# 210+ Electronic Circuits 1 Solved MCQs

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101.

## For the circuit shown below express v0 as a function of v1 and v2.

A. v0 = v1 + v2
B. v0 = v2 – v1
C. v0 = v1 – v2
D. v0 = -v1 – v2
Answer» B. v0 = v2 – v1
Explanation: considering the fact that the potential at the input terminals are identical and proceeding we obtain the given result.
102.

## Determine Ad and Ac for the given circuit.

A. ac = 0 and ad = 1
B. ac ≠ 0 and ad = 1
C. ac = 0 and ad ≠ 1
D. ac ≠ 0 and ad ≠ 1
Explanation: consider the fact that the potential at the input terminals are identical and obtain the values of v1 and v2. thus obtain the value of vd and vc.
103.

## Determine the voltage gain for the given circuit known that R1 = R3 = 10kΩ abd R2 = R4 = 100kΩ.

A. 1
B. 10
C. 100
D. 1000
Explanation: voltage gain is 100/10.
104.

## What is trans-conductance?

A. ratio of change in drain current to change in collector current
B. ratio of change in drain current to change in gate to source voltage
C. ratio of change in collector current to change in drain current
D. ratio of change in collector current to change in gate to source voltage
Answer» B. ratio of change in drain current to change in gate to source voltage
Explanation: the change in drain current which is resulted due to change in gate to source voltage in a fet is measured by trans- conductance. this is termed as trans because it provides relationship between input and output quantity.
105.

## What is the value of trans conductance?

A. 1
B. 2
C. 0.1
D. 0.01
Explanation: trans-conductance= change in drain current/ change in gate to source voltage
106.

## 002. What is the value ofgm?

A. 1
B. 2
C. 0.002
D. 0
Explanation: gm = change in drain current/ change in gate to source voltage gm = slope of vgs vs id gm = 0.002.
107.

## Find the maximum value of gm for FET with IDSS=10mA, Vp=-2V, VGS=5V?

A. 10ms
B. 20ms
C. 1ms
D. 0
Explanation: gm0=2idss/
108.

A. 10ms
B. 20ms
C. 5ms
D. 14ms
109.

## Determine the value of output impedance for JFET, if the value of gm =1mS?

A. 1kohm
B. 0
C. 100kohm
D. 5kohm
Explanation: output impedance=inverse of trans conductance
110.

## In a small signal equivalent model of an FET, What does gm VGS stand for?

A. a pure resistor
B. voltage controlled current source
C. current controlled current source
D. voltage controlled voltage source
Answer» B. voltage controlled current source
Explanation: for fet, the voltage is applied across the gate and source to control the drain current, hence while writing small signal model of an fet, on the output side gm vgs represents a current source which can be controlled by the input voltage vgs.
111.

## Given yfs = 3.6mS and yos = 0.02mS, determine r0?

A. 100kohm
B. 50mohm
C. 50kohm
D. 20kohm
Explanation: r0=1/yos
112.

## Which of the following statement is incorrect?

A. output of ce amplifier is out of phase with respect to its input
B. cc amplifier is a voltage buffer
C. cb amplifier is a voltage buffer
D. ce amplifier is used as an audio (low frequency) amplifier
Answer» C. cb amplifier is a voltage buffer
Explanation: the output of the ce amplifier has a phase shift of 180o with respect to the input. the cc amplifier has av≅1, thus it is a voltage buffer. however, the cb amplifier has a large voltage gain, and its current gain ai≅1, thus it is a current buffer. ce amplifier has an application has an audio amplifier.
113.

## Given hfe = 60, hie=1000Ω, hoe = 20μ Ω–, hre = 2 * 10-4. Find the current gain of the BJT, correct up to two decimal points.

