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Up To Date Electrical Engineering MCQs – Inductance in a DC Circuit MCQs ( Electrical Engineering ) MCQs 

Up To Date Electrical Engineering MCQs – Inductance in a DC Circuit MCQs ( Electrical Engineering ) MCQs 

Latest Electrical Engineering MCQs

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Up To Date Electrical Engineering MCQs – Inductance in a DC Circuit MCQs ( Electrical Engineering ) MCQs 

The most occurred mcqs of Inductance in a DC Circuit MCQs ( Electrical Engineering ) in past papers. Past papers of Inductance in a DC Circuit MCQs ( Electrical Engineering ) Mcqs. Past papers of Inductance in a DC Circuit MCQs ( Electrical Engineering ) Mcqs . Mcqs are the necessary part of any competitive / job related exams. The Mcqs having specific numbers in any written test. It is therefore everyone have to learn / remember the related Inductance in a DC Circuit MCQs ( Electrical Engineering ) Mcqs. The Important series of Inductance in a DC Circuit MCQs ( Electrical Engineering ) Mcqs are given below:

Inductive and Non-Inductive Circuits

1. In case of Inductive circuit, Frequency is ______________ to the inductance.
a) Directly proportional
b) Inversely proportional
c) