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# Multiple choice question for engineering

## Set 1

1. 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 current

Answer: a [Reason:] Presence of inductor usually dampens the AC signal. Due to self induction induces opposing EMF or changes in the current.

2. The ripple factor (ϒ) of inductor filter is_________
a) ϒ = RZ3/√2ωL
b) ϒ = RZ/3√2ωL
c) ϒ = RZ3√2/ωL
d) ϒ = RZ3/√2ωL

Answer: b [Reason:] Ripple factor will decrease when L is increased and RL. Inductor has a higher dc resistance. It depends on property of opposing the change of direction of current.

3. The inductor filter gives a smooth output because_________
a) It offers infinite resistance to ac components
b) It offers infinite resistance to dc components
c) Pulsating dc signal is allowed
d) The ac signal is amplified

Answer: a [Reason:] The inductor does not allow the ac components to pass through the filter. The main purpose of using an inductor filter is to avoid the ripples. By using this property, the inductor offers an infinite resistance to ac components and gives a smooth output.

4. 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) 20mA

Answer: c [Reason:] For a rectifier with an inductor filter, VDC=2Vm/π, Idc=VDC/RL=2Vm/RLπ IDC=2*250/(3.14*15*103)=10.6mA.

5. 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 components

Answer: c [Reason:] 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.

6. A dc voltage of 380V with a peak ripple voltage not exceeding 7V is required to supply a 500Ω load. Find out the inductance required.
a) 10.8henry
b) 30.7henry
c) 28.8henry
d) 15.4henry

Answer: c [Reason:] Given the ripple voltage is 7V. So, 7=1.414VRMS ϒ=VRMS/VDC=4.95/380=0.0130. ϒ=1/3√2(RL/Lω) So, L=28.8henry.

7. The inductor filter should be used when RL is consistently small because_________
a) Effective filtering takes place when load current is high
b) Effective filtering takes place when load current is low
c) Current lags behind voltage
d) Current leads voltage

Answer: a [Reason:] When RLis infinite, the ripple factor is 0.471. This value is close to that of a rectifier. So, the resistance should be small.

8. The output voltage VDC for a rectifier with inductor filter is given by_________
a) (2Vm/π)-IDCR
b) (2Vm/π)+IDCR
c) (2Vmπ)-IDCR
d) (2Vmπ)+IDCR

Answer: a [Reason:] The inductor with high resistance can cause poor voltage regulation. The choke resistance, the resistance of half of transformer secondary is not negligible.

9. What causes to decrease the sudden rise in the current for a rectifier?
a) the electrical energy
b) The ripple factor
c) The magnetic energy
d) Infinite resistance

Answer: c [Reason:] When the output current of a rectifier increases above a certain value, magnetic energy is stored in the inductor. This energy tends to decrease the sudden rise in the current. This also helps to prevent the current to fall down too much.

10. 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 ripple factor (ϒ).
a) 0.1
b) 0.6
c) 0.5
d) 0.4

Answer: d [Reason:] ϒ=IAC/IDC, IAC=2√2Vm/3π(RL2+4ω2L2)1/2 By putting the values, IAC=4.24Ma. VDC=2Vm/π, IDC=VDC/RL=2Vm/RL π IDC=2*250/(3.14*15*103)=10.6mA. ϒ=4.24/10.6=0.4.

## Set 2

1. The other name for Miller Circuit is
a) Non-Inverting Integrator
b) Inverting Integrator
c) Non-Inverting Differentiator
d) Inverting Differentiator

Answer: b [Reason:] Miller Circuit is also called Inverting integrator.

2. The slope of the frequency response of an integrator is
a) Linear with negative slope
b) Linear with positive slope
c) Exponential increase
d) Exponential decrease

Answer: a [Reason:] The slope is linear and negative.

3. The integrating transfer function has the value of
a) jωCR
b) –jωCR
c) 1 / jωCR
d) -1 / jωCR

Answer: d [Reason:] Standard mathematical expression for the transfer function.

