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# 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

Answer: b [Reason:] A mesh is defined as a loop which does not contain any other loop within it.

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

Answer: a [Reason:] In Mesh-Current method, b-(n-1) equations are required to describe the circuit. b= the number of branches and n= the number of essential nodes.

3. Find the current flowing through 5Ω resistor in the given circuit. a) 0.57A
b) 0.64A
c) 0.78A
d) 0.89A

Answer: c [Reason:] There are 3 meshes in the given circuit. Assuming currents I1, I2, I3 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

Answer: b [Reason:] A Super Mesh analysis could be done when there is a common current source between any two loops.

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

Answer: a [Reason:] Loop2 and loop3 forms a supermesh. Supermesh: I3-I2=4 Loop1: 11I1-10I2=2 KVL at Supermesh: -2I1+3I2+3I3=0 Solving these gives the currents flowing in the circuit and current through 10Ω resistor is either I1-I2 or I2-I1.

6. Find the power delivered by the voltage source in the network given below. a) 65Watts
b) 72Watts
c) 63Watts
d) 76Watts

Answer: c [Reason:] 3 loops and a supermesh between loop1 and loop3. Using KVL currents are found out. I1=9A, I2=2.5A, I3=2A. As voltage source is in 1st loop, Power delivered by voltage source=V*I1.

7. The Mesh-Current method is applicable only for ___________
a) Non-linear networks
b) Equivalent networks
c) Non-planar networks
d) Planar networks

Answer: d [Reason:] The Mesh-Current method is applicable only for Planar networks. A network is said to be planar if there are no crossovers in it and it can be drawn freely on a plane surface.

8. Find the value of VX in the circuit given below. a) –0.8A
b) +0.8A
c) -4.8A
d) +4.8A

Answer: a [Reason:] Applying KVL, currents could be found out. I1=0.4A, I2=2.4A. VX=-I1R1.

9. A Supermesh is formed between two loops which share a common voltage source.
a) True
b) False

Answer: b [Reason:] Meshes that share a current source with other meshes, none of which contains a current source in the outer loop, forms a supermesh.

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

Answer: b [Reason:] Standard formulae: b-(n-1) Given b-(n-1) =4 -> b=4+ (n-1) =n+3.

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

Answer: a [Reason:] In Mesh-Current method, b-(n-1) equations are required to describe the circuit. b=the number of branches and n= the number of essential nodes.

12. Determine the current through 3Ω resistor in the network given below. a) 2A
b) 3A
c) 4A
d) -2A

Answer: c [Reason:] Mesh1 and Mesh2 form a super mesh. Assuming currents I1 and I2 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

Answer: a [Reason:] Mesh analysis is best suitable for Current sources.

## 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.

Answer: a [Reason:] In series, voltage is the difference and current same.

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.

Answer: b [Reason:] In parallel voltage is same and current is the difference.

3. If there are 3 Resistors R1, R2 and R3 in series and V is total voltage and I is total current then Voltage across R2 is
a) V R3/ R1 + R2 + R3
b) V R2/ R1 + R2 + R3
c) V R1/R1 + R2 + R3
d) V

Answer: b [Reason:] V2 =I R2 = V R2/ R1 + R2 + R3.

4. Calculate Voltage across 2Ω Resistor where supply v= 10volts.
a) 2V
b) 3V
c) 10V
d) 4V

Answer: d [Reason:] I = 10/5 = 2A V2 = 10(2) V2 = I.R2 = 2(2) 4V.

5. Calculate i =?
a) -1A
b) +2A
c) 8A
d) -5A

Answer: b [Reason:] i = 1/1+3(8) = 2A.

6. For a parallel connected resistor R1, R2 and a voltage of V volts. Current across the first resistor is given by
a) I R1
b) I R2
c) I R1 / R1 + R2
d) I R2 / R1 + R2

Answer: d [Reason:] I1 = V / R1 R = R1. R2 / R1 + R2 = I . R1. R2 / R1 . R1 + R2 I1 = I R2 / R1 + R2.

7. R1 = 1Ω, R2 = 3Ω, R3 = 5Ω and R4 = 7Ω connected in series. Total voltage = 20V, Current I, V2 =?
a) I = 1.23, V2 = 3.75
b) I = 1.25, V2 = 3.75
c) I = 1.15, V2= 3.73
d) I = 1.16, V2 = 3.72

Answer: b [Reason:] I = 20/ 1 + 3 + 5 + 7 = 1.25A V2 = V. R2 / R1 + R2 + R3 + R4 = 20(3)/16 = 3.75V.

