# Multiple choice question for engineering

## Set 1

1. The loop which does not contain any other inner loop is known as _____________

a) A node

b) A mesh

c) A branch

d) A super mesh

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2. If there are 6 branches and 4 essential nodes, how many equations are required to describe a circuit in mesh-current method?

a) 3

b) 6

c) 4

d) 2

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3. Find the current flowing through 5Ω resistor in the given circuit.

a) 0.57A

b) 0.64A

c) 0.78A

d) 0.89A

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_{1}, I

_{2}, I

_{3}in the 3 meshes and by applying KVL, equations will be obtained which on solving gives the respective currents flowing in the circuits.

4. A Super Mesh analysis could be done when there is a common _____________ between any two loops.

a) Voltage source

b) Current source

c) Resistor

d) Both voltage and current source

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5. Calculate the current flowing through 10Ω resistor in the circuit shown below.

a) ±0.435A

b) ±0.985A

c) 1.217A

d) 2.782A

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_{3}-I

_{2}=4 Loop1: 11I

_{1}-10I

_{2}=2 KVL at Supermesh: -2I

_{1}+3I

_{2}+3I

_{3}=0 Solving these gives the currents flowing in the circuit and current through 10Ω resistor is either I

_{1}-I

_{2}or I

_{2}-I

_{1}.

6. Find the power delivered by the voltage source in the network given below.

a) 65Watts

b) 72Watts

c) 63Watts

d) 76Watts

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_{1}=9A, I

_{2}=2.5A, I

_{3}=2A. As voltage source is in 1st loop, Power delivered by voltage source=V*I

_{1}.

7. The Mesh-Current method is applicable only for ___________

a) Non-linear networks

b) Equivalent networks

c) Non-planar networks

d) Planar networks

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8. Find the value of V_{X} in the circuit given below.

a) –0.8A

b) +0.8A

c) -4.8A

d) +4.8A

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_{1}=0.4A, I

_{2}=2.4A. V

_{X}=-I

_{1}R

_{1}.

9. A Supermesh is formed between two loops which share a common voltage source.

a) True

b) False

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10. If 4 equations are required to describe a circuit by Mesh-Current method and there are n nodes. How many branches are there in the network?

a) n+5

b) n+3

c) n

d) n-1

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11. If there are 16 branches and 5 essential nodes, how many equations are required to describe a circuit in mesh-current method?

a) 12

b) 16

c) 21

d) 9

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12. Determine the current through 3Ω resistor in the network given below.

a) 2A

b) 3A

c) 4A

d) -2A

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_{1}and I

_{2}and applying KVL, the current through required resistor is found out.

13. Mesh analysis is best suitable for _____________

a) Current sources

b) Voltage sources

c) Complex elements

d) Unilateral elements

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## Set 2

1. Where voltage division problem arises

a) Series connected resistors

b) Parallel connected resistors

c) When resistors are equal

d) Both series and parallel resistors.

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2. Where current division problem arises

a) Series connected resistors

b) Parallel connected resistors

c) When resistors are equal

d) Both series and parallel resistors.

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3. If there are 3 Resistors R_{1}, R_{2} and R_{3} in series and V is total voltage and I is total current then Voltage across R_{2} is

a) V R_{3}/ R_{1} + R_{2} + R_{3}

b) V R_{2}/ R_{1} + R_{2} + R_{3}

c) V R_{1}/R_{1} + R_{2} + R_{3}

d) V

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_{2}=I R

_{2}= V R

_{2}/ R

_{1}+ R

_{2}+ R

_{3}.

4.

Calculate Voltage across 2Ω Resistor where supply v= 10volts.

a) 2V

b) 3V

c) 10V

d) 4V

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_{2}= 10(2) V

_{2}= I.R

_{2}= 2(2) 4V.

5.

Calculate i =?

a) -1A

b) +2A

c) 8A

d) -5A

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6. For a parallel connected resistor R_{1}, R_{2} and a voltage of V volts. Current across the first resistor is given by

a) I R_{1}

b) I R_{2}

c) I R_{1} / R_{1} + R_{2}

d) I R_{2} / R_{1} + R_{2}

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_{1}= V / R

_{1}R = R

_{1}. R

_{2}/ R

_{1}+ R

_{2}= I . R

_{1}. R

_{2}/ R

_{1}. R

_{1}+ R

_{2}I

_{1}= I R

_{2}/ R

_{1}+ R

_{2}.

7. R_{1} = 1Ω, R_{2} = 3Ω, R_{3} = 5Ω and R_{4} = 7Ω connected in series. Total voltage = 20V, Current I, V2 =?

a) I = 1.23, V_{2} = 3.75

b) I = 1.25, V_{2} = 3.75

c) I = 1.15, V_{2}= 3.73

d) I = 1.16, V_{2} = 3.72

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_{2}= V. R

_{2}/ R

_{1}+ R

_{2}+ R

_{3}+ R

_{4}= 20(3)/16 = 3.75V.

8. R_{1} = 1Ω, R_{2} = 3Ω, R_{3} = 5Ω and R_{4} = 7Ω connected in parallel. Total Current = 23A. Then V, I_{1} , I_{2} =?

a) 12.26v, 1.725, 2.875

b) 12.23v, 2.875, 1.725

c) 11.26v, 1.95, 1.74

d) 11.23v, 1.74, 1.95

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_{1}+ R

_{2}) R

_{1}R

_{2}= 12.26v I1 = IR

_{2}/ R

_{1}+ R

_{2}= 1.725A I

_{2}= IR

_{1}/ R

_{1}+ R

_{2}= 2.875A.

9. Voltage division is necessary for parallel resistance networks

a) True

b) False

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10. Why is current division necessary?

a) In series current is the same

b) In parallel current differs

c) Because the voltage is also different

d) Because of Kirchhoff’s laws.

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## Set 3

1. The symbol used for inductance is __________

a)

b)

c)

d)

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2. The symbol used for capacitance is _____________

a)

b)

c)

d)

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3. The formula used to find the capacitance C is __________

a) Q/v

b) Qv

c) Q-v

d) Q + v

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4. The capacitor doesn’t allow sudden changes in ___________

a) Voltage

b) Current

c) Resistance

d) Capacitance

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5. The Inductor doesn’t allow sudden changes in ___________

a) Voltage

b) Current

c) Resistance

d) Inductance

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6. The expression for energy of an inductor ____________

a) ½ LI

b) L/2I

c) ½ L^{2}I

d) ½ LI^{2}

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^{2}.

7. The units for inductance is _________ and capacitance is ___________

a) Faraday, Henry

b) Coulomb, Faraday

c) Henry, Faraday

d) Henry, Coulomb

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8. The voltage applied to a pure capacitor of 50*10^{-6} F is as shown in figure. Calculate the current for 0-1msec.

a) 5A

b) 1A

c) -5A

d) -1A

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^{-3}*m →m= 1*10

^{5}→ V (t) = 1*10

^{5}t Current I (t) = c. d (v (t))/dt = 50*10

^{-6}* (d (1*10

^{5}t)/dt) = 5A.

9. If a capacitor of capacitance 9.2F has a voltage of 22.5V across it. Calculate the energy of the capacitor.

a) 5062.5W

b) 506.25W

c) 50.625W

d) 50625W

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^{2}.

