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# Multiple choice question for engineering

## Set 1

1. Op-amp was introduced by __________
a) Fairchild
b) Maxwell
c) Rutherford
d) Sahani

Answer: a [Reason:] Op-amp was introduced by Fairchild semiconductor in 1968.

2. The number of terminals in an Op-amp ______________
a) 6
b) 2
c) 5
d) 3

Answer: c [Reason:] Inverting input, the Non-inverting input, Output, Positive power supply, Negative power supply.

3. The Op-amp is a type of ___________
a) Differential amplifier
b) Integrated amplifier
c) Isolation amplifier
d) Feedback amplifier

Answer: a [Reason:] The Op-amp is a type of differential amplifier.

4. In the circuit of Op-amp given V- stands for _________

a) Non-inverting input
b) Non-inverting output
c) Inverting input
d) Inverting output

Answer: c [Reason:] V+: Non-inverting input and V- : Inverting input.

5. When the input voltage difference is small in magnitude, the Op-amp behaves as ____________
a) Non-linear device
b) Linear device
c) Complex device
d) Bipolar device

Answer: b [Reason:] When │vp-vn│is small then Op-amp acts as a linear device as the output voltage is a linear function of input voltages.

6. If the output voltage is not a linear function of input voltage then ____________
a) Op-amp acts a linear device
b) Op-amp acts as a non-linear device
c) Op-amp acts a polar device
d) Op-amp acts as an inverter

Answer: b [Reason:] If output voltage is not a linear function of input voltage then Op-amp acts as a non-linear device.

7. The negative feedback causes the input voltage difference to ____________
a.) 1
b) Increase
c) Decrease
d) 0

Answer: c [Reason:] Negative feedback means a signal is fed back from output terminals to the non-inverting input terminals and this results in a decrease in input voltage difference.

8. Find the gain for the following circuit.

a) -2
b) 2
c) -1
d) 1

Answer: a [Reason:] In this circuit, the only node is at the negative terminal of the Op-amp (say Vn) and by ideal rules of Op-amp, Vn= Vp =0(in this circuit). Gain= Vout/Vin= -R2/R1.

9. Calculate the gain for the Op-amp given.

a) 0.719
b) 2.572
c) 1.390
d) 1.237

Answer: c [Reason:] Gain= Vout/Vin= -R2/R1 = -5.98*10-3/4.3*10-3.

10. Given Op-amp is ideal. Calculate vo if va=1v and vb=0v.

a) -4v
b) -2.5v
c) 4v
d) 2.5v

Answer: b [Reason:] In the given circuit, a negative feedback exists between Op-amp’s output and its inverting input (voltage here is 0, as vp =vb=0 and vn=vp). Node-voltage equation is i50=i125=i0. i50= (va-vn)/50 =1/50 mA. I125= (v0-vn)/125 = v0/125 mA. 1/50 + v0/125 = 0. v0 is -2.5volts.

## Set 2

1) The symbol used for representing Independent sources
a) Diamond
b) Square
c) Circle
d) Triangle

Answer: c [Reason:] Independent sources are represented by circle Dependent sources are represented by Diamond.

2. Controlled sources are also known as
a) Independent sources
b) Dependent sources
c) Ideal sources
d) Voltage sources

Answer: a [Reason:] Voltage V = dw/dq and its SI unit is Volt.

3.

I3 = α vx .This is
a) Voltage control voltage source
b) Current control voltage source
c) Voltage control current source
d) Current control current source

Answer: b [Reason:] i3 =αvx means ix value depends on vx Controlled voltage is vx.

4. Inductor is _______________ element.
a) Active
b) Passive
c) Linear
d) Polar

Answer: b [Reason:] Passive element means it could not generate electricity.

5.

Which of the above is valid?
a) 1
b) 2
c) Both
d) Neither 1 nor 2

6.

The above circuit is valid.
a) False
b) True

Answer: b [Reason:] Independent current source supplied current through terminals a and b. Dependent source supplies voltage across the same pair of terminals and an ideal current source supplies same current regardless of voltage, similarly an ideal voltage source supplies same voltage irrespective of current, so this is an allowable connection.

7. The opposing capacity of materials against the current flow is
a) Conductance
b) Inductance
c) Susceptance
d) Resistance

Answer: d [Reason:] The opposing capacity of materials against the current flow is resistance.

8. The conductance of a 923Ω resistance is
a) 1.08 * 10-3 mho
b) 1.08 * 10-4 mho
c) 1.02 * 10-3 mho
d) 1.02 * 10-4 mho

Answer: a [Reason:] c=1/R =1/923 = 1.08 * 10-3 mho

9. The current passing through a circuit is 7.2A and the power at the terminals is 27 watts.
Existence is ___________ ohms.
a) 0.5402
b) 0.5208
c) 0.5972
d) 0.5792

Answer: b [Reason:] p = vi = (iR) i = i2 R R = P/i2 =27/ (7.2)2 = 0.5208Ω.

10. Relation between power, voltage and conductance
a) V = P2.G
b) V = P2/G
c) P = v2/G
d) P = V2 G

Answer: d [Reason:] P = vi = v (v/R) = v2/R = v2G.

