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# Multiple choice question for engineering

## Set 1

1. Find the value of the currents I1, I2 and I3 flowing clockwise in the first, second and third mesh respectively.

a) 1.54A, -0.189A, -1.195A
b) 2.34A, -3.53A, -2.23A
c) 4.33A, 0.55A, 6.02A
d) -1.18A, -1.17A, -1.16A

Answer: a [Reason:] The three mesh equations are: -3I1+2I2-5=0 2I1-9I2+4I3=0 4I2-9I3+10=0 Solving the equations, we get I1= 1.54A, I2=-0.189 and I3= -1.195A.

2. Find the value of the currents I1 and I2 flowing clockwise in the first and second mesh respectively.

a) 5A, 0A
b) -5A, 0A
c) 0A, 5A
d) 0A, -5A

Answer: b [Reason:] The two mesh equations are: 2I1-3I2=10 -3I1+7I2=-15 Solving the equations simultaneously, we get: I1=-5A and I2=0A.

3. Find the value of V if the current in the 3 ohm resistor=0.

a) 3.5V
b) 6.5V
c) 7.5V
d) 8.5V

Answer: c [Reason:] Taking the mesh currents in the three meshes as I1, I2 and I3, the mesh equations are: 3I1+0I2+0V=5 -2I1-4I2+0V=0 0I1+9I2+V=0 Solving these equations simultaneously and taking the value of I2=0, we get V=7.5V.

4. Find the value of V1 if the current through the 1 ohm resistor=0A.

a) 83.33V
b) 78.89V
c) 87.87V
d) 33.33V

Answer: a [Reason:] Taking I1, I2 and I3 as the currents in the three meshes and taking I3=0 since it is the current across the 1 ohm resistor, the three mesh equations are: 15I1-5I2=V1 -5I1+10I2+0V1=0 0I1-3I2+0V1=10 Solving these equations simultaneously we get V1= 83.33V.

5. Calculate the mesh currents I1 and I2 flowing in the first and second meshes respectively.

a) 1.75A, 1.2A
b) 0.5A, 2.5A
c) 2.3A, 0.3A
d) 3.2A, 6.5A

Answer: a [Reason:] In this circuit, we have a super mesh present. The two mesh equations are: I2-I1=3 -5I1-3I2=5 Solving these equations simultaneously, we get I1=1.75A and I2= 1.2A.

6. I1 is the current flowing in the first mesh. I2 is the current flowing in the second mesh and I3 is the current flowing in the top mesh. If all three currents are flowing in the clockwise direction, find the value of I1, I2 and I3.

a) 8A, 10A, 2A
b) 10A, 8A, 2A
c) 8A, 9A, 2A
d) 3A, 6A, 2A

Answer: a [Reason:] The two meshes which contain the 3A current is a super mesh. The three mesh equations therefore are: I3=2A I2-I1=2 -2I1-I2=-26 Solving these equations simultaneously we get: I1=8A, I2=10A and I3=2A.

7. Calculate the mesh currents.

a) 7A, 6A, 6.22A
b) 2A, 1A, 0.57A
c) 3A, 4A, 5.88A
d) 6A, 7A, 8,99A

Answer: b [Reason:] The two meshes which contain the 3A source, act as a supper mesh. The mesh equations are: I1-I2=3 -11I1-4I2+14I3=-10 10I1+4I2-28I3=0 Solving these equations simultaneously, we get the three currents as 2A, 1A and 0.57A.

8. Mesh analysis employs the method of ___________
a) KVL
b) KCL
c) Both KVL and KCL
d) Neither KVL nor KCL

Answer: a [Reason:] KVL employs mesh analysis to find the different mesh currents by finding the IR products in each mesh.

9. Mesh analysis is generally used to determine _________
a) Voltage
b) Current
c) Resistance
d) Power

Answer: b [Reason:] Mesh analysis uses Kirchhoff’s Voltage Law to find all the mesh currents. Hence it is a method used to determine current.

10. Mesh analysis can be used for __________
a) Planar circuits
b) Non-planar circuits
c) Both planar and non-planar circuits
d) Neither planar nor non-planar circuits

Answer: a [Reason:] If the circuit is not planar, the meshes are not clearly defined. In planar circuits, it is easy to draw the meshes hence the meshes are clearly defined.

## Set 2

1. The phenomenon due to which there is an induced current in one coil due to current in a neighbouring coil is?
a) Electromagnetism
b) Susceptance
c) Mutual inductance

Answer: c [Reason:] When there is a current induced in a coil, due to the magnetic field caused by the current there is current induced in the neighbouring coil as well. This is known as mutual inductance.