A. – 58.44
B. -59.21
C. – 60.10
D. – 60.00
Explanation: current gain, ai = – hf / (1 +
114.

## Consider the circuit. Given hfe = 50, hie = 1200Ω. Find voltage gain.

A. – 278
B. -277.9
C. – 300
D. – 280
Explanation: voltage gain = av = - hferl’/hie
115.

## Given that IB = 5mA and hfe = 55, find load current.

A. 28ma
B. 280ma
C. 2.5a
D. 2a
Explanation: in given circuit, which is an emitter follower, current gain = 1 + hfe
116.

## Consider the following circuit, where source current = 10mA, hfe = 50, hie = 1100Ω, then for the transistor circuit, find output resistance RO and input resistance RI.

A. ro = 0, ri = 21Ω
B. ro = ∞, ri = 0Ω
C. ro = ∞, ri = 21Ω
D. ro = 10, ri = 21Ω
Answer» C. ro = ∞, ri = 21Ω
Explanation: since hoe is not given, we can consider it to be small; i.e 1/hoe is neglected, open circuited. hence output resistance ro =
117.

## For the given circuit, input resistance RI = 20Ω, hfe = 50. Output resistance = ∞. Find the new values of input and output resistance, if a base resistance of 2kΩ is added to the circuit.

A. ri = 20Ω, ro = ∞
B. ri = 20Ω, ro = 2kΩ
C. ri = 59Ω, ro = ∞
D. ri = 59Ω, ro = 2kΩ
Answer» C. ri = 59Ω, ro = ∞
Explanation: ri = 20k = hie/(1+hfe) = hie/51 hie =1020 Ω
118.

## If source resistance in an amplifier circuit is zero, then voltage gain (output to input voltage ratio) and source voltage gain (output to source voltage ratio) are the same.

A. true
B. false
Explanation: when a source resistance rs is present, the voltage gain with respect to source becomes
119.

## The Differential output of the difference amplifier is the amplification of

A. difference between the voltages of input signals
B. difference between the output of the each transistor
C. difference between the supply and the output of the each transistor
D. all of the mentioned
Answer» A. difference between the voltages of input signals
Explanation: none.
120.

## The inputs to the differential amplifier are applied at

A. at x and y
B. at the gates of m1 and m2
C. all of the mentioned
D. none of the mentioned
Answer» B. at the gates of m1 and m2
Explanation: none.
121.

## In Common Mode Differential Amplifier, the outputs Vout1 and Vout2 are related as:

A. vout2 is in out of phase with vout1 with same amplitude
B. vout2 and vout1 have same amplitude but the phase difference is 90 degrees
C. vout1 and vout2 have same amplitude and are in phase with each other and their respective inputs
D. vout1 and vout2 have same amplitude and are in phase with each other but out of phase with their respective inputs
Answer» D. vout1 and vout2 have same amplitude and are in phase with each other but out of phase with their respective inputs
Explanation: none.
122.

## In a small signal differential gain vs input CM level graph, the gain decreases after V2 due to:

A. as the input voltage increases, the output will be clipped
B. when the input voltage to the transistors are high, the transistor enters saturation region and increases the current, which inturn decreases the output voltage = vdd – rd.iss
C. when common mode voltage is greater than or equal to v2, the input transistors enter triode region, the gain begins to fall
D. increasing the input voltage beyond v2 causes the gate oxide to conduct and the gain is reduced
Answer» C. when common mode voltage is greater than or equal to v2, the input transistors enter triode region, the gain begins to fall
Explanation: none.
123.

## Consider a voltage amplifier having a frequency response of the low-pass STC type with a dc gain of 60 dB and a 3-dB frequency of 1000 Hz. Then the gain db at

A. f = 10 hz is 55 db
B. f = 10 khz is 45 db
C. f = 100 khz is 25 db
D. f = 1mhz is 0 db
Answer» D. f = 1mhz is 0 db
Explanation: use standard formulas for frequency response and voltage gain.
124.

## STC networks can be classified into two categories: low-pass (LP) and high-pass (HP). Then which of the following is true?