8. R1 = 1Ω, R2 = 3Ω, R3 = 5Ω and R4 = 7Ω connected in parallel. Total Current = 23A. Then V, I1 , I2 =?
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

Answer: a [Reason:] V = I/R V = I (R1 + R2) R1 R2 = 12.26v I1 = IR2/ R1 + R2 = 1.725A I2 = IR1/ R1 + R2 = 2.875A.

9. Voltage division is necessary for parallel resistance networks
a) True
b) False

Answer: b [Reason:] In parallel, connection voltage is same so no division is required.

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.

Answer: b [Reason:] In parallel current differs.

## Set 3

1. The symbol used for inductance is __________
a) b) c) d) Answer: c [Reason:] is the symbol used to represent inductance.

2. The symbol used for capacitance is _____________
a) b) c) d) Answer: b [Reason:] is the symbol used to represent capacitance.

3. The formula used to find the capacitance C is __________
a) Q/v
b) Qv
c) Q-v
d) Q + v

Answer: a [Reason:] Q=cv. Q-charge, V-voltage, c-capacitance.

4. The capacitor doesn’t allow sudden changes in ___________
a) Voltage
b) Current
c) Resistance
d) Capacitance

Answer: a [Reason:] Any small change in voltage occurs within zero time across the gives an infinite current which is practically impossible. So, in a fixed capacitor, the voltage cannot change abruptly.

5. The Inductor doesn’t allow sudden changes in ___________
a) Voltage
b) Current
c) Resistance
d) Inductance

Answer: b [Reason:] Any small change in current occurs within zero time across the gives an infinite voltage which is practically impossible. So, in a fixed inductor, the voltage cannot change abruptly.

6. The expression for energy of an inductor ____________
a) ½ LI
b) L/2I
c) ½ L2I
d) ½ LI2

Answer: d [Reason:] E=∫p dt =∫ LI*(dI/dt).dt = L∫I dI = ½ LI2.

7. The units for inductance is _________ and capacitance is ___________
d) Henry, Coulomb

Answer: c [Reason:] The unit for inductance is ‘Henry’ and capacitance is ‘Faraday’.

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

Answer: a [Reason:] For 0≤t≤1msec, V (t) =m*t (y=mx form) →100= 1*10-3*m →m= 1*105 → V (t) = 1*105t Current I (t) = c. d (v (t))/dt = 50*10-6* (d (1*105t)/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

Answer: b [Reason:] E= ½ cv2.

10. The voltage applied to the 212mH inductor is given by v(t)= 15e-5tv. Calculate the current.
a) 16.782e-10t
b) 15.75e-5t
c) 11.27e-10t
d) 14.15e-5t

Answer: d [Reason:] Current I(t)= 1/L 0∫t v*dt.

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

Answer: a [Reason:] i= C*(dv/dt).

12. If the voltage across a capacitor is constant, then current passing through it is ________
a) 1
b) 0
c) -1
d) Infinity

Answer: b [Reason:] I= c*(dv/dt).

13. An Inductor works as a ___________ circuit for DC supply.
a) Open
b) Short
c) Polar
d) Non-polar

Answer: b [Reason:] Induced voltage across an inductor is zero if the current flowing through it is constant. I.e. Inductor works as a short circuit for DC supply.

14. The insulating medium between the two plates of capacitor is known as __________
a) Electrode
b) Capacitive medium
c) Conducting medium
d) Dielectric

Answer: d [Reason:] The conducting surfaces are called electrodes and the insulating medium is called Dielectric.

15. If the current flowing through an inductor of inductance 0.3Henry is 5.3t2+4.7t. Calculate the power.
a) 0
b) 1
c) 16.854t3+22.41t2+6.62t
d) 15.3t3+27.8t2+19

Answer: c [Reason:] P= L*i*(di/dt).

## Set 4

1. The opamp in the Inverting circuit is in __________
a) Linear region
b) Saturation
c) Cut-off region
d) Non-linear region

Answer: a [Reason:] We assume that the opamp is in linear region.

2. In an Inverting Amplifier circuit, the output voltage vo is expressed as a function of ____________
a) Input current
b) Output current
c) Source voltage
d) Source current

Answer: c [Reason:] The goal of an inverting circuit is to express output voltage vo 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

Answer: a [Reason:] The gain is also known as scaling factor and it is the ratio of Rf/Rs in case of an Inverting amplifying circuit.

4. If VCC = 12V and vs=1mV, then Rf/Rs is _____________
a) >12000
b) <12000
c) 12000
d) 1

Answer: b [Reason:] Rf/Rs ≤ │VCC/vs│.