2. If the current in one coil becomes steady, the magnetic field is?
a) Zero
b) Infinity
c) Doubles
d) Halves

Answer: a [Reason:] A magnetic field is produced when there is a changing electric field. Hence the magnetic field is zero when the current is steady.

3. If the current in one coil is steady, what happens to the mutual inductance?
a) Zero
b) Infinity
c) Doubles
d) Halves

Answer: a [Reason:] A magnetic field is produced when there is a changing electric field. Hence the magnetic field is zero when the current is steady. When the magnetic field is zero there is no current induced in the other coil, thus electric field is zero.

4. What is the SI unit of mutual inductance?
a) Ohm
b) Henry
c) Volt
d) Siemens

Answer: b [Reason:] Mutual inductance is the inductance between the two neighbouring coils. Since it is a type of inductance, its unit is that of inductance, that is, henry.

5. Which, among the following, is the correct expression for mutual inductance?
a) M=N2φ2/I2
b) M=N2φ2/I1
c) M=N1φ2/I2
d) M=N1φ1/I1

Answer: b [Reason:] Mutual inductance is the product of the number of turns in one coil and the flux linkages of that coil, divided by the current in the other coil. Hence M=N2φ2/I1 is the correct expression.

6. If the flux linkage in coil 1 is 3Wb-t and it has 500 turns and the current in coil 2 is 2A, calculate the mutual inductance.
a) 750H
b) 500H
c) 450H
d) 900H

Answer: a [Reason:] We know that mutual inductance is the product of the number of turns in one coil and the flux linkages of that coil, divided by the current in the other coil. M=3*500/2=750H.

7. The flux linkage in coil 1 is 3Wb-t and it has x turns and the current in coil 2 is 2A, calculate the value of x if the mutual inductance is 750H.
a) 300
b) 400
c) 500
d) 700

Answer: c [Reason:] We know that mutual inductance is the product of the number of turns in one coil and the flux linkages of that coil, divided by the current in the other coil. N=750*2/3= 300 turns.

8. The flux linkage in coil 1 is x Wb-t and it has 500 turns and the current in coil 2 is 2A, calculate the value of x if the mutual inductance is 750H.
a) 1Wb-t
b) 2Wb-t
c) 3Wb-t
d) 4Wb-t

Answer: c [Reason:] We know that mutual inductance is the product of the number of turns in one coil and the flux linkages of that coil, divided by the current in the other coil. φ=750*2/500= 3Wb-t.

9. The flux linkage in coil 1 is 3 Wb-t and it has 500 turns and the current in coil 2 is xA, calculate the value of x if the mutual inductance is 750H.
a) 1A
b) 2A
c) 3A
d) 4A

Answer: b [Reason:] We know that mutual inductance is the product of the number of turns in one coil and the flux linkages of that coil, divided by the current in the other coil. I=3*500/750= 2A.

10. Practical application of mutual inductance is ____________
a) DC generator
b) AC generator
c) Transformer
d) Capacitor

Answer: c [Reason:] A transformer is a device made of two or more inductors, one of which is powered by AC, inducing an AC voltage across the second inductor.

## Set 3

1. Find the value of the node voltage V.

a) 60V
b) 50V
c) 40V
d) 30V

Answer: a [Reason:] The node equation is: -2+8+V/10=0 => 6 + v/10 = 0 => v = 10*6=>60v Solving this equation, we get V=60V.

2. Calculate the node voltages V1 and V2.

a) 12V, 13V
b) 13V, 15V
c) 14V, 16V
d) 16V, 18V

Answer: c [Reason:] The nodal equations are: 2V1-V2=12 -4V1+5V2=24 Solving these equations simultaneously, we get V1=14V and V2=16V.

3. Find the node voltage V.

a) 1V
b) 2V
c) 3V
d) 4V

Answer: d [Reason:] The nodal equation is: (V-10)/2+(V-7)/3+V/1=0 Solving for V, we get V=4V.

4. Calculate the node voltages.

a) 24.32V, 4.09V, 7.04V
b) 32.34V, 7.87V, 8.78V
c) 34.34V, 8.99V, 8.67V
d) 45.44V, 6.67V, 7.77V

Answer: a [Reason:] The nodal equations, considering V1, V2 and V3 as the first, second and third node respectively, are: -8+(V1-V2)/3-3+(V1_V3)/4=0 3+V2+(V2-V3)/7+(v2-V1)/3=0 2.5+(V3-V2)/7+(V3-V1)/4+V3/5=0 Solving the equations simultaneously, we get V1=24.32V, V2=4.09V and V3=7.04V.