A. hp network passes dc and low frequencies and attenuate high frequency and opposite for lp network
B. lp network passes dc and low frequencies and attenuate high frequency and opposite for hp network
C. hp network passes dc and high frequencies and attenuate low frequency and opposite for lp network
D. lp network passes low frequencies only and attenuate high frequency and opposite for hp network
Answer» B. lp network passes dc and low frequencies and attenuate high frequency and opposite for hp network
Explanation: by definition a lp network allows dc current (or low frequency current) and an lp network does the opposite, that is, allows high frequency ac current.
125.

## The signal whose waveform is not effected by a linear circuit is

A. triangular waveform signal
B. rectangular waveform signal
C. sine/cosine wave signal
D. sawtooth waveform signal
Explanation: only sine/cosine wave are not affected by a linear circuit while all other waveforms are affected by a linear circuit.
126.

## Which of the following is not a classification of amplifiers on the basis of their frequency response?

A. capacitively coupled amplifier
B. direct coupled amplifier
C. bandpass amplifier
D. none of the mentioned
Answer» D. none of the mentioned
Explanation: none of the options provided are correct.
127.

## General representation of the frequency response curve is called

A. bode plot
B. miller plot
C. thevenin plot
D. bandwidth plot
Explanation: general representation of frequency response curves are called bode plot. bode plots are also called semi logarithmic plots since they have logarithmic values values on one of the axes.
128.

## Under what condition can the circuit shown be called a compensated attenuator.

A. c1r1 = c2r2
B. c1r2 = c2r1
C. c1c2 = r1r2
D. r1 = 0
Explanation: standard condition of a compensated attenuator. here is the derivation for the same.
129.

## Which of the following is true?

A. coupling capacitors causes the gain to fall off at high frequencies
B. internal capacitor of a device causes the gain to fall off at low frequencies
C. all of the mentioned
D. none of the mentioned
Answer» D. none of the mentioned
Explanation: both the statements are false.
130.

## Which of the following is true?

A. monolithic ic amplifiers are directly coupled or dc amplifiers
B. televisions and radios use tuned amplifiers
C. audio amplifiers have coupling capacitor amplifier
D. all of the mentioned
Answer» D. all of the mentioned
Explanation: these all are practical applications of different types of amplifiers.
131.

## During high frequency applications of a B.J.T., which parasitic capacitors arise between the base and the emitter?

A. cje and cb
B. ccs
C. cb
D. ccs and cb
Explanation: there are two capacitors which arise between bases and emitter. one is cje due to depletion region associated between base and emitter. cb is another capacitor which arises due to the accumulation of electrons in the base which further results into the concentration gradient within the base of the transistor.
132.

## During high frequency applications of a B.J.T., which parasitic capacitors arise between the collector and the emitter?

A. no capacitor arises
B. ccs
C. cb
D. ccs and cb
Explanation: the emitter and the collector are far away from each other when the b.j.t. is being constructed. hence, we find that they don’t share a common junction where charges can accumulate. thus, no such parasitic capacitors appear.
133.

## During high frequency applications of a B.J.T, which parasitic capacitors arise between the collector and the base?

A. cje and cb
B. ccs
C.
D.
Explanation: only one capacitor up between the base and the collector. this is due to the depletion region present between the base and the collector region.
134.

## 3 BJT FREQUENCY RESPONSE

A. cje and cb
B. ccs and cµ
C. cb
D. ccs and cb
Explanation: there are two capacitors attached to the collector terminal. the collector-base junction provides a depletion capacitance (cµ) while the collector substrate junction provides a certain capacitance (ccs).
135.

## Which parasitic capacitors do not affect the frequency response of the C.E. stage, of the B.J.T.?

A. cje and cb
B. ccs and cµ
C. cb and cµ
D. no parasitic capacitor gets deactivated
Answer» D. no parasitic capacitor gets deactivated
Explanation: while observing the frequency response of a c.e. stage, we find that all the parasitic capacitances of the b.j.t. end up slowing the speed of the b.j.t. the frequency response of this stage is affected by all the parasitic capacitors.
136.

