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Multiple choice question for engineering

Set 1

(Q.1-Q.2) For the circuit given below i1 = 4 sin(2t) and i2 = 0
electronic-devices-circuits-questions-answers-magnetically-coupled-circuits-q1
1. v1 = ?
a) -16 cos 2t V
b) 16 cos 2t V
c) 4 cos 2t V
d) -4 cos 2t V

View Answer

Answer: b [Reason:] v1 = 2 d(i1)/dt + d(i2)/dt.

2. v2 = ?
a) 2 cos 2t V
b) -2 cos 2t V
c) 8 cos 2t V
d) -8 cos 2t V

View Answer

Answer: c [Reason:] v2 = d(i1)/dt + d(i2)/dt.

Circuit for Q.3-Q.4
electronic-devices-circuits-questions-answers-magnetically-coupled-circuits-q3
3. If i1 = 0 and i2 = 2 sin(4t), the voltage v1 is
a) -24 cos(4t) V
b) 24 cos(4t) V
c) 1.5 cos(4t) V
d) -1.5 cos(4t) V

View Answer

Answer: b [Reason:] v1 = 3 d(i1)/dt – 3 d(i2)/dt.

4. If i1 = e(-2t) and i2 = 0, the voltage v1 is
a) -6e(-2t) V
b) 6e(-2t) V
c) 1.5e(-2t) V
d) -1.5e(-2t) V

View Answer

Answer: c [Reason:] v1 = 4 d(i1)/dt – 3 d(i2)/dt.

Circuit for Q.5-Q.8
electronic-devices-circuits-questions-answers-magnetically-coupled-circuits-q5
5. If i1 = 3 cos(4t) and and i2 = 0. Find v1.
a) -24 sin(4t) V
b) 24 sin(4t) V
c) 1.5 sin(4t) V
d) -1.5 sin(4t) V

View Answer

Answer: a [Reason:] v1 = 2 d(i1)/dt – 2 d(i2)/dt.

6. If i1 = 3 cos(4t) and and i2 = 0. Find v2.
a) -24 sin(4t) V
b) -36 sin(4t) V
c) sin(4t) V
d) -sin(4t) V

View Answer

Answer: a [Reason:] v2 = -3 d(i1)/dt + 2 d(i2)/dt.

7. If i1 = 4 cos(3t) and and i1 = 0. Find v2.
a) 12 cos(3t) V
b) -12 cos(3t) V
c) -24 cos(3t) V
d) 24 cos(3t) V

View Answer

Answer: c [Reason:] v1 = 2 d(i1)/dt – 2 d(i2)/dt.

8. If i2 = 4 cos(3t) and and i1 = 0. Find v2.
a) -12 cos(3t) V
b) -24 cos(3t) V
c) -36 cos(3t) V
d) -48 cos(3t) V

View Answer

Answer: c [Reason:] v2 = 3 d(i1)/dt + 2 d(i2)/dt.

9. Leq = ?
electronic-devices-circuits-questions-answers-magnetically-coupled-circuits-q9
a) 4 H
b) 6 H
c) 7 H
d) 0 H

View Answer

Answer: c [Reason:] Leq = L1 + L2 – 2M.

10. Leq = ?
electronic-devices-circuits-questions-answers-magnetically-coupled-circuits-q10
a) 2 H
b) 4 H
c) 6 H
d) 8 H

View Answer

Answer: a [Reason:] Leq = L1 + L2 – 2M.

Set 2

1. V1 = ?
electronic-devices-circuits-questions-answers-method-analysis-q1
a) 0.4 Vg
b) 1.5 Vg
c) 0.67 Vg
d) 2.5 Vg

View Answer

Answer: b [Reason:] Apply nodal analysis, V1 = Vs(4/6 + 1/3)/(1/6 + 1/3 + 1/6) = 1.5 Vs.

2. Va = ?
electronic-devices-circuits-questions-answers-method-analysis-q2
a) -11 V
b) 11 V
c) 3 V
d) -3 V

View Answer

Answer: b [Reason:] Va = 2(3+1) + 3 = 11 V.

3. V1 = ?
electronic-devices-circuits-questions-answers-method-analysis-q3
a) 120 V
b) -120 V
c) 90 V
d) -90 V

View Answer

Answer: d [Reason:] -V1/60 – V1/60 + 6 = 9 or V1 = -90 V.

