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# Multiple choice question for engineering

## Set 1

1. The network function (s2 + 4s)/(s + 1)(s + 2)(s + 3) represents
a) RC impedance
b) RL impedance
c) LC impedance
d) None of the mentioned

Answer: d [Reason:] Poles and zero does not interlace on negative real axis so it is not a immittance function.

2. The network function (3s2 + 8s)/(s + 1)(s + 3) represents
a) RC impedance
b) RL impedance
c) LC impedance
d) None of the mentioned

Answer: c [Reason:] The singularity nearest to origin is a zero. So it may be RL impedance or RC admittance function. Because of (D) option it is required to check that it is a valid RC admittance function. The poles and zeros interlace along the negative real axis. The residue of Yrc(s)/s is positive.

3. The network function (s + 1)(s + 4)/s(s + 2)(s + 5) represents
a) RC impedance
b) RL impedance
c) LC impedance
d) All of the mentioned

Answer: b [Reason:] The singularity nearest to origin is a pole. So it may be RC impedance or RL admittance function.

4. The network function s^2 + 10s + 24/s2 + 8s + 15 represents
a) RC impedance
b) RL impedance
c) LC impedance
d) None of the mentioned

Answer: d [Reason:] The singularity is near to origin is pole. So it may be RC impedance or RL admittance function.

5. A valid immittance function is
a) (s + 4)(s + 8)/(s + 2)(s – 5)
b) s(s + 1)/(s + 2)(s + 5)
c) s(s + 2)(s + 3)/(s + 1)(s + 4)
d) s(s + 2)(s + 6)/(s + 1)(s + 4)

Answer: d [Reason:] a) pole lie on positive real axis b) poles and zero does not interlace on axis. c) poles and zero does not interlace on axis. d) is a valid immittance function.

6. The network function (s2 + 8s +15)/(s2 + 6s + 8) is
d) All of the mentioned

Answer: a [Reason:] The singularity nearest to origin is a pole. So it may be a RL admittance or RC impedance function.

7. The voltage response of a network to a unit step input is Vo(s) = 10/s(s2 + 8s + 16). The response is
a) under damped
b) over damped
c) critically damped
d) can’t be determined

Answer: c [Reason:] The characteristic equation has real and repeated roots (-4, -4). Hence it is critically damped.

8. The current response of a network to a unit step input is Io(s) = 10(s + 2)/s(s2 + 11s + 30). The response is
a) Under damped
b) Over damped
c) Critically damped
d) None of the mentioned

Answer: b [Reason:] The roots are real and unequal (-6, -5) for the characteristic equation. Hence it is over damped.

Circuit for q.9-Q.10

9. The ratio of the transfer function Io/Is is
a) s(s + 4)/(s2 + 3s + 4)
b) s(s + 4)/(s + 1)(s + 3)
c) (s2 + 3s + 4)/s(s + 4)
d) (s + 1)(s + 3)/s(s + 4)

Answer: b [Reason:] Io/Is = (s + 4)/ (s + 4 + 3/s) = s(s + 4)/(s + 1)(s + 3).

10. The response is
a) Over damped
b) Under damped
c) Critically damped
d) can’t be determined

Answer: a [Reason:] The characteristic roots are real and unequal (-1, -3), therefore it is over damped.

## Set 2

1. Buffer amplifier needs to have
a) Low input resistance and low output resistance
b) High Input resistance and high output resistance
c) Low input resistance and high output resistance
d) High input resistance and low output resistance

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

2. The ideal values for the input resistance (Ri) and the output resistance (Ro) of a transconductance amplifier are
a) Ri = 0 and Ro = 0
b) Ri = ∞ and R0 = ∞
c) Ri = 0 and R0 = ∞
d) Ri = ∞ and Ri = 0

Answer: b [Reason:] It is a desired characteristics of transconductance amplifier ideally.

3. An amplifier has a voltage gain of 40db. The value of AVO is
a) 10
b) 100
c) 20
d) 200

Answer: b [Reason:] The expression is given by 10 log AVO = 40. Solving for Avo gives 100 as the answer.

4. The output voltage of a voltage amplifier has been found to decrease by 20% when a load resistance of 1 kΩ is connected. What is the value of the amplifier output resistance?
a) 50Ω
b) 200Ω
c) 250Ω
d) 350Ω

Answer: c [Reason:] 250 / (1000 + 250) X 100% = 20%. Hence the output resistance is 250 ohm.

5. Which of the following is a transresistance amplifier?
a)
b)
c)
d)

Answer: d [Reason:] Figure d is the correct representation rest are voltage, current, transconductance amplifiers.

6. The signal to be amplified is current signal and the output desired is a voltage signal. Which of the following amplifier can perform this task?
a) Voltage amplifier
b) Current amplifier
c) Transconductance amplifier
d) Transresistance amplifier

Answer: d [Reason:] It is a characteristic of transconductance amplifier.