## Which parasitic capacitors don’t affect the frequency response of the C.B. stage of the B.J.T.?

A. none of the parasitic capacitances
B. all the parasitic capacitances
C. some of the coupling capacitors
D. ccs and cb
Answer» B. all the parasitic capacitances
Explanation: all the parasitic capacitors of a
137.

## Which parasitic capacitors don’t affect the frequency response of the C.C. stage of the B.J.T.?

A. ccs
B. ccs and cb
C. cb
D. ccs and cµ
Explanation: in the follower stage, the load is present at the emitter. the parasitic capacitors present between the collector and the substrate i.e. cµ gets deactivated. this is observed from the small signal analysis where both the terminals of this capacitor get shorted to a.c. ground.
138.

## If the transconductance of the B.J.T increases, the transit frequency

A. increases
B. decreases
C. doesn’t get affected
D. doubles
Explanation: the transit frequency is directly proportional to the transconductance of the
139.

## If the total capacitance between the base and the emitter increases by a factor of 2, the transit frequency

A. reduces by 2
B. increases by 2
C. reduces by 4
D. increases by 4
Explanation: the transit frequency is almost inversely proportional to the total capacitance between the base and the emitter of the b.j.t. hence, the transit frequency will approximately reduce by 2 and the correct option becomes reduces by 2.
140.

## Which effect plays a critical role in producing changes in the frequency response of the B.J.T.?

A. thevenin’s effect
B. miller effect
C. tellegen’s effect
D. norton’s effect
Explanation: the miller effect results in a change in the capacitance seen between the base and the collector. this is why, it affects the frequency response of the b.j.t. deeply by changing the poles and affecting the high frequency voltage gain stage.
141.

## If a C.E. stage has a load Rl and transconductance gm, what is the factor by which the capacitance between the base and the collector at the input side gets multiplied?

A. 1 + gmrl
B. 1 – gmrl
C. 1 + 2*gmrl
D. 1 – 2*gmrl
Explanation: the low frequency gain of the
142.

## If a C.E. stage has a load Rl and transconductance gm, what is the factor by which the capacitance between the base and the collector at the output side gets multiplied?

A. 1 + 1/gmrl
B. 1 – 1/gmrl
C. 1 + 2/gmrl
D. 1 – 2/gmrl
Explanation: the low frequency response of
143.

## Given that transition capacitance is 5 pico F and diffusion capacitance is 80 pico F, and base emitter dynamic resistance is 1500 Ω, find the β cut-off frequency.

Explanation: the frequency in radians is calculated by
144.

## 8 x 106.

A. 200
B. 100
C. 141.42
D. 440.2
Explanation: the current gain for the ce circuit is a =          β
145.

## Gain bandwidth frequency is GBP= 3000 Mhz. The cut-off frequency is f=10Mhz. What is the CE short circuit current gain at the β cutoff frequency?

A. 212
B. 220
C. 300
D. 200
Explanation: ft = 3000mhz βfβ = 3000mhz
146.

## Which of the statement is incorrect?

A. at unity gain frequency the ce short circuit current gain becomes 1
B. unity gain frequency is the same as gain
Answer» A. at unity gain frequency the ce short circuit current gain becomes 1
Explanation: gain bandwidth product for any mosfet is ft = gm/2π(cgs+cgd) thus gbp is approximately 5.9 mhz.
147.

## In an RC coupled CE amplifier, when the input frequency increases, which of these are incorrect?

A. reactance csh decreases
B. voltage gain increases
C. voltage gain decreases due to shunt capacitance
D. an rc coupled amplifier behaves like a low pass filter
Explanation: when frequency increases, shunt reactance decreases. the voltage drop across shunt capacitance decreases and net voltage gain decrease. rc coupled amplifier acts as a low pass filter at high frequencies.
148.

## In Miller’s theorem, what is the constant K?

A. total voltage gain
B. internal voltage gain
C. internal current gain
D. internal power gain
Explanation: the constant k=v2/v1, which is the internal voltage gain of the network.
149.

