conduction starts is called cut-in point. The instant at which the conduction stops is called cut-out point.

Discussion

83.	In a general rectifier voltmeter, the meter has low sensitivity because of ___________ of the diode.(a) low forward resistance(b) high forward resistance(c) low reverse impedance(d) high reverse impedanceThe question was posed to me in a national level competition.My doubt is from The Rectifier Voltmeter in division Application of Diodes of Electronic Devices & Circuits
Answer» Correct choice is (b) HIGH forward resistance The EXPLANATION is: The diode offers high resistance when forward biased. Thus, offering more resistance to the current flow in the CIRCUIT. Thus making the meter LESS sensitive.

Discussion

84.	If the peak voltage on a centre tapped full wave rectifier circuit is5V and diode cut in voltage is 0.7. The peak inverse voltage on diode is_________(a) 4.3V(b) 9.3V(c) 5.7V(d) 10.7VI have been asked this question in an online quiz.Question is taken from Full-wave Rectifier in chapter Application of Diodes of Electronic Devices & Circuits
Answer» The correct choice is (b) 9.3V The explanation: PIV is the maximum REVERSE bias VOLTAGE that can be appeared ACROSS a DIODE in the given circuit, if PIV rating is less than this value of breakdown of diode will occur. For a RECTIFIER, PIV=2Vm-Vd = 10-0.7 = 9.3V.

Discussion

85.	If input frequency is 50Hz for a full wave rectifier, the ripple frequency of it would be _________(a) 100Hz(b) 50Hz(c) 25Hz(d) 500HzI got this question in examination.I would like to ask this question from Full-wave Rectifier in section Application of Diodes of Electronic Devices & Circuits
Answer» RIGHT answer is (a) 100HZ For explanation I would say: In the output of the centre tapped rectifier, one of the half cycle is REPEATED. The FREQUENCY will be twice as that of INPUT frequency. So, it’s 100Hz.

Discussion

86.	A full wave rectifier delivers 50W to a load of 200Ω. If the ripple factor is 2%, calculate the AC ripple across the load.(a) 2V(b) 5V(c) 4V(d) 1VI had been asked this question in exam.This intriguing question originated from Full-wave Rectifier topic in chapter Application of Diodes of Electronic Devices & Circuits
Answer» CORRECT CHOICE is (a) 2V To elaborate: We KNOW that, PDC=VDC^2/RL. So, VDC=(PDCRL)^1/2=10000^1/2=100V. Here, ϒ=0.02 ϒ=VAC/VDC=VAC/100.So, VAC=0.02100=2V.

Discussion

87.	A full wave rectifier supplies a load of 1KΩ. The AC voltage applied to diodes is 220V (rms). If diode resistance is neglected, what is the ripple voltage?(a) 0.562V(b) 0.785V(c) 0.954V(d) 0.344VThe question was posed to me in an interview for internship.Enquiry is from Full-wave Rectifier topic in portion Application of Diodes of Electronic Devices & Circuits
Answer» The CORRECT answer is (c) 0.954V The explanation is: The ripple voltage is(Vϒ)RMS=ϒVDC /100. VDC=0.636VRMS √2=0.636220 √2=198V and ripple factor ϒ for full wave rectifier is 0.482. Hence, (Vϒ)RMS=0.482*198 /100=0.954V.

Discussion

88.	In a half wave rectifier, the sine wave input is 200sin300t. The average value of output voltage is?(a) 57.876V(b) 67.453V(c) 63.694V(d) 76.987VI had been asked this question by my school principal while I was bunking the class.Origin of the question is Half-Wave Rectifier topic in portion Application of Diodes of Electronic Devices & Circuits
Answer» Correct answer is (C) 63.694V The EXPLANATION: Comparing with the STANDARD equation, Vm=200V. AVERAGE value is given by, Vavg=Vm/π. So, 200/π=63.694.

Discussion

Explore topic-wise InterviewSolutions in .

The ripple factor for an l section filter is_______(a) ϒ= 1/6√2ω^2LC(b) ϒ= 6√2ω^2LC(c) ϒ= 6√3ω^2LC(d) ϒ= 1/6√3ω^2LCThe question was asked in final exam.The origin of the question is L-Section Filter topic in chapter Application of Diodes of Electronic Devices & Circuits

Transformer utilization factor of bridge full wave rectifier _________(a) 0.623(b) 0.812(c) 0.693(d) 0.825I have been asked this question during an online exam.This interesting question is from Bridge Rectifier topic in section Application of Diodes of Electronic Devices & Circuits

The resistance of an ideal voltmeter is __________(a) low(b) infinite(c) zero(d) highI have been asked this question during an interview for a job.My doubt is from The Rectifier Voltmeter topic in chapter Application of Diodes of Electronic Devices & Circuits

If input frequency is 50Hz then ripple frequency of bridge full wave rectifier will be equal to_________(a) 200Hz(b) 50Hz(c) 45Hz(d) 100HzI have been asked this question in unit test.My enquiry is from Bridge Rectifier topic in portion Application of Diodes of Electronic Devices & Circuits

Transformer utilisation factor of a half wave rectifier is _________(a) 0.234(b) 0.279(c) 0.287(d) 0.453This question was addressed to me in an interview for job.Origin of the question is Half-Wave Rectifier in chapter Application of Diodes of Electronic Devices & Circuits

Efficiency of a half wave rectifier is(a) 50%(b) 60%(c) 40.6%(d) 46%The question was posed to me in an online interview.The above asked question is from Half-Wave Rectifier in section Application of Diodes of Electronic Devices & Circuits

Which of the instruments is most accurate?(a) PMMC(b) Moving iron(c) Thermo couple(d) Induction typeThis question was posed to me in an internship interview.I'd like to ask this question from The Rectifier Voltmeter topic in section Application of Diodes of Electronic Devices & Circuits

The condition for the regulation curve in a choke filter is_________(a) LC≥RL/3ω(b) LC≤RL/3ω(c) L≥RL/3ω(d) LC≥RL3ωThis question was posed to me during an online interview.Query is from L-Section Filter topic in portion Application of Diodes of Electronic Devices & Circuits

In reverse bias, rectifier behaves as a___________(a) Resistor(b) Capacitor(c) Inductor(d) AmplifierThe question was asked during a job interview.This is a very interesting question from The Rectifier Voltmeter topic in section Application of Diodes of Electronic Devices & Circuits

AC voltmeter consists of_________(a) half wave rectifier(b) full wave rectifier(c) center tap rectifier(d) bridge wave rectifierThis question was posed to me in my homework.The query is from The Rectifier Voltmeter in section Application of Diodes of Electronic Devices & Circuits

If input frequency is 50Hz for a full wave rectifier, the ripple frequency of it would be _________(a) 100Hz(b) 50Hz(c) 25Hz(d) 500HzI got this question in examination.I would like to ask this question from Full-wave Rectifier in section Application of Diodes of Electronic Devices & Circuits