## Set 3

1. Which among the following chemical bond were described by Kossel and Lewis?
a) Metallic bond
b) Polar covalent bond
c) Coordinate bond
d) Ionic and Covalent bond

Answer: d [Reason:] Both Ionic and Covalent bond arise from the tendency of atoms to attain stable configuration of electrons.

2. Which among the following is not a property of Ionic bond?
a) Losing of electrons
b) Gain of electrons
c) Sharing of electrons
d) Transfer of electrons

Answer: c [Reason:] Ionic bond results from a), b) and d). But the transfer of electrons is a property of Covalent bond.

3. Which among the following formation is not an example of Covalent bond?
a) LiF
b) NH3
c) CF4
d) HF

Answer: a [Reason:] LiF (Lithium Fluoride) is an example of Ionic bond, as the formation takes place by transfer of electrons and not by sharing.

4. State whether the given statement is true or false “Ionic bonds are non-directional”
a) True
b) False

Answer: a [Reason:] The ionic bond breaks up when dissolved in water and allow the charged particles to move freely.

5. If a bond is made up of a large number of organic compound, then the bond is termed as?
a) Ionic bond
b) Metallic bond
c) Covalent bond
d) Dipolar bond

Answer: c [Reason:] This is because they have the tendency to transfer electrons.

6. Which among the following is not an example of hydrogen bond?
a) H20
b) Liquid HCl
c) NH3
d) CHCl3

Answer: b [Reason:] There is no hydrogen bond in liquid HCl since the bond breaks up when dissolved in water.

7. Atoms undergo bonding in order to ?
a) Attain stability
b) Lose stability
c) Move freely
d) increase energy

Answer: a [Reason:] Atoms undergo bonding to attain stable electronic configuration and to gain energy.

8. An atom differs from its ion in which among the following ?
a) Mass number
b) Atomic number
c) Neutrons
d) Number of protons

Answer: d [Reason:] When an atom loses or gains electrons its forms its corresponding ion, and hence it differs in proton number.

9. Which among the following is both a molecule and a compound?
a) C6 H12 O6
b) H2O
c) CO2
d) NaCl

Answer: b Water H2O is both a molecule and a compound. This is because the atoms which make them are not the same.

10. Bond energy and the corresponding bond length vary directly with each other. comment whether the statement is true or false.
a) True
b) False

Answer: b [Reason:] Smaller the bond energy, greater will be the bond length and vice versa and hence the two vary inversely with each other.

## Set 4

1. A single phase full bridge inverter has RLC load with R = 4 Ω, Xl = 11 Ω and Xc = 20.54 Ω. The dc input voltage is 230 V. Find the value of fundamental load power.
a) 1633 W
b) 1603 W
c) 1576 W
d) none of the mentioned

Answer: b [Reason:] Average value of fundamental output voltage = 4Vs/π = 292.85 V Z = [R2 + (Xl – Xc)2]1/2 I = V/Z = 28.31 I(rms)(fundamental) = V/Z√2 = (292.85)/(1.414 x 10.345) = 20.02 A Fundamental Load power = (20.02)2 x R = (20.02)2 x 4 = 1603.2 Watts.

2. A single phase full bridge inverter has RLC load with R = 4 Ω, L = 35 mH and C = 155 μF. The dc input voltage is 230 V and the output frequency is 50 Hz. Find the angle by which the third harmonic current will lead/lag the third harmonic output voltage.
a) 67.25°
b) 96.4°
c) 49.87°
d) 81.3°

Answer: d [Reason:] XL = 2 x 3.14 x 50 x 0.035 = 10.99 Ω XC = 1/(2 x 3.14 x 50 x 155 x 10-6) = 20.54 Ω For the third harmonic component XL(3rd harmonic) = 10.99 x 3 = 33 Ω (approx.) XC(3rd harmonic) = 20.54/3 = 6.846 Ω R = 4 Ω P = tan-1 (XL – XC)/R = 81.3°.

3. A single phase half bridge inverter has RLC load. The dc input voltage (Vs/2) = 115 V and the output frequency is 50 Hz. The expression for the load voltage up to the fifth harmonic will be given by
a) 146 sin 314t + 48.81 sin 314t + 58.57 sin 318t + 28.31 sin 318t + 3.686 sin 318t
b) 146 sin 314t + 48.81 sin (3 x 314t) + 29.28 sin (5 x 318t)
c) 146 sin 314t + 48.81 sin (2 x 314t) + 58.57 sin (3 x 318t)
d) none of the mentioned

Answer: b [Reason:] In a single phase HALF bridge inverter only odd harmonics are present. i.e. 1,3,5 etc. Vo = (2Vs/π) sin ωt + (2Vs/3π) sin 3ωt + (2Vs/5π) sin 5ωt . . . (2Vs/π) = 146 V ωt = 2 x f x π x t = 2 x 3.14 x 50 x t = 314t.

4. Let Vs be the amplitude of the output voltage and P be the output power for a single-phase half bridge inverter. Then the corresponding values for a full bridge inverter would be
a) 2Vs, 4P
b) 2Vs, 2P
c) Vs, P
d) 2Vs, P

Answer: a [Reason:] The voltage is doubled (4Vs/π) in full and (2Vs/π) in half bridge configuration. As the power is proportional to the square of the current which is proportional to the voltage it is 4 times that obtained by a half ridge configuration.