## When applying miller’s theorem to resistors, resistance R1 is for node 1 and R2 for node 2. If R1>R2, then for same circuit, then for capacitance for which the theorem is applied, which will be larger, C1 or C2?

A. c1
B. c2
C. both are equal
D. insufficient data
Explanation: given r1>r2
150.

## Find net voltage gain, given hfe = 50 and hie = 1kΩ.

A. 27.68
B. -22
C. 30.55
D. -27.68
Explanation: apply millers theorem to resistance between input and output.
151.

## Given that capacitance w.r.t the input node is 2pF and output node is 4pF, find capacitance between input and output node.

A. 0.67 pf
B. 1.34pf
C. 0.44pf
D. 2.2pf
Explanation: c1=c(1-k), c2=c(1-k-1) c1=2pf
152.

## Consider an RC coupled amplifier at low frequency. Internal voltage gain is -120. Find the voltage gain magnitude, when given that collector resistance = 1kΩ, load = 9kΩ, collector capacitance is 0. is 0.1μF, and input frequency is 20Hz.

A. 120
B. 12
C. 15
D. -12
Explanation: av = -120
153.

## 15Hz

A. 22.73 hz
B. 612 hz
C. 673hz
D. 317 hz
Explanation: rc = 2kΩ, rl = 5kΩ, cc = 1μf, cb = 10μf, ce = 20μf, rs = 2 kΩ
154.

## Find the 3-dB frequency given that the gain of RC coupled amplifier is 150, the low frequency voltage gain is 100 and the input frequency is 50Hz.

A. 50.8 hz
B. 55.9 hz
C. 60hz
D. 100hz
Explanation: avm = 150 avl = 100
155.

## What is the phase shift in RC coupled CE amplifier at lower 3dB frequency?

A. 180°
B. 225°
C. 270°
D. 100°
Explanation: total phase shift = 180°+ tan- 1(fl/f)
156.

## We cannot use h-parameter model in high frequency analysis because

A. they all can be ignored for high frequencies
B. junction capacitances are not included in it
C. junction capacitances have to be included in it
D. ac analysis is difficult for high frequency using it
Answer» B. junction capacitances are not included in it
Explanation: the effect of smaller capacitors is considerable in high frequency analysis of analog circuits, and hence they cannot be ignored as. instead of h-parameter model, we use π-model.
157.

## Which of the statement is incorrect?

A. at unity gain frequency the ce short circuit current gain becomes 1
B. unity gain frequency is the same as gain bandwidth product of bjt
C. gain of bjt decreases at higher frequencies due to junction capacitances
D. β- cut-off frequency is one where the ce short circuit current gain becomes β/2
Answer» A. at unity gain frequency the ce short circuit current gain becomes 1
Explanation: gain bandwidth product for any mosfet is ft = gm/2π(cgs+cgd) thus gbp is approximately 5.9 mhz.
158.

## The collector current will not reach the steady state value instantaneously because of

A. stray capacitances
B. resistances
C. input blocking capacitances
D. coupling capacitance
Explanation: when a pulse is given, the collector current will not reach the steady state value instantaneously because of stray capacitances. the charging and discharging of capacitance makes the current to reach a steady state value after a given time constant.
159.

## For the BJT, β=∞, VBEon=0.7V VCEsat=0.7V. The switch is initially closed. At t=0, it is opened. At which time the BJT leaves the active region?

A. 20ms
B. 50ms
C. 60ms
D. 70ms
Explanation: at t < 0, the bjt is off in cut off region. ib=0 as β=∞, so ic=ie. when t > 0, switch opens and bjt is on. the voltage across capacitor increases. from the input loop, -5-vbe-i(4.3k)+10=0 and gives i=1ma. ic1=1-0.5=0.5ma. vc1=0.7+4.3+10=-5v. ic1=c1dvc1/dt. from
160.