5. In the expression vo= -Avn, A is called ______________
a) Closed loop gain
b) Closed loop fault
c) Open loop fault
d) Open loop gain

Answer: d [Reason:] A is called open loop gain.

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

Answer: b [Reason:] Inverting amplifying circuit- Rs, Rf. Non-Inverting amplifying circuit – Rs, Rf, Rg.

7. The condition for a Non-inverting amplifying circuit to operate in linear region operation _____________
a) (Rs+Rf)/Rs < │VCC/vg
b) (Rs+Rf)/Rs ≠ │VCC/vg
c) (Rs+Rf)/Rs > │VCC/vg
d) (Rs+Rf)/Rs = │VCC/vg

Answer: a [Reason:] Assume that opamp is ideal. The condition for the linear region operation in a Non-inverting amplifying circuit is (Rs+Rf)/Rs <│VCC/vg│.

8. If Rs= 3Ω, Rf= 6Ω then the relation between vo and vg in case of a Non-Inverting amplifying circuit.
a) vo= 9vg
b) vo= 6vg
c) vo= 3vg
d) vo= vg

Answer: c [Reason:] vo= ((Rs+Rf)/Rs) *vg.

9. If Rs= 5Ω, Rf= 25Ω and -2.5V ≤ vg ≤ 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

Answer: d [Reason:] vo= ((Rs+Rf)/Rs) *vg. By substituting the values, we have vo=6vg. vo=6(-2.5) = -15 vo=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

Answer: b [Reason:] Gain= Rf/Rs= 10 and vo= (-Rf/Rs)*vs. → vo= -10vs and given -12V≤ vo ≤ 12V. → -15= -10vs. So, vs= 1.5V → 15=-10vs. So, vs=-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

Answer: a [Reason:] Gain= Rf/Rs= 13 and vo= (-Rf/Rs)*vs. → vo= -13vs and given -22V≤ vo ≤ 22V. → -22= -13vs. So, vs=1.692 V → 22=-13vs. So, vs=-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

Answer: b [Reason:] The name itself indicates it is an Inverting amplifier. So, the input applied is inverted and is given as output. Suppose the input applied is sinusoidal then, the output is 13. In R1=10kΩ, Rf=100kΩ, v1=1V. A load of 25kΩ is connected to the output terminal. Calculate i1 and vo. a) 0.5mA, 10V
b) 0.1mA, 10V
c) 0.1mA, -10V
d) 0.5mA, -10V

Answer: c [Reason:] i1= v1/R1 = 1V/10kΩ = 0.1mA V0= -(Rf/R1)*v1 = -(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.

Answer: b [Reason:] Since the node is not a circuit element, any charge which enters node must leave immediately.

2. Relation between currents according to KCL is a) i1=i2=i3=i4=i5
b) i1+i4+i3=i5+i2
c) i1-i5=i2-i3-i4
d) i1+i5=i2+i3+i4

Answer: d [Reason:] According to KCL, entering currents=leaving currents.

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

Answer: c [Reason:] According to KVL, the sum of voltages around the closed path in a network is zero.

4. Calculate potential difference between x and y
a) 4.275v
b) -4.275v
c) 4.527v
d) -4.527v

Answer: b [Reason:] I1 = 3/3+5 = 3/8 = 0.375Ω I2 = 4/5 = 0.8Ω Vxy = vx – vy Vx + 5I1 + 4 – 2I2 – vy = 0 Vx – vy = 2I2 – 4 – 5I1 = -4.275Ω

5. Find R
a) 17.5 Ω
b) 17.2 Ω
c) 17.4 Ω
d) 17.8 Ω

Answer: a [Reason:] KVL: 70 – 5I – 7(I – 2) = 0 I = 7A KVL to 2nd loop: 7(I – 2) – 2R = 0 R=17.5Ω

6. Determine currents I1 , I2 and I3. 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

Answer: c [Reason:] I1 = I1 – I2 + 8 + I3 + 3 I2 – I3 = 11 -> 1 And -11 I1 – 7(I1 – I2) = 0 -18 I1 + 7 I2 = 0 -> 2 And -11 I1 – 15 I3 =0 -> 3 Solving I1 = 3.32A I2 = 8.5A I3 = -2.4A.

7. All _____________ are loops but _______________ are not meshes
a) Loops, Meshes
b) Meshes, loops
c) Branches, loops
d) Nodes, Branches

Answer: b [Reason:] A mesh cannot be divided further in loops.

8. Solve for I.
a) -0.5A
b) 0.5A
c) -0.2A
d) 0.2A

Answer: a [Reason:] Veq = 10 + 5 -20 = -5u Req = 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