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# Multiple choice question for engineering

## Set 1

1. The waveform of the emf generated undergoes one complete cycle when?
a) Conductors move past north pole
b) Conductors move past south pole
c) Conductors move past north and south poles
d) Conductors are stationary

Answer: c [Reason:] The waveform of the e.m.f. generated in an a.c. generator undergoes one complete cycle of variation when the conductors move past a N and a S pole.

2. When is the shape of the negative half of the emf waveform equal to the positive half?
a) When the conductors move past north pole
b) When conductors move past south pole
c) When conductors move past both north and south pole
d) When conductors are stationary

Answer: c [Reason:] The waveform of the e.m.f. generated in an a.c. generator undergoes one complete cycle of variation when the conductors move past a N and a S pole and the shape of the wave over the negative half is exactly the same as that over the positive half.

3. Which is the correct formula for frequency in an ac generator?
a) f=pn
b) f=p/n
c) f=n/p
d) f=n2p

Answer: a [Reason:] The frequency in an ac generator is pn, where p is pairs of poles and speed is n revolutions per second.

4. What will happen to the frequency if the number of revolutions increases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

Answer: a [Reason:] We know that: f=pn, therefore, as n increases, also increases.

5. What happens to the frequency if the number of pairs of poles increases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

Answer: a [Reason:] We know that: f=pn, therefore, as p increases, also increases.

6. Calculate the frequency if the number of revolutions in 300 and the paired poles are 50.
a) 15kHz
b) 150kHz
c) 1500kHz
d) 150Hz

Answer: a [Reason:] We know that: f=pn Substituting the values from the given question, we get f=15kHz.

7. Calculate the number of revolutions if the frequency is 15kHz and the paired poles are 50.
a) 100
b) 200
c) 300
d) 400

Answer: c [Reason:] We know that: f=pn Substituting the values from the given question, we get n=300.

8. Calculate the number of paired poles if the frequency id 15kHz and the number of revolutions is 300.
a) 10
b) 30
c) 50
d) 70

Answer: c [Reason:] We know that: f=pn Substituting the values from the given question, we get n=50.

9. What is the frequency of a two pole machine having n=50.
a) 100Hz
b) 200Hz
c) 50Hz
d) 25Hz

Answer: c [Reason:] For a two pole machine, p=1. f=pn= 1*50= 50Hz.

10. What is the minimum number of poles that a machine must have __________
a) 1
b) 2
c) 4
d) 10

Answer: b [Reason:] The minimum number of poles that a machine must have is 2 because a machine must have at least one pair of poles= 2 poles.

## Set 2

1. Is n/p=ni2 is a correct formula?
a) True
b) False

Answer: b [Reason:] The correct formula is n*p=ni2.

2. Calculate the number of electrons is the number of holes are 15*1010?
a) 15*1010
b) 1.5*108
c) 1.5*109
d) 1.5*1010

Answer: c [Reason:] n*p=(1.5*1010)2 n*15*1010=1.5*1.5*1010*1010 n=1.5*109 electrons.

3. For which type of material, the number of free electron concentration is equal to the number of donor atoms?
a) P type semiconductor
b) Metal
c) N-type semiconductor
d) Insulator

Answer: c [Reason:] The n-type semiconductor has equal concentration of free electron and donor atoms.

4. Identify the correct condition for a semiconductor to be electrically neutral.
a) Nd+p=Na+n
b) Nd-p=Na+n
c) Nd+p=Na-n
d) Nd-p=Na-n

Answer: a [Reason:] The sum of the number of donors and the holes is equal to the sum of the number of the acceptors and the electrons.

5. Do the Fermi energy level changes in a semiconductor?
a) True
b) False

Answer: a [Reason:] The Fermi energy level changes as the electron and hole concentrations change because of the formula which defines the position of the Fermi level depending on the concentration of holes and electrons.

6. Consider a silicon wafer having Nc=2.8*1019cm-3 and the Fermi energy is .25eV below the conduction band. Calculate the equilibrium concentrations of electrons at T=300K?
a) 18*1016cm-3
b) 1.8*1016cm-3
c) 1.8*1014cm-3
d) 180*1016cm-3

7. If Ef>Efi, then what is the type of the semiconductor?
a) n-type
b) P-type
c) Elemental
d) Compound

Answer: a [Reason:] For n-type, the Fermi energy level is greater than the intrinsic Fermi energy level because in an energy band, Fermi level of donors is always greater than that of the acceptors.