## The technique used to quickly switch off a transistor is by

A. reverse biasing its emitter to collector junction
B. reverse biasing its base to collector junction
C. reverse biasing its base to emitter junction
D. reverse biasing any junction
Answer» C. reverse biasing its base to emitter junction
Explanation: the technique used to quickly switch off a transistor is by reverse biasing its base to collector junction. it is demonstrated in a high voltage switching circuit. the advantage of this circuit is that it is not necessary to have high voltage control signal.
161.

## Which of the following circuits helps in the applications of switching times?

A. c)
B. d)
Explanation: this is an inverter, in which the transistor in the circuit is switched between cut off and saturation. the load, for example, can be a motor or a light emitting diode or any other electrical device.
162.

## Which of the following helps in reducing the switching time of a transistor?

A. a resistor connected from base to ground
B. a resistor connected from emitter to ground
C. a capacitor connected from base to ground
D. a capacitor connected from emitter to ground
Answer» A. a resistor connected from base to ground
Explanation: connecting a resistor connected from base of a transistor to ground/negative voltage helps in reducing the switching the switching time of the transistor. when transistor saturate, there is stored charge in the base that must be removed before it turns off.
163.

## The time taken for a transistor to turn from saturation to cut off is

A. inversely proportional to charge carriers
B. directly proportional to charge carriers
C. charging time of the capacitor
D. discharging time of the capacitor
Answer» B. directly proportional to charge carriers
Explanation: when sufficient charge carriers exist, the transistor goes into saturation.
164.

## Switching speed of P+ junction depends on

A. mobility of minority carriers in p junction
B. life time of minority carriers in p junction
C. mobility of majority carriers in n junction
D. life time of minority carriers in n junction
Answer» D. life time of minority carriers in n junction
Explanation: switching leads to move holes in p region to n region as minority carriers. removal of this accumulation determines switching speed. p+ regards to a diode in which the p type is doped excessively.
165.

## Fixed voltage regulators and adjustable regulators are often called as

A. series dissipative regulators
B. shunt dissipative regulators
C. stray dissipative regulators
D. all the mentioned
Explanation: series dissipative regulators simulate a variable resistance between the input voltage & the load and hence functions in a linear mode.
166.

## Linear series regulators are suited for application with

A. high current
B. medium current
C. low current
D. none of the mentioned
Explanation: in series dissipative regulator, conversion efficiency decreases as the input or output voltage differential increases and vice versa. so, linear series regulators are suited for medium current application with a small voltage differential.
167.

## A series switching regulators

A. improves the efficiency of regulators
B. improves the flexibility of switching
C. enhance the response of regulators
D. all of the mentioned
Answer» A. improves the efficiency of regulators
Explanation: a series switching regulators is constructed such that, a series pass transistor is used as a switch rather than as a variable resistance in linear mode.
168.

## The switching regulators can operate in

A. step up
B. step down
C. polarity inverting
D. all the mentioned
Explanation: the switching regulators can operate in any one of the three modes depending on the way in which the components are connected.
169.

## It may be required to store energy for a specified amount of time during power failure especially if the system is designed for a computer power supply.

A. 1 and 3
B. 1,2,3 and 4
C. 2,3 and 4
D. 1,3 and 4
Explanation: a voltage source must satisfy the entire given requirement to be used in switching regulator.
170.

## Which among the following act as a switch in switching regulator?

A. rectifiers
B. diode
C. transistors
D. relays
Explanation: a transistor is connected as power switch and is operated in the saturated mode.thus, the pulse generator output alternatively turns the switch on and off in switching regulator.
171.

## What should be the frequency range of pulse generator?

A. 250 khz
B. 40 khz
C. 120 khz
D. 20 khz
Explanation: the most effective frequency range for pulse generator for optimum efficiency and component size is 20khz.
172.

## Filter used in switching regulator’s are also as called

A. dc – ac transformers
B. ac – dc transformers
C. dc transformer
D. ac transformer
Explanation: filter converts the pulse waveform from the output of the switch into a dc voltage. since this switching mechanism allows a conversion similar to transformers, the switching regulators is often referred to as a dc transformer.
173.