5. Find the value of V1 and V2.

a) 87.23V, 29.23V
b) 23.32V, 46.45V
c) 64.28V, 16.42V
d) 56.32V, 78, 87V

Answer: c [Reason:] The nodal equations are: 0.3V1-0.2V2=16 -V1+3V2=-15 Solving these equations simultaneously, we get V1=64.28V and V2=16.42V.

6. Nodal analysis is generally used to determine_______
a) Voltage
b) Current
c) Resistance
d) Power

Answer: a [Reason:] Nodal analysis uses Kirchhoff’s Current Law to find all the node voltages. Hence it is a method used to determine voltage.

7. KCL is associated with_________
a) Mesh analysis
b) Nodal analysis
c) Both mesh and nodal
d) Neither mesh nor nodal

Answer: b [Reason:] KCL employs nodal analysis to find the different node voltages by finding the value if current in each branch.

8. If there are 10 nodes in a circuit, how many equations do we get?
a) 10
b) 9
c) 8
d) 7

Answer: b [Reason:] The number of equations we get is always one less than the number of nodes in the circuit, hence for 10 nodes we get 9 equations.

9. Nodal analysis can be applied for________
a) Planar networks
b) Non-planar networks
c) Both planar and non-planar networks
d) Neither planar nor non-planar networks

Answer: c [Reason:] Nodal analysis can be applied for both planar and non-planar networks since each node, whether it is planar or non-planar, can be assigned a voltage.

10. How many nodes are taken as reference nodes in nodal analysis?
a) 1
b) 2
c) 3
d) 4

Answer: a [Reason:] In nodal analysis one node is treated as the reference node and the voltage at that point is taken as 0.

## Set 4

1. The Norton current is the_______
a) Short circuit current
b) Open circuit current
c) Both open circuit and short circuit current
d) Neither open circuit nor short circuit current

Answer: a [Reason:] Norton current is the short circuit current. It is the current through the specified load resistance. It is not the open circuit current because open circuit current is equal to zero.

2. Norton resistance is found by?
a) Shorting all voltage sources
b) Opening all current sources
c) Shorting all voltage sources and opening all current sources
d) Opening all voltage sources and shorting all current sources

Answer: c [Reason:] Current sources have infinite internal resistance hence behave like an open circuit whereas ideal voltage sources have 0 internal resistances hence behave as a short circuit.

3. Norton’s theorem is true for __________
a) Linear networks
b) Non-Linear networks
c) Both linear networks and nonlinear networks
d) Neither linear networks nor non-linear networks

Answer: a [Reason:] Norton’s theorem works for only linear circuit elements and not non-linear ones such as BJT, semiconductors etc.

4. In Norton’s theorem Isc is__________
a) Sum of two current sources
b) A single current source
c) Infinite current sources
d) 0

Answer: b [Reason:] Norton’s theorem states that a combination of voltage sources, current sources and resistors is equivalent to a single current source Ith and a single parallel resistor R.

5. Isc is found across the ____________ terminals of the network.
a) Input
b) Output
c) Neither input nor output
d) Either input or output

Answer: b [Reason:] According to Norton’s theorem, Isc is found through the output terminals of a network and not the input terminals.

6. Can we use Norton’s theorem on a circuit containing a BJT?
a) Yes
b) No
c) Depends on the BJT
d) Insufficient data provided

Answer: b [Reason:] We can use Norton’s theorem only for linear networks. BJT is a non-linear network hence we cannot apply Norton’s theorem for it.

7. Calculate the Norton’s equivalent voltage for the following circuit if 5 ohm is the load resistance.

a) 10 ohm
b) 11 ohm
c) 12 ohm
d) 13 ohm

Answer: c [Reason:] Shorting all voltage sources and opening all current sources we have: RN=(3||6)+10= 12 ohm.

8. Calculate the short circuit current is the 5 ohm resistor is the load resistance.

a) 0.72A
b) 0.32A
c) 0.83A
d) 0.67A

Answer: a [Reason:] Since the 5 ohm is the load resistance, we short it and find the resistance through the short. If we apply source transformation between the 6 ohm resistor and the 1A source, we get a 6V source in series with a 6 ohm resistor. Now we have two meshes. Let us consider I1 flowing in the first mesh and I2 flowing in the second mesh. The mesh equations are: 9I1-6I2=4 -6I1+16I2=6 On solving these equations simultaneously, we get I2=0.72A, which is the short circuit current.