7. You are given two amplifiers, A and B, to connect in cascade between a 10-mV, 100-kΩ source (S) and a 100-Ω load (L). The amplifiers have voltage gain, input resistance, and output resistance as follows: for A, 100 V/V, 10 kΩ, 10 kΩ, respectively; for B, 1 V/V, 100 kΩ, 100 Ω, respectively. Your problem is to decide how the amplifiers should be connected so that the voltage gain is maximum.
a) SABL
b) SBAL
c) Both have the same voltage gain
d) None as neither combination is able to amplify the input signal

8. A transconductance amplifier with Ri = 2 kΩ, Gm = 40 mA/V, and Ro = 20 kΩ is fed with a voltage source having a source resistance of 2 kΩ and is loaded with a 1-kΩ resistance. Find the voltage gain realized.
a) 18.05 V/V
b) 19.05 V/V
c) 20.05 V/V
d) 21.05V/V

9. The ratio of the short circuit current gain of a current amplifier (Ai) to the open circuit voltage gain of a voltage amplifier (AV), given that both amplifiers have the same value of the input resistance (Ri) and output resistance (R0), is
a) Ri
b) Ro
c) Ri / R0
d) Ro / Ri

Answer: c [Reason:] It is a standard mathematical relation.

10. The ratio of the open circuit voltage of a voltage amplifier (AV) to the short circuit transconductance of a (Gm) of a transconductance amplifier, given that both have the same value of the internal resistance (Ri) and the output resistance (R0), is
a) Ri
b) R0
c) 1/Ri
d) 1/R0

Answer: b [Reason:] It is a standard mathematical relation.

## Set 3

1. 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 respectively

Answer: b [Reason:] 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.

2. Major part of the filtering is done by the first capacitor in a CLC filter because _________
a) The capacitor offers a very low reactance to the ripple frequency
b) The capacitor offers a very high reactance to the ripple frequency
c) The inductor offers a very low reactance to the ripple frequency
d) The inductor offers a very high reactance to the ripple frequency

Answer: a [Reason:] The CLC filters are used when high voltage and low ripple frequency is needed than L section filters. The capacitor in a CLC filter offers very low reactance to the ripple frequency. So, maximum of the filtering is done by the first capacitor across the L section part.

3. At f=50Hz, the ripple factor of CLC filter is_________
a) ϒ=5700RL / (LC1C2)
b) ϒ=5700/ (LC1C2RL)
c) ϒ=5700LC1/ (C2RL)
d) ϒ=5700C1C2/ (LRL)

Answer: b [Reason:] The ripple factor of a rectifier is the measure of disturbances produced in the output. It’s the effectiveness of a power supply filter to reduce the ripple voltage. The ratio of ripple voltage to DC output voltage is ripple factor which is 5700 / (LC1C2RL) at 50Hz.

4. A single phase full wave rectifier makes use of pi section filter with 10µF capacitors and a choke of 10henry. The secondary voltage is 280V and the load current is 100mA. Determine the dc output voltage when f=50Hz.
a) 345V
b) 521V
c) 243V
d) 346V

Answer: d [Reason:] Given, VRMS=280V So, V¬m = 1.414*280=396V. From theory of capacitor filter, VDC = Vm –IDC/4fC=396-0.1/ (4*50*10*10-6)=346V.

5. 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%

Answer: a [Reason:] We have, ϒ=5700 / (LC1C2RL) =5700 / (10*100*10-12*3460) =0.165%.

6. 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 component

Answer: a [Reason:] 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.

7. The voltage in case of a full wave rectifier in a CLC filter is_________
a) Vϒ = IDC/2fC
b) Vϒ = IDC fC
c) Vϒ = IDC/fC
d) Vϒ = 2IDCfC

Answer: a [Reason:] T he filter circuit is a combination of capacitors and inductors. The RMS value depends on the peak value of charging and discharging magnitude, VPEAK.

8. The advantages of a pi-filter is_________
a) low output voltage
b) low PIV
c) low ripple factor
d) high voltage regulation

Answer: c [Reason:] Due to the use of two capacitors with an inductor, an improved filtering action is provided. This leads to decrement in ripple factor. A low ripple factor signifies regulated and ripple free DC voltage.

9. What is the relation between time constant and load resistance?
a) They don’t depend on each other
b) They are directly proportional
c) They are inversely proportional
d) Cannot be predicted

Answer: c [Reason:] If the load resistance value is large, the discharge time constant will be of a high value. Thus the capacitors time to discharge will get over soon. This lowers the amount of ripples in the output and increases the output voltage. If the load resistance is small, the discharge time constant will be more with decrease in output voltage.

10. The output waveform of CLC filter is superimposed by a waveform referred to as_________
a) Square wave
b) Triangular wave
c) Saw tooth wave
d) Sine wave

Answer: c [Reason:] Since the rectifier conducts current only in the forward direction, any energy discharged by the capacitor will flow into the load. This result in a DC voltage upon which is superimposed a waveform referred to as a saw tooth wave.

## Set 4

1. What is the SI unit of conductivity?
a) Ωm
b) (Ωm)-1
c) Ω
d) m

Answer : b [Reason:] The formula of the conductivity is the σ=1/ρ. So, the unit of resistivity is Ωm. Now, the unit of conductivity becomes the inverse of resistivity.