## Which of the following is considered to be the most important components of the switching regulator?

A. rc or rlc filter
B. rl or rlc filter
C. orc or rl filter
D. rc, rlc or rl filter
Answer» B. rl or rlc filter
Explanation: rl or rlc filter is the most important components of the switching regulator, because there are several areas that are affected by the choke of inductor including energy storage for the regulators output ripple, transient, response etc.
174.

## Which is the most commonly used low voltage switching regulators?

A. powdered permalloy toroids
B. fermite ei, u and toroid cores
C. silicon steel ei butt stacks
D. none of the mentioned
Answer» C. silicon steel ei butt stacks
Explanation: the silicon steel ei butt stack exhibits high permeability high flux density and ease of construction and mounting therefore, it is most commonly used in low voltage switching regulators.
175.

## The diode in a half wave rectifier has a forward resistance RF. The voltage is Vmsinωt and the load resistance is RL. The DC current is given by

A. vm/√2rl
B. vm/(rf+rl)π
C. 2vm/√π
D. vm/rl
Explanation: for a half wave rectifier, the idc=iavg=im/π
176.

## What is the output as a function of the input voltage (for positive values) for the given figure. Assume it’s an ideal op-amp with zero forward drop (Di=0)

A. 0
B. -vi
C. vi
D. 2vi
Explanation: when the input of the inverted mode op-amp is positive, the output is negative.
177.

## In a half wave rectifier, the sine wave input is 50sin50t. If the load resistance is of 1K, then average DC power output will be?

A. 3.99v
B. 2.5v
C. 5.97v
D. 6.77v
Explanation: the standard form of a sine wave is vmsinωt. by comparing the given
178.

## Efficiency of a half wave rectifier is

A. 50%
B. 60%
C. 40.6%
D. 46%
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.
179.

## If peak voltage for a half wave rectifier circuit is 5V and diode cut in voltage is 0.7, then peak inverse voltage on diode will be?

A. 5v
B. 4.9v
C. 4.3v
D. 6.7v
Explanation: piv is the maximum reverse bias voltage that can be appeared across a diode in the given circuit, if the piv rating is less than this value of breakdown of diode will occur. for a rectifier, piv=vm-vd=5- 0.7=4.3v.
180.

## Transformer utilisation factor of a half wave rectifier is

A. 0.234
B. 0.279
C. 0.287
D. 0.453
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.
181.

## If the input frequency of a half wave rectifier is 100Hz, then the ripple frequency will be

A. 150hz
B. 200hz
C. 100hz
D. 300hz
Explanation: the ripple frequency of the output and input is same. this is because, one half cycle of input is passed and other half cycle is seized. so, effectively the frequency is the same.
182.

## Efficiency of a centre tapped full wave rectifier is

A. 50%
B. 46%
C. 70%
D. 81.2%
Explanation: efficiency of a rectifier is the effectiveness to convert ac to dc. it’s obtained by taking ratio of dc power output to maximum ac power delivered to load. it’s usually expressed in percentage. for centre tapped full wave rectifier, it’s 81.2%.
183.

## 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. 1v
Explanation: we know that, p =v 2/r .
184.

## 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%
Explanation: the ripple factor ϒ= [(irms/iavg)2 – 1]1/2. irms =im /
185.

## 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%.

A. 0.623
B. 0.678
C. 0.693
D. 0.625
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.
186.

## If the peak voltage on a centre tapped full wave rectifier circuit is 5V 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. 10.7v
Explanation: piv is the maximum reverse bias voltage that can be appeared across a
187.

## What is done in switching regulators to minimize its power dissipation during switching?

A. uses external transistor
B. uses 1mh choke
C. uses external transistor and 1mh choke
D. none of the mentioned
Answer» C. uses external transistor and 1mh choke
Explanation: to minimize power dissipation during switching, the external transistor must be a switching power transistor and a 1mh choke smooth out the current pulses delivered to the load.
188.