4. Va = ?
electronic-devices-circuits-questions-answers-method-analysis-q4
a) 4.33 V
b) 4.09 V
c) 8.67 V
d) 8.18 V

View Answer

Answer: c [Reason:] (Va – 10)/4 + Va/2 = 4 or Va = 8.67 V.

5. V2 = ?
electronic-devices-circuits-questions-answers-method-analysis-q5
a) 0.5 V
b) 1.0 V
c) 1.5 V
d) 2.0 V

View Answer

Answer: d [Reason:] V2/20 + V2+10/30 = 0.5 V or V2 = 2.0 V.

6. Ib = ?
electronic-devices-circuits-questions-answers-method-analysis-q5
a) 0.6 A
b) 0.5 A
c) 0.4 A
d) 0.3 A

View Answer

Answer: b [Reason:] electronic-devices-circuits-questions-answers-method-analysis-q6.

7. i1 = ?
electronic-devices-circuits-questions-answers-method-analysis-q7
a) 0.6 A
b) 2.1 A
c) 1.7 A
d) 1.1 A

View Answer

Answer: a [Reason:] i1 = 10/36+64 + 0.5 => i1 = 0.6 A.

8. i1 = ?
electronic-devices-circuits-questions-answers-method-analysis-q8
a) 1 mA
b) 1.5 mA
c) 2 mA
d) 2.5 mA

View Answer

Answer: b [Reason:] 75 = 90k (i1 – 7.5m) => 150 = 100ki1 => i = 1.5 mA.

9. i1 = ?
electronic-devices-circuits-questions-answers-method-analysis-q9
a) 4 A
b) 3 A
c) 6 A
d) 5 A

View Answer

Answer: b [Reason:] 3 – 5i1 – 12 => i1 = 3A.

10. i1 = ?
electronic-devices-circuits-questions-answers-method-analysis-q10
a) 20 mA
b) 15 mA
c) 10 mA
d) 5 mA

View Answer

Answer: b [Reason:] 45 = 2ki1 + 500(i1 + 15m) => i1 = 15 mA.

Set 3

1. Which of the following is method to model a diode’s forward characteristics?
a) Iteration method
b) Graphical method
c) Constant-voltage drop model
d) All of the mentioned

View Answer

Answer: d [Reason:] All of the mentioned are different methods used to model a diode’s forward characteristics.

2. A voltage regulator needs to provide a constant voltage in spite of the fact that there may be
a) Change in the load current drawn from the terminals of the regulator
b) Change of the DC power supply that feeds the regulator circuit
c) None of the mentioned
d) All of the mentioned

View Answer

Answer: d [Reason:] Voltage regulators are required in both of the situations mentioned.

(Q.3-Q.4) For the circuit shown below calculate
electronic-devices-circuits-questions-answers-modelling-diode-forward-characteristic-q3
3. Calculate the %age change in the regulated voltage caused by a change of ±10% in the input voltage. (RL is not connected to the circuit)
a) ± 0.5%
b) ± 1%
c) ± 5%
d) ± 10%

View Answer

Answer: a [Reason:] Current I in the circuit = (10 – 2.1) / 1 or 7.9 mA Incremental resistance of each diode is rd = 25mV/7.9mA at room temperature or 3.2Ω Total resistance of the diode connection = r = 9.6Ω The series combination of R and r acts as a voltage divider, hence ▲V0 = ±r/r+R or 9.5 mA or ± 0.5%.

4. Calculate the change in the voltage when RL is connected as shown
a) -10 mA
b) -15 mA
c) -20 mA
d) -25 mA

View Answer

Answer: c [Reason:] When RL is connected it draws approx. 2.1v/1000 A current or 2.1 mA. Hence the current in the diode decreases by approximately 2.1 mA. The voltage change across the diode is thus -2.1mA * 9.6 Ω or -20mA.

5. The value of the diode small-signal resistance rd at bias currents of 0.1 mA is
a) 250 Ω
b) 25 Ω
c) 2.5 Ω
d) 0.25 Ω

View Answer

Answer: a [Reason:] Thermal voltage at room temperature is 25mV and the current is 0.1mA. rd = 25mV/0.1mA or 250 Ω.

6. In many applications, a conducting diode is modeled as having a constant voltage drop, usually approximately
a) 1 V
b) 0.1 V
c) 0.7 V
d) 7V

View Answer

Answer: c [Reason:] Normally a silicon diode has a voltage drop of 0.7V.