8. The 1-fF (E) increases in which of the following band for n type semiconductor?
a) Conduction band
b) Donor band
c) Acceptor band
d) Valence band

Answer: d [Reason:] For an n-type semiconductor, the probability of fF (E) decreases in the valence band. The probability of finding the electron in the conduction band is more.

9. The fF (E) decreases in which of the following band for p-type semiconductor?
a) Conduction band
b) Donor band
c) Acceptor band
d) Valence band

Answer: a [Reason:] The probability of finding the electron in the conduction band decreases for a p-type semiconductor because in a p-type semiconductor, the holes will be in conduction band rather than the electrons.

10. Do the intrinsic Fermi energy level changes with the addition of dopants and acceptors?
a) True
b) False

Answer: b [Reason:] The intrinsic Fermi energy level always remains constant because it is an imaginary level taken to distinguish between the Fermi level of the types of semiconductor.

## Set 3

1. When a forward biased is applied to a diode, the electrons enter to which region of the diode?
a) P-region
b) N-region
c) P-n junction
d) Metal side

Answer: a [Reason:] When the forward biased is applied, the electrons enter to the p-region and the holes enter to the n-region so that holes can flow from p-region to n-region. Whereas, the electrons can travel from n-region to p-region.

2. The number of injected minority carriers falls off linearly with the increase in the distance from the junction. Is it true or false?
a) True
b) False

Answer: b [Reason:] The number of minority carriers fall off exponentially rather than linearly with the increase in the distance from the junction.

3. What is the total current in a diode when x=0?
a) I = Ipn (0) – Inp (0)
b) I = Ipn (0) + Inp (0)
c) I = -Ipn (0) – Inp (0)
d) I = -Ipn (0) + Inp (0)

Answer: b [Reason:] At junction, the total current is equal to the minority hole current plus the minority electron current.

4. The current in the diode is
1. Unipolar
2. Bipolar
a) I only
b) II only
c) I and II both
d) Neither I nor II

Answer: b [Reason:] The current in the diode consists of both the electrons and holes. So, it is bipolar.

5. The current is constant throughout the device. Is it true or false?
a) True
b) False

Answer: a [Reason:] The current in the device is constant but the proportion due to the electrons and holes varies with distance.

6. Which of the following statements is correct under forward biased p-n diode?
a) current enters n side as hole current and leaves p side as electron current
b) current enters n side as electron current and leaves p side as hole current
c) current enters p side as hole current and leaves n side as electron current
d) current enters p side as hole current and leaves p side as electron current

Answer: c [Reason:] When the current flows in a p-n diode, the current enters p side as hole current and leaves n side as electron current.

7. Calculate the total current when Ipn (0)=1mA and Inp (0)=2mA.
a) 1mA
b) -1mA
c) 0
d) 3mA

Answer: d [Reason:] I=Ipn (0)+Inp (0) =1mA+2mA =3mA.

8. What is the hole current in the p region of the diode?
a) Ipp (x)=I-Inp (x)
b) Ipp (x)=I+Inp (x)
c) Ipp (x)=-I-Inp (x)
d) Ipp (x)=-I+Inp (x)

Answer: a [Reason:] The hole current in the p region is equal to the total current minus the minority electrons in the p region.

9. What does Inp represent?
a) Hole current in n region
b) Hole current in p region
c) Electron current in n region
d) Electron current in p region

Answer: d [Reason:] Inp constitutes of the electron current in the p region. It is the minority electron carrier in the p region.

10. Deep into the p side the current is a drift current Ipp of holes sustained by the small electric field in the semiconductor. Is the statement true or false?
a) True
b) False

Answer: a [Reason:] In the p region, the drift current is sustained into the p region by the small electric field which is formed at the junction in the semiconductor. So, the above statement is true.

## Set 4

1. The donor ions is represented by a positive plus sign. Is it True or False?
a) True
b) False

Answer: a [Reason:] The donor atom donates the extra ion to the semiconductor. Therefore, it is represented by the positive plus sign.

2. Initially, the p-type carriers are located to the____________of the semiconductor.
a) Right
b) Left
c) Middle
d) Top

Answer: b [Reason:] The p-type carriers are nominally located to the left of the junction and n-type carriers are to the right.

3. The displacement of the charges results in
a) Magnetic field
b) Electric field
c) Rust
d) Hall effect

Answer: The flow of carriers in a semiconductor results in the electric field across the junction. The electric field thus makes the current flow in the device.