5. In VSI (voltage source inverters)
a) both voltage and current depend on the load impedance
b) only voltage depends on the load impedance
c) only current depends on the load impedance
d) none of the mentioned

Answer: c [Reason:] In VSIs the voltage is independent on the load impedance Z.

6. The harmonic factor of nth harmonic is given by
a) Vn
b) V1/Vn
c) Vn/V1
d) None of the mentioned

Answer: c [Reason:] It is the ratio of rms value of the nth harmonic voltage component to the rms value of the fundamental voltage component (V1).

7. ____________ is the measure of the contribution of any individual harmonic to the inverter output voltage.
a) THD
b) Distortion Factor
c) Harmonic Factor
d) TUF

Answer: c [Reason:] The HF or Harmonic factor is the ratio of the nth harmonic voltage component to the fundamental voltage component. Hence it shows how much a particular harmonic is contributing in the total output of the circuit.

8. The Total Harmonic Distortion (THD) is the ratio of
a) rms value of all the harmonic components to the rms value of the fundamental component
b) average value of all the harmonic components to the rms value of the fundamental component
c) rms value of all the third harmonic component to the rms value of the fundamental component
d) rms value of all the fundamental component to the rms value of all the harmonic components

Answer: a [Reason:] THD = Vo/V1 V1 = Fundamental component Vo = value of all the harmonic components expect the fundamental.

9. The HF (Harmonic factor of nth harmonic) is the ratio of
a) an average and a rms value
b) a rms and an average value
c) two volt-ampere values
d) two rms values

Answer: d [Reason:] HF is the ratio of rms value of the nth harmonic voltage component to the rms value of the fundamental voltage component.

10. The total harmonic distortion (THD) is the measure of
a) input vs output power factor
b) temperature sensitivity
c) waveform distortion
d) contribution of each harmonic to the total output

Answer: c [Reason:] Lower the value of THD, closer is the waveform to a sine-wave.

## Set 5

1. For a certain SCR configuration, the below shown waveform is obtained. Find the value of the average output voltage with α = 30° & Vs = 240 V.

a) 50.27 V
b) 100.8 V
c) 140 V
d) 120 V

Answer: b [Reason:] Vm = √2 Vs = 339.41 V Vavg = ∫ Vm Sin ωt d(ωt) where the integration runs from α to π Vavg = Vm/2π (1+cos α).

2. A single-phase half wave circuit has Vs = 230 V with a R load of 100 Ω. Find the average load current at α = 30°.
a) 1.45 A
b) 0.57 A
c) 0.96 A
d) 2.3 A

Answer: c [Reason:] Io = Vo/R Vo = (Vm/2π) x (1+cosα).

3. A single phase 230 V, 1 kW heater is connected across a 1-phase 230 V, 50 Hz supply through an SCR. The firing angle of the SCR is adjusted to give a rms voltage of 155 V. Find the power absorbed by the heater element.
a) 250 watts
b) 1 kW
c) 454 watts
d) 378 watts

Answer: c [Reason:] R (Heater element resistance) = 2302/1kW = 52.9 Ω Now, P = Vrms2/R = 1552/52.9.

4. For a single phase half-wave thyristor circuit with R load, the power delivered to the resistive load is
b) (rms supply voltage)2/R

Answer: c [Reason:] P = I2.R V = IR I = V/R Hence, P = V2/R.

5. For a single phase half-wave thyristor circuit with R load, the input power factor is given by
a) rms source voltage/total rms line current
b) rms input power/power delivered to the load
c) cos α
d) power delivered to load/input VA

6. For the below shown circuit,

The load voltage waveform from ωt = π to ωt = π+α = β (extinction angle) is
b) going from positive to negative
d) zero

Answer: c [Reason:] In this case, the Inductor L is force conducting the SCR & supply current to the source the current is gradually decaying as the inductor is discharging.

7.For the below shown circuit,

The SCR is gated at α. At ωt = π
b) only load voltage is zero
c) only load current is zero
d) both are non zero & non negative

Answer: b [Reason:] At π the L force conducts the SCR & current starts gradually decaying whereas at π the voltage is zero as it goes from positive to negative at π.

8. In case of a single-phase half-wave circuit with RL load, with firing angle α and extinction angle β, the conduction angle γ can be written as
a) γ = β+α
b) γ = β-α
c) γ = β/α
d) γ = α/β

Answer: b [Reason:] Extinction angle is some angle > π at which load current reduces to zero. It depends upon the inductor value.

9. For a half-wave single phase thyristor rectifier circuit with any type of load the _______ + _________ waves must ideally give the supply voltage waveform.

d) supply current, thyristor voltage

Answer: b [Reason:] The supply voltage either appears across at the thyristor or the load.

10. In case of a single-phase half-wave circuit with RL load, with firing angle α and extinction angle β, the thyristor is reversed biased from
a) β to α
b) β to 2π+α
c) β to 2π
d) β to 2β