4. The expression for the integration frequency is
a) CR
b) 1/CR
c) R/C
d) C/R

Answer: b [Reason:] Standard mathematical expression for the integrator frequency.

5. Determine the expression for the transfer function for the circuit shown below. a) (Rf/R)/(1+jωCRfC)
b) (Rf/R)/(1-jωCRfC)
c) – (Rf/R)/(1+jωCRfC)
d) – (Rf/R)/(1-jωCRfC)

Answer: c [Reason:] It is a standard expression.

6. The frequency transfer function of a differentiator is given by
a) jωCR
b) 1/jωCR
c) – jωCR
d) – 1/jωCR

Answer: a [Reason:] Standard mathematical expression for the transfer function of a differentiator.

7. The slope of the frequency response of a differentiator is
a) Linear with negative slope
b) Linear with positive slope
c) Exponential increase
d) Exponential decrease

Answer: b [Reason:] The slope is linear with a positive slope.

8. The phase in the integrator and differentiator circuit respectively are
a) +90 degrees and +90 degrees
b) -90 degrees and -90 degrees
c) -90 degrees and +90 degrees
d) +90 degrees and -90 degrees

Answer: d [Reason:] These are the characteristics of the integrators and differentiators circuits respectively.

9. Consider a symmetrical square wave of 20-V peak-to-peak, 0 average, and 2-ms period applied to a Miller integrator. Find the value of the time constant CR such that the triangular waveform at the output has a 20-V peak-to-peak amplitude.
a) 0.25ms
b) 0.50ms
c) 2.5ms
d) 5.0ms

Answer: b [Reason:] According to the question 1/CR = 2.

10. The expression for the differentiator time constant is
a) CR
b) 1/CR
c) R/C
d) C/R

Answer: a [Reason:] Standard mathematical expression for the time constant for the differentiators.

## Set 3

1. When does a resistance provide a negative feedback to an amplifier?
a) Resistance is connected between the positive input terminal and the output terminal
b) Resistance is connected between the negative input terminal and the output terminal
c) Resistance is connected between the input terminals
d) Resistance is connected between the negative input terminal and ground

Answer: b [Reason:] An op amp is said to have a negative feedback when a resistance is connected between the input and output terminals respectively.

2. The effect of the inverting configuration is
a) The output signal and the input signal are out of phase by 180o
b) The output signal and the input signal are in phase
c) The output phase is leading the input phase by 90o
d) The output phase is lagging behind the input phase by 90o

Answer: a [Reason:] Inverting introduces a phase shift of 180o or it ‘inverts’ a peak.

3. For an ideal negative feedback configuration which of the following is true?
a) There is a virtual open circuit between the input terminals
b) The closed loop gain for a negative feedback does not depend only on the external parameters
c) There is a virtual short circuit between the input terminals
d) There is a virtual ground at the negative input terminal

Answer: c [Reason:] There is always a virtual short circuit in this type of case. There will be a virtual ground if and only if one of the terminals is grounded.

4. The negative feedback causes
a) The voltage between the two input terminals to the very small, ideally zero
b) The voltage between the two input resistance very high, ideally infinite
c) Current flow through the positive input terminal and no current flows through the negative input terminal
d) Both a and c

Answer: a [Reason:] Ideally the input terminals are at the same potential but in real practice there is a very small potential between the two terminals.

5. The non-inverting closed loop configuration features a high resistance. Therefore in many cases unity gain follower called buffer amplifier is often used to
a) Connect a high resistance source to high resistance load
b) Connect low resistance source to low resistance load
c) Connect low resistance source to a high resistance source
d) Connect high resistance source to a low resistance load

Answer: d [Reason:] Buffer amplifiers are required to connect a high resistance load to a low input resistance output.

6. The advantage of a weighted summer operational amplifier is
a) It is capable of summing various input voltages together
b) Each input signal may be independently adjusted by adjusting the corresponding input resistance
c) If one needs both sign of a voltage signal then two operational amplifiers are needed
d) All of the mentioned

Answer: d [Reason:] All of the mentioned are characteristics of a weighted summer operational amplifier over the traditional amplifier.