9. Find the current in the 5 ohm resistance using Norton’s theorem.

a) 1A
b) 1.5A
c) 0.25A
d) 0.5A

Answer: d [Reason:] From Q8 and Q7 we have found the values of the Isc and RN respectively. Connecting the current source in parallel to RN which is in turn connected in parallel to the load resistance=5ohm, we get the Norton’s equivalent circuit. Using current divider: I=0.72*12/(12+5)= 0.5 A.

10. Which of the following is also known as the dual of Norton’s theorem?
a) Thevenin’s theorem
b) Superposition theorem
c) Maximum power transfer theorem
d) Millman’s theorem

Answer: a [Reason:] Thevenin’s theorem is also known as the dual of Norton’s theorem because in Norton’s theorem we find short circuit current which is the dual of open circuit voltage-what we find in Thevenin’s theorem.

## Set 5

1. Ohm’s law for magnetic circuits is _________
a) F=ϕS
b) F=ϕ/S
c) F=ϕ2S
d) F=ϕ/S2

Answer: a [Reason:] Ohm’s law for magnetic circuits states that the MMF is directly proportional to the magnetic flux where reluctance is the constant of proportionality.

2. What happens to the MMF when the magnetic flux decreases?
a) Increases
b) Decreases
c) Remains constant
d) Becomes zero

Answer: b [Reason:] Ohm’s law for magnetic circuit’s states that the MMF is directly proportional to the magnetic flux hence as the magnetic flux decreases, the MMF also decreases.

3. Calculate the MMF when the magnetic flux is 5Wb and the reluctance is 3A/Wb.
a) 10At
b) 10N
c) 15N
d) 15At

Answer: d [Reason:] We know that: F=ϕS Substituting the given values from the question, we get MMF= 15At.

4. A ring having a cross-sectional area of 500 mm2, a circumference of 400 mm and ϕ=800microWb has a coil of 200 turns wound around it. Calculate the flux density of the ring.
a) 1.6T
b) 2.6T
c) 3.6T
d) 4.6T

Answer: a [Reason:] From the given question: Flux density= 800*10-6/500*106=1.6 Wb/m2.

5. A ring having a cross-sectional area of 500 mm2, a circumference of and ϕ=800microWb 400 mm has a coil of 200 turns wound around it. Calculate the reluctance.
a) 1.68 * 10-4A/Wb
b) 1.68 * 104 A/Wb
c) 1.68 * 106 A/Wb
d) 1.68 * 10-6A/Wb

Answer: c [Reason:] From the given question: Flux density= 800*10-6/500*106=1.6 Wb/m2. Reluctance= 0.4/(380*4*pi*10-7*10-4*5)=1.68 * 106 A/Wb.

6. A ring having a cross-sectional area of 500 mm2, a circumference of 400 mm and ϕ=800microWb has a coil of 200 turns wound around it. Calculate the magnetomotive force.
a) 1442At
b) 1342At
c) 1432At
d) 1344At

Answer: b [Reason:] We know that: F=ϕS Substituting the given values from the question, we get F= 1342At.

7. A ring having a cross-sectional area of 500 mm2 , a circumference of 400 mm and ϕ=800microWb has a coil of 200 turns wound around it. Calculate the magnetising current.
a) 6.7A
b) 7.7A
c) 7.6
d) 6.1A

Answer: a [Reason:] We know that: F=ϕS Substituting the given values from the question, we get F= 1342At. The magnetic current is: I=F/N Substituting the values from the question, we get I=6.7A.

8. Can we apply Kirchhoff’s was to magnetic circuits?
a) Yes
b) No
c) Depends on the circuit
d) Insufficient information provided

Answer: a [Reason:] Magnetic circuits have an equivalent to the potential difference of electric circuits. This is the magnetic potential difference which allows us to apply Kirchhoff’s laws to magnetic circuit analysis.

9. What is MMF?
a) Magnetic Machine Force
b) Magnetomotive Force
c) Magnetic Motion Force
d) Magnetomotion Force

Answer: b [Reason:] MMF stands for magnetomotive force. It is the sum of the magnetizing forces along a circuit.

10. The equivalent of the current I in magnetic ohm’s law is?
a) Flux
b) Reluctance
c) MMF
d) Resistance