7. The graphical method of modeling a diode characteristics is based on
a) Iteration method
b) Constant voltage drop method
c) Small signal approximation
d) Exponential method

View Answer

Answer: d [Reason:] Graphical method uses equations obtained in the exponential method and then the Answer of these equations is obtained by plotting them on a graph.

8. For small-signal operation around the dc bias point, the diode is modeled by a resistance equal to the
a) Slope of the i-v characteristics of the diode at the bias point
b) Square root of the slope of the i-v characteristics of the diode at the bias point
c) Inverse of the slope of the i-v characteristics of the diode at the bias point
d) Square root of the inverse of the slope of the i-v characteristics of the diode at the bias point

View Answer

Answer: d [Reason:] It is a required characteristic for small-ground operations.

9. The other name for bias point is
a) Quiescent point
b) Node point
c) Terminal point
d) Static point

View Answer

Answer: a [Reason:] Quiescent point is also called base point. It is due to this name that is also called Q–point.

10. In iteration method for modelling a diode the answer obtained by each subsequent iteration is
a) Is close to the true value
b) Is close to the value obtained by exponential method
c) All of the mentioned
d) None of the mentioned

View Answer

Answer: c [Reason:] True value is obtained by the exponential method and therefore both of the statements are true.

Set 4

(Q.1 & Q.2) Various measurements are made on an NMOS amplifier for which the drain resistor RD is 20 kΩ. First, DC measurements show the voltage across the drain resistor, VRD, to be 2 V and the gate-to-source bias voltage to be 1.2 V. Then, ac measurements with small signals show the voltage gain to be −10 V/V.

1. What is the value of Vt for this transistor?
a) 0.7V
b) 0.8V
c) 0.9V
d) 1.0V

View Answer

Answer: b [Reason:] electronic-devices-circuits-questions-answers-mosfet-amplifier-design-q1

2. If the process transconductance parameter is 50μA/V2, what is the MOSFET’s W/L?
a) 25
b) 50
c) 75
d) 100

View Answer

Answer: a [Reason:] electronic-devices-circuits-questions-answers-mosfet-amplifier-design-q1

(Q.3-Q.5) Consider the amplifier below for the case VDD = 5 V, RD = 24 kΩ, (W/L) = 1 mA/V2, and Vt = 1 V.
electronic-devices-circuits-questions-answers-mosfet-amplifier-design-q3

3. If the amplifier is biased to operate with an overdrive voltage VOV of 0.5 V, find the incremental gain at the bias point.
a) -3 V/V
b) -6 V/V
c) -9 V/V
d) -12 V/V

View Answer

Answer: d [Reason:] electronic-devices-circuits-questions-answers-mosfet-amplifier-design-q3a

4. For amplifier biased to operate with an overdrive voltage of 0.5V, and disregarding the distortion caused by the MOSFET’s square-law characteristic, what is the largest amplitude of a sine-wave voltage signal that can be applied at the input while the transistor remains in saturation?
a) 1.61 V
b) 1.5 V
c) 0.11 V
d) 3.11 V

View Answer

Answer: b [Reason:] Use the standard mathematical formula to obtain the result.

5. For the input signal of 1.5V what is the value of the gain value obtained?
a) -12.24 V/V
b) -12.44 V/V
c) -12.64 V/V
d) -12.84 V/V

View Answer

Answer: c [Reason:] The amplitude of the output voltage signal that results is approximately equal to Voq – VOB = 2 – 0.61 = 1.39v. The gain implied by amplitude is Gain = -1.39/0.11 = -12.64 V/V.

6. Which of the following is the fastest switching device?
a) JEFT
b) Triode
c) MOSFET
d) BJT

View Answer

Answer: c [Reason:] MOSFET is the fastest switching device among the given four options.