7. The following is a circuit of weighted summer capable of summing coefficients of both sign. The expressions for the output voltage v0 is a) v0 = v1 (R1/Ra ) (Rc/Rb ) + v1 (Ra/R2 ) (Rc/Rb ) – v1 (Rc/R3 ) – v1 (Rc/R4 )
b) v0 = – v1 (R1/Ra ) (Rc/Rb ) – v1 (Ra/R2 ) (Rc/Rb ) + v1 (Rc/R3 ) + v1 (Rc/R4 )
c) v0 = v1 (Ra/R1 ) (Rc/Rb ) + v1 (Ra/R2 ) (Rc/Rb ) – v1 (Rc/R3 ) – v1 (Rc/R4 )
d) v0 = – v1 (Ra/R1 ) (Rc/Rb ) – v1 (Ra/R2 ) (Rc/Rb ) + v1 (Rc/R3 ) + v1 (Rc/R4 )

Answer: c [Reason:] The voltages are increased first by the left side of the portion and then are also magnified by the right side of the circuit. There are four inputs given out of which two are magnified twice and the other are magnified only once.

8. You are provided with an ideal op amp and three 10kΩ resistors. Using series and parallel resistor combinations, how many different inverting-amplifier circuit topologies are possible?
a) 2
b) 3
c) 4
d) 5

Answer: c [Reason:] Consider series and parallel combination of the resistances provided and arrange then in the feedback region and as output resistance.

9. The loop gain for an ideal operational amplifier with R1 = 10kΩ and R2(negative feedback) = 1MΩ is
a) 20 db
b) 40 db
c) 60 db
d) 80 db

Answer: b [Reason:] Loop gain in this case is given by 20 log (1000000/10000).

10. In an inverting op-amp circuit for which the gain is −4 V/V and the total resistance used is 100 kΩ. Then the value of R1 and R2 (negative feedback)
a) R1 = 20KΩ and R1 = 80KΩ
b) R1 = 80KΩ and R1 = 20KΩ
c) R1 = 40KΩ and R1 = 60KΩ
d) R1 = 50KΩ and R1 = 50KΩ

Answer = a [Reason:] Solve R1 + R2 = 100 R2/R1 = 4 for R1 and R2 respectively.

## Set 4

1. The advantages over the vacuum triode for a junction transistor is_________
a) high power consumption
b) high efficiency
c) large size
d) less doping

Answer: b [Reason:] A junction transistor is an analogous to a vacuum triode. The main difference between them is that a transistor is a current device while a vacuum triode is a voltage device. The advantages of a transistor over a vacuum triode are long life, high efficiency, light weight, smaller in size, less power consumption.

2. What is the left hand section of a junction transistor called?
a) base
b) collector
c) emitter
d) depletion region

Answer: c [Reason:] The main function of this section is to supply majority charge carriers to the base. Hence it is more heavily doped in comparison to other regions. This forms the left hand section of the transistor.

3. In an NPN transistor, the arrow is pointed towards_________
a) the collector
b) the base
c) depends on the configuration
d) the emitter

Answer: d [Reason:] As regards to the symbols, the arrow head is always at the emitter. The direction indicates the conventional direction of current flow. In case of PNP transistor, it is from base to emitter.

4. Which of the following is true in construction of a transistor?
a) the collector dissipates lesser power
b) the emitter supplies minority carriers
c) the collector is made physically larger than the emitter region
d) the collector collects minority charge carriers

Answer: c [Reason:] In most of the transistors, the collector is made larger than emitter region. This is due to the fact that collector has to dissipate much greater power. The collector and emitter cannot be interchanged.

5. In the operation of an NPN transistor, the electrons cross which region?
a) emitter region
b) the region where there is high depletion
c) the region where there is low depletion
d) P type base region

Answer: d [Reason:] The electrons in the emitter region are repelled by the negative terminal of the battery towards the emitter junction. The potential barrier at the junction is reduced due to forward bias and base region is very thin and lightly doped, electrons cross the P type base region.