4. What is the value of 1 micron?
a) 10-6cm
b) 10-5cm
c) 10-4cm
d) 10-3cm

5. Which of the following results when the equilibrium established in a semiconductor?
a) Restrain the process of diffusion
b) Electric field becomes very high
c) Both a and b
d) None of these

Answer: c [Reason:] As the electric field is very high, the flow of the carries will be restricted and the equilibrium will be obtained.

6. Which of the following options doesn’t defined for the necessity for the existence of the potential barrier?
a) Contact
b) Potential
c) Diffusion
d) Fermi dirac

Answer: d [Reason:] The potential barrier is formed at the junction of the semiconductor. It’s necessity of the potential barrier is known as the contact, potential or diffusion.

7. Under the open-circuited conditions the net hole current must be zero. Is this statement is True or false?
a) True
b) False

Answer: a [Reason:] The net hole current is zero because if this wasn’t true, the hole density at one end of the semiconductor would continue to increase indefinitely with time, a situation which is obviously physically impossible.

8. The un-neutralised ions in the neighbourhood of the junction are known as
a) Depletion charges
b) Uncovered charges
c) Mobile ions
d) Counter ions

Answer: b [Reason:] The un-neutralised ions in the neighbourhood of the junction are known as uncovered ions because they are not mobile.

9. Which of the following doesn’t defines for the junction which is depleted of mobile charges?
a) Depletion region
b) Uncovered region
c) Space charge region
d) Transition region

Answer: b [Reason:] The junction which is depleted of mobile charges is known as depletion region or space charge region and transition region.

10. Convert 10 micron to meters.
a) 10-5m
b) 107m
c) 10-6m
d) 10-4m

Answer: a [Reason:] Since, 1 micron=10-6m 10 micron =10-5m.

## Set 5

1. An internally compensated op amp is specified to have an open-loop dc gain of 106 dB and a unity gain bandwidth of 3 MHz. Find fb and the open-loop gain at fb.
a) 15Hz and 103 db
b) 30Hz and 103 db
c) 15 Hz and 51.5 db
d) 30 Hz and 51.5 db

Answer: a [Reason:] Use the equations below.

2. A single-pole model has __________ db/decade roll-off of the gain.

Answer: d [Reason:] It is a standard characteristic of a single-pole model.

3. Single-pole model is also known as
a) Frequent pole
b) Stable pole
c) Dominant pole
d) Responsive pole

Answer: c [Reason:] Single-pole model is also called dominant pole.

4. An op amp having a 106-dB gain at dc and a single-pole frequency response with ft = 2 MHz is used to design a non-inverting amplifier with nominal dc gain of 100. The 3-dB frequency of the closed-loop gain is
a) 10 kHz
b) 20 kHz
c) 30 kHz
d) 40 kHz

Answer: b [Reason:] Use the equation below to obtain a frequency response curve and proceed further.

5. An internally compensated op amp has a dc open-loop gain of 106 V/V and an AC open-loop gain of 40 dB at 10 kHz. Estimate its gain–bandwidth product and its expected gain at 1 kHz.
a) 0.1 MHz and -60 db
b) 10 MHz and -60 db
c) 10 MHz and 60 db
d) 1 MHz and 60 db

Answer: d [Reason:]s: Use the following results.

6. An inverting amplifier with nominal gain of −20 V/V employs an op amp having a dc gain of 104 and a unity-gain frequency of 106 Hz. What is the 3-dB frequency f3dB of the closed-loop amplifier?
a) 2π 23.8 kHz
b) 2π 47.6 kHz
c) 2π 71.4 kHz
d) 2π 95.2 kHz

7. cascading two identical amplifier stages, each having a low-pass STC frequency response with a 3dB frequency f1, results in an overall amplifier with a 3dB frequency given by
a) √(√2+1) f1
b) √(√3-1) f1
c) √(√2-1) f1
d) √(√3+1) f1

8. Find the ft required for internally compensated op amps to be used in the implementation of the closed loop amplifiers with dc gain of +100 V/V and 3db bandwidth of 100kHz?
a) 1 kHz
b) 10 kHz
c) 100 kHz
d) 1 MHz

9. A particular op amp, characterized by a gain–bandwidth product of 20 MHz, is operated with a closed-loop gain of +100 V/V. What 3-dB bandwidth results? At what frequency does the closed-loop amplifier exhibit a −6° phase shift?
a) 21 kHz
b) 31.5 kHz
c) 42 kHz
d) 52.5 kHz