## 33/ Ipk

A. 1- iii , 2- i , 3- ii
B. 1- i , 2- ii , 3- iii
C. 1- iii , 2- ii , 3- i
D. 1- iii , 2- ii , 3- i
Answer» B. 1- i , 2- ii , 3- iii
Explanation: characteristics and design formula for step up, step down and converting mode of switching regulator.
189.

## Choose the correct statement

A. mosfet is a unipolar, voltage controlled, two terminal device
B. mosfet is a bipolar, current controlled, three terminal device
C. mosfet is a unipolar, voltage controlled, three terminal device
D. mosfet is a bipolar, current controlled, two terminal device
Answer» C. mosfet is a unipolar, voltage controlled, three terminal device
Explanation: mosfet is a three terminal device, gate, source & drain. it is voltage controlled unlike the bjt & only electron current flows.
190.

## The arrow on the symbol of MOSFET indicates

A. that it is a n-channel mosfet
B. the direction of electrons
C. the direction of conventional current flow
D. that it is a p-channel mosfet
Answer» B. the direction of electrons
Explanation: the arrow is to indicate the direction of electrons (opposite to the direction of conventional current flow).
191.

## The controlling parameter in MOSFET is

A. vds
B. ig
C. vgs
D. is
Explanation: the gate to source voltage is the controlling parameter in a mosfet.
192.

## In the internal structure of a MOSFET, a parasitic BJT exists between the

A. source & gate terminals
B. source & drain terminals
C. drain & gate terminals
D. there is no parasitic bjt in mosfet
Answer» B. source & drain terminals
Explanation: examine the internal structure of a mosfet, notice the n-p-n structure between the drain & source. a p-channel mosfet will have a p-n-p structure.
193.

## In the transfer characteristics of a MOSFET, the threshold voltage is the measure of the

A. minimum voltage to induce a n-channel/p- channel for conduction
B. minimum voltage till which temperature is constant
C. minimum voltage to turn off the device
D. none of the above mentioned is true
Answer» A. minimum voltage to induce a n-channel/p- channel for conduction
Explanation: it is the minimum voltage to induce a n-channel/p-channel which will allow the device to conduct electrically through its length.
194.

## The output characteristics of a MOSFET, is a plot of

A. id as a function of vgs with vds as a parameter
B. id as a function of vds with vgs as a parameter
C. ig as a function of vgs with vds as a parameter
D. ig as a function of vds with vgs as a parameter
Answer» B. id as a function of vds with vgs as a parameter
Explanation: it is id vs vds which are plotted for different values of vgs (gate to source voltage).
195.

## SMPS is used for

A. obtaining controlled ac power supply
B. obtaining controlled dc power supply
C. storage of dc power
D. switch from one source to another
Answer» B. obtaining controlled dc power supply
Explanation: smps (switching mode power supply) is used for obtaining controlled dc power supply.
196.

## Choose the incorrect statement.

A. smps is less sensitive to input voltage variations
B. smps is smaller as compared to rectifiers
C. smps has low input ripple
D. smps is a source of radio interference
Answer» C. smps has low input ripple
Explanation: smps has higher output ripple and its regulation is worse.
197.

## is used for critical loads where temporary power failure can cause a great deal of inconvenience.

A. smps
B. ups
C. mps
D. rccb
Explanation: uninterruptible power supply is used where loads where temporary power failure can cause a great deal of inconvenience.
198.

## is used in the rotating type UPS system to supply the mains.

A. dc motor
B. self excited dc generator
C. alternator
D. battery bank
Explanation: when the supply is gone, the diesel engine is started, which runs the alternator and the alternator supplies power to the mains. non-rotating type ups are not used anymore.
199.

## Static UPS requires

A. only rectifier
B. only inverter
C. both inverter and rectifier
D. none of the mentioned
Answer» C. both inverter and rectifier
Explanation: rectifier to converter the dc from the battery to ac. inverter to charge the battery from mains.
200.

A. nc
B. li-on