6. Which of the following are true for a PNP transistor?
a) the emitter current is less than the collector current
b) the collector current is less than the emitter current
c) the electrons are majority charge carriers
d) the holes are the minority charge carriers

Answer: b [Reason:] The 2 – 5% of holes is lost in recombination with electrons in the base region. The majority charge carriers are holes for a PNP transistor. Thus the collector current is slightly less than the emitter current.

7. In the saturated region, the transistor acts like a_________
a) poor transistor
b) amplifier
c) open switch
d) closed switch

Answer: d [Reason:] In saturated mode, both emitter and collector are forward biased. The negative of the battery is connected to emitter and similarly the positive terminals of batteries are connected to the base. The transistor now acts like a closed switch.

8. When does the transistor act like an open switch?
a) cut off region
b) inverted region
c) saturated region
d) active region

Answer: a [Reason:] In cut off region, both the junctions are reverse biased. The transistor has practically zero current because the emitter does not emit charge carriers to the base. So, the transistor acts as open switch.

9. If the emitter-base junction is forward biased and the collector-base junction is reverse biased, what will be the region of operation for a transistor?
a) cut off region
b) saturated region
c) inverted region
d) active region

Answer: d [Reason:] When the emitter-base junction is forward biased and the collector-base junction is reverse biased, the transistor is used for amplification. A battery is connected to collector base circuit. The positive terminal is connected to the collector while the negative is connected to the base.

10. The transfer of a signal in a transistor is_________
a) low to high resistance
b) high to low resistance
c) collector to base junction
d) emitter to base junction

Answer: a [Reason:] A forward biased emitter base junction has a low resistance path. A reversed biased junction has a high resistance path. The weak signal is introduced in a low resistance circuit and the output is taken from the high resistance circuit.

## Set 5

1. An L section filter with L=2henry and C= 49µF is used in the output of a full wave single phase rectifier that is fed from a 40-0-40 V peak transformer. The load current is 0.2A. Calculate the output dc voltage.
a) 20.76V
b) 24.46V
c) 34.78V
d) 12.67V

Answer: b [Reason:] Given, VL= 40V. VDC=2/π*VL=2/π*40=25.46V.

2. Calculate LC for a full wave rectifier which provides 10V dc at 100mA with a maximum ripple of 2%. Input ac frequency is 50Hz.
a) 40*10-6
b) 10*10-6
c) 30*10-6
d) 90*10-6

Answer: a [Reason:] LC=1/6√2ω2ϒ ω=2πf=314 By putting the values, LC=1/6√2(314)2 0.02=40*10-6.

3. 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 inductance

Answer: c [Reason:] 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.

4. 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≥RL

Answer: a [Reason:] 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 relation LC≥RL/3ω.

5. The ripple factor for an l section filter is_______
a) ϒ= 1/6√2ω2LC
b) ϒ= 6√2ω2LC
c) ϒ= 6√3ω2LC
d) ϒ= 1/6√3ω2LC

Answer: a [Reason:] 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.

6. 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 – IDCR

Answer: d [Reason:] 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.

7. The rms value of ripple current for an L section filter is_______
a) IRMS=√2/3*XL*VDC
b) IRMS=√2/3*XL*VDC
c) IRMS=√2/3*XL *VDC
d) IRMS=√2/3*XL*VDC

Answer: a [Reason:] The ac current through L is determined primarily by XL=2ωL. It is directly proportional to voltage produced and indirectly proportional to the reactance.

8. What makes the load in a choke filter to bypass harmonic components?
a) capacitor
b) inductor
c) resistor
d) diodes

Answer: a [Reason:] When the capacitor is shunted across the load, it bypasses the harmonic components. This is because it offers low reactance to ac ripple component. In another hand the inductor offers high impedance to harmonic terms.

9. The ripple to heavy loads by a capacitor is_______
a) high
b) depends on temperature
c) low
d) no ripple at all