2. Which of the following expressions doesn’t represent the correct formula for Drift current density?
a) J=σE
b) J=qnµE
c) J=µE
d) None

Answer : c [Reason:] : The following formulae represent the correct expression for drift current density, J=σE And J=qnµE.

3. Does a semiconductor satisfy the ohm’s law?
a) True
b) False

Answer : a [Reason:] : V=IR J=σE I/A=σ(V/L) V=(L/ σA)*I=(ρL)*I/A=IR Thus, above equation satisfies Ohm’s law.

4. In which range of temperature, freeze out point begins to occur?
a) Higher range
b) Lower range
c) Middle range
d) None

Answer : b [Reason:] : At lower range of temperature, the concentration and conductivity decreases with lowering of the temperature.

5. Which of the following expression represents the correct formula for the conductivity in an intrinsic material?
a) ρ=e(μn+μp )ni
b) σ=e(μn+μp )ni
c) σ=1/(e(μn+μp )ni)
d) ρ=1/(e(μn+μp )ni)

Answer : b [Reason:] Option b is the correct formula.

6. What is the voltage difference if the current is 1mA and length and area is 2cm and 4cm2 respectively?(ρ=2Ωm)
a) 0.025V
b) 25V
c) 0.25V
d) None

Answer : d [Reason:] : V=IR R=ρl/(A)=2*2/4=100Ω V=1mA*100 =0.1V.

7. Is resistivity is a function of temperature?
a)True
b)False

Answer : a [Reason:] : Resistance depends on the temperature and the resistivity depends on the resistance, so now the resistivity depends on the temperature.

8. What is the electric field when the voltage applied is 5V and the length is 100cm?
a) 0.5V/m
b) 5V/m
c) 50V/m
d) None

Answer : c [Reason:] : E=V/L=5/100cm=5V/m.

9. Calculate the average random thermal energy at T=300K?
a) 0.038eV
b) 3.8eV
c) 38eV
d) 0.38eV

Answer : a [Reason:] : Average random thermal energy=3/2*k*T=0.038eV.

10.

In the above figure, a semiconductor having an area ‘A’ and length ‘L’ and carrying current ‘I’ applied a voltage of ‘V’ volts across it. Calculate the relation between V and A?
a) V= ((ρ*L)/A)*I
b) V= ((ρ*A)/L)*I
c) V= ((ρ*I)/(A*L))
d) V=((ρ*I*A*L)

Answer : a [Reason:] Option A, satisfies the Ohm’s law which is V=IR where R=(ρl)/A.

## Set 5

(Q.1-Q.2) The number of cars arriving at ICICI bank drive-in window during 10-min period is Poisson random variable X with b=2.

1. The probability that more than 3 cars will arrive during any 10 min period is
a) 0.249
b) 0.143
c) 0.346
d) 0.543

Answer: b [Reason:] Evaluate 1 – P(x = 0) – P(x = 1) – P(x = 2) – P(x = 3).

2. The probability that no car will arrive is
a) 0.516
b) 0.459
c) 0.246
d) 0.135

Answer: d [Reason:] Evaluate P(x = 0).

(Q.3-Q.5) Delhi averages three murder per week and their occurrences follow a Poisson distribution.

3. The probability that there will be five or more murder in a given week is
a) 0.1847
b) 0.2461
c) 0.3927
d) 0.4167

Answer: a [Reason:] P(5 or more) = 1 – P(0) – P(1) – P(2) – P(3) – P(4) = 0.1847.

4. On the average, how many weeks a year can Delhi expect to have no murders ?
a) 1.4
b) 1.9
c) 2.6
d) 3.4

Answer: c [Reason:] P(0) = 0.0498. Hence average number of weeks per year with no murder is 52 x P(0) = 2.5889 week.

5. How many weeds per year (average) can the Delhi expect the number of murders per week to equal or exceed the average number per week?
a) 15
b) 20
c) 25
d) 30

Answer: d [Reason:] P(3 or more) = 1 – P(0) – P (1) – P(2) = 0.5768. Therefore average number of weeks per year = 52 x 0.5768 or 29.994 weeks.

(Q.6-Q.8) The random variable X is defined by the density f(x) = 0.5u(x) e(0.5x)

6. The expect value of g(x) = X3 is
a) 48
b) 192
c) 36
d) 72

Answer: a [Reason:] Solve E[g(x)] = E[X3].

7. The mean of the random variable x is
a) 1/4
b) 1/6
c) 1/3
d) 1/5

Answer: a [Reason:] Solve integral (x f(x) dx) from negative infinity to x.

8. The variance of the random variable x is
a) 1/10
b) 3/80
c) 5/16
d) 3/16

Answer: b [Reason:] Variance is given by E[X230] – 1/16.

(Q.9-Q.10) A joint sample space for two random variable X and Y has four elements (1,1), (2,2), (3,3) and (4,4). Probabilities of these elements are 0.1, 0.35, 0.05 and 0.5 respectively.

9. The probability of the event{X 2.5, Y 6} is
a) 0.45
b) 0.50
c) 0.55
d) 0.60