7. Bias point is also referred by the name
a) DC Operating Point
b) Quiescent Point
c) None of the mentioned
d) All of the mentioned

View Answer

Answer: d [Reason:] Bias point is called dc operating point as the MOSFET functions best at this point. Also since at the bias point no signal component is present it is called quiescent point (he reason why it is represented by the symbol ‘Q’)

(Q.8 –Q.10) Consider the amplifier circuit shown below. The transistor is specified to have Vt = 0.4 V, kn = 0.4 mA/V2, W/L = 10 and λ = 0. Also, let VDD = 1.8V, RD = 17.5kΩ, VGS = 0.6V and vgs = 0V.
electronic-devices-circuits-questions-answers-mosfet-amplifier-design-q8
8. Find ID.
a) 0.08 mA
b) 0.16 mA
c) 0.4 mA
d) 0.8 mA

View Answer

Answer: a [Reason:] electronic-devices-circuits-questions-answers-mosfet-amplifier-design-q9

9. Find VDS.
a) 0.1V
b) 0.2 V
c) 0.4 V
d) 0.8 V

View Answer

Answer: c [Reason:] electronic-devices-circuits-questions-answers-mosfet-amplifier-design-q9

10. Find Av.
a) -12 V/V
b) -14 V/V
c) -16 V/V
d) -18 V/V

View Answer

Solution: b [Reason:] Av = – kn Vov RD = -0.4 * 10 * 0.2 * 17.5 = – 14.4v

Set 5

1. If a MOSFET is to be used in the making of an amplifier then it must work in
a) Cut-off region
b) Triode region
c) Saturation region
d) Both cut-off and triode region can be used

View Answer

Answer: c [Reason:] Only in the saturation region a MOSFET can operate as an amplifier.

2. For MOSFET is to be used as a switch then it must operate in
a) Cut-off region
b) Triode region
c) Saturation region
d) Both cut-off and triode region can be used

View Answer

Answer: d [Reason:] In both regions it can perform the task of a switch.

(Q.3 & Q.4) Using the circuit shown below,
electronic-devices-circuits-questions-answers-mosfets-current-voltage-characterisitcs-q3
3. Determine the conditions in which the MOSFET is operating in the triode region.
i. VGD > Vt (Threshold voltage)
ii. VDS > VOV
iii. ID ∝ (VOV – 0.5VDS)VDS

a) i, ii, and iii are correct
b) i and iii are correct
c) i and ii are correct
d) ii and iii are correct

View Answer

Answer: b [Reason:] Only the points I and iii are correct and ii is false.

4. Determine the conditions in which the MOSFET is operating in the saturation region
i. VGD > Vt (Threshold voltage)
ii. VDS > VOV
iii. ID ∝ (VOV)2

a) i, ii, and iii are correct
b) i and iii are correct
c) i and ii are correct
d) ii and iii are correct

View Answer

Answer: d [Reason:] i is false and ii and iii are true.

5. In the saturation region of the MOSFET the saturation current is
a) Independent of the voltage difference between the source and the drain
b) Depends directly on the voltage difference between the source and the drain
c) Depends directly on the overdriving voltage
d) Depends directly on the voltage supplied to the gate terminal

View Answer

Answer: a [Reason:] Saturation current does not depends on the voltage difference between the source and the drain in the saturation region of a MOSFET.

6. An n-channel MOSFET operating with VOV=0.5V exhibits a linear resistance = 1 kΩ when VDS is very small. What is the value of the device transconductance parameter kn?
a) 2 mA/V2
b) 20 mA/V2
c) 0.2 A/V2
d) 2 A/V2

View Answer

Answer: a [Reason:] Use the standard mathematical expression to determine the value of kn.

7. An NMOS transistor is operating at the edge of saturation with an overdrive voltage VOV and a drain current ID. If is VOV is doubled, and we must maintain operation at the edge of saturation, what value of drain current results?
a) 0.25ID
b) 0.5ID
c) 2ID
d) 4ID

View Answer

Answer: c [Reason:] I0 is directly proportional to VOS.

(Q.8-Q.10) Using the circuit below answer the question
electronic-devices-circuits-questions-answers-mosfets-current-voltage-characterisitcs-q8
8. Which of the following is true for the triode region?
a. VDG > Vtp
b. VSD < VOV
c. ID ∝ VOV
d. None of the mentioned

View Answer

Answer: d [Reason:] VDG > |Vtp| and VSD < |VOV|.

9. Which of the following is true for the saturation region?
a) VDG ≤ |Vtp|
b) VSD ≤| VOV|
c) VDG < |Vtp|
d) VSD <| VOV|

View Answer

Answer: a [Reason:] It is a characteristic for the saturation region.

10. The current iD
a) Depends linearly on VOV in the saturation region
b) Depends on the square of VOV in the saturation region
c) Depends inversely on VOV in the triode region
d) None of the mentioned

View Answer

Answer: b [Reason:] Use the standard mathematical expressions for i0 in different regions.