10. The voltage applied to the 212mH inductor is given by v(t)= 15e^{-5t}v. Calculate the current.

a) 16.782e^{-10t}

b) 15.75e^{-5t}

c) 11.27e^{-10t}

d) 14.15e^{-5t}

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11. A voltage across a capacitor of 0.5F is defined by

V (t) = [0, t<0

2t, 0<t<2s

4e^{-(t-2)}, t>2s]

Find i (t).

a) -2e^{-(t-2)} A

b) -4e^{-(t-2)} A

c) -20e^{-(t-2)} A

d) -12e^{-(t-2)} A

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12. If the voltage across a capacitor is constant, then current passing through it is ________

a) 1

b) 0

c) -1

d) Infinity

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13. An Inductor works as a ___________ circuit for DC supply.

a) Open

b) Short

c) Polar

d) Non-polar

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14. The insulating medium between the two plates of capacitor is known as __________

a) Electrode

b) Capacitive medium

c) Conducting medium

d) Dielectric

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15. If the current flowing through an inductor of inductance 0.3Henry is 5.3t^{2}+4.7t. Calculate the power.

a) 0

b) 1

c) 16.854t^{3}+22.41t^{2}+6.62t

d) 15.3t^{3}+27.8t^{2}+19

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## Set 4

1. The opamp in the Inverting circuit is in __________

a) Linear region

b) Saturation

c) Cut-off region

d) Non-linear region

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2. In an Inverting Amplifier circuit, the output voltage v_{o} is expressed as a function of ____________

a) Input current

b) Output current

c) Source voltage

d) Source current

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_{o}as a function of source voltage vs.

3. The other name for Gain is ____________

a) Scaling factor

b) Output

c) Amplifying factor

d) Scaling level

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_{f}/R

_{s}in case of an Inverting amplifying circuit.

4. If V_{CC} = 12V and vs=1mV, then R_{f}/R_{s} is _____________

a) >12000

b) <12000

c) 12000

d) 1

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_{f}/R

_{s}≤ │V

_{CC}/vs│.

5. In the expression v_{o}= -Av_{n}, A is called ______________

a) Closed loop gain

b) Closed loop fault

c) Open loop fault

d) Open loop gain

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6. The circuits of an inverting and Non-Inverting amplifying comprises of __________ and _______ number of resistors.

a) 3, 2

b) 2, 3

c) 2, 2

d) 3, 3

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_{s}, R

_{f}. Non-Inverting amplifying circuit – R

_{s}, R

_{f}, R

_{g}.

7. The condition for a Non-inverting amplifying circuit to operate in linear region operation _____________

a) (R_{s}+R_{f})/R_{s} < │V_{CC}/v_{g}│

b) (R_{s}+R_{f})/R_{s} ≠ │V_{CC}/v_{g}│

c) (R_{s}+R_{f})/R_{s} > │V_{CC}/v_{g}│

d) (R_{s}+R_{f})/R_{s} = │V_{CC}/v_{g}│

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_{s}+R

_{f})/R

_{s}<│VCC/vg│.

8. If R_{s}= 3Ω, Rf= 6Ω then the relation between v_{o} and v_{g} in case of a Non-Inverting amplifying circuit.

a) v_{o}= 9v_{g}

b) v_{o}= 6v_{g}

c) v_{o}= 3v_{g}

d) v_{o}= v_{g}

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_{o}= ((R

_{s}+R

_{f})/R

_{s}) *v

_{g}.

9. If R_{s}= 5Ω, R_{f}= 25Ω and -2.5V ≤ v_{g} ≤ 2.5V. What are the smallest power supply voltages that could be applied and still have opamp in linear region?

a) ±9V

b) ±2.5V

c) ±6V

d) ±15V

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_{s}+R

_{f})/R

_{s}) *v

_{g}. By substituting the values, we have v

_{o}=6v

_{g}. v

_{o}=6(-2.5) = -15 v

_{o}=6(2.5) =15.

10. If an inverting amplifying circuit has a gain of 10 and ±15V power supplies are used. The values of input for which opamp would be in the linear region?

a) ±1.25

b) ±1.5V

c) ±2.25

d) ±0.5

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_{f}/R

_{s}= 10 and v

_{o}= (-R

_{f}/R

_{s})*v

_{s}. → v

_{o}= -10v

_{s}and given -12V≤ v

_{o}≤ 12V. → -15= -10v

_{s}. So, v

_{s}= 1.5V → 15=-10v

_{s}. So, v

_{s}=-1.5V.

11. If the gain of an inverting amplifying circuit is 13 and ±22V power supplies are used. What range of input values allows the opamp to be in linear region?

a) ±1.69

b) ±1.35V

c) ±2.28

d) ±0.5

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_{o}= (-R

_{f}/R

_{s})*v

_{s}. → v

_{o}= -13v

_{s}and given -22V≤ v

_{o}≤ 22V. → -22= -13v

_{s}. So, v

_{s}=1.692 V → 22=-13v

_{s}. So, v

_{s}=-1.692V.

12. The input applied to an Inverting amplifier is ______________

a) Equal to output

b) Equal to Inverted output

c) Not equal to output

d) Output is equal to input

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13. In R_{1}=10kΩ, R_{f}=100kΩ, v_{1}=1V. A load of 25kΩ is connected to the output terminal. Calculate i_{1} and v_{o}.

a) 0.5mA, 10V

b) 0.1mA, 10V

c) 0.1mA, -10V

d) 0.5mA, -10V

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_{1}= v

_{1}/R

_{1}= 1V/10kΩ = 0.1mA V

_{0}= -(R

_{f}/R

_{1})*v

_{1}= -(100kΩ/10kΩ)*1V = -10V.

## Set 5

1.KCL is based on the fact that

a) There is a possibility for a node to store energy.

b) There cannot be an accumulation of charge at a node.

c) Charge accumulation is possible at node

d) Charge accumulation may or may not be possible.

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2. Relation between currents according to KCL is

a) i_{1}=i_{2}=i_{3}=i_{4}=i_{5}

b) i_{1}+i_{4}+i_{3}=i_{5}+i_{2}

c) i_{1}-i_{5}=i_{2}-i_{3}-i_{4}

d) i_{1}+i_{5}=i_{2}+i_{3}+i_{4}

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3. The algebraic sum of voltages around any closed path in a network is equal to ____________

a) Infinity

b) 1

c) 0

d) Negative polarity

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4.

Calculate potential difference between x and y

a) 4.275v

b) -4.275v

c) 4.527v

d) -4.527v

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_{1}= 3/3+5 = 3/8 = 0.375Ω I

_{2}= 4/5 = 0.8Ω V

_{xy }= v

_{x}– v

_{y}V

_{x}+ 5I

_{1}+ 4 – 2I

_{2}– v

_{y}= 0 V

_{x}– v

_{y}= 2I2 – 4 – 5I

_{1}= -4.275Ω

5.

Find R

a) 17.5 Ω

b) 17.2 Ω

c) 17.4 Ω

d) 17.8 Ω

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6. Determine currents I_{1} , I_{2} and I_{3}.

a) -3.3A, -8.5A, 2.4A

b) 3A, -8A, 2A

c) 3.3A, 8.5A, -2.4A

d) 3.2A, 8.6A, 2.3A

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_{1}= I

_{1}– I

_{2}+ 8 + I

_{3}+ 3 I

_{2}– I

_{3}= 11 -> 1 And -11 I1 – 7(I

_{1}– I

_{2}) = 0 -18 I

_{1}+ 7 I

_{2}= 0 -> 2 And -11 I

_{1}– 15 I

_{3}=0 -> 3 Solving I

_{1}= 3.32A I

_{2}= 8.5A I

_{3}= -2.4A.

7. All _____________ are loops but _______________ are not meshes

a) Loops, Meshes

b) Meshes, loops

c) Branches, loops

d) Nodes, Branches

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8.

Solve for I.

a) -0.5A

b) 0.5A

c) -0.2A

d) 0.2A

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_{eq}= 10 + 5 -20 = -5u R

_{eq}= 5 + 2 + 3 = 10Ω I = V/R = -5/10 = -0.5A.

9. The basic laws for analyzing an electric circuit are :-

a) Einstein’s theory

b) Newtons laws

c) Kirchhoff’s laws

d) Faradays laws

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10. A junction whell two (or) more than two network elements meet is known as a ______________

a) Node

b) Branch

c) Loop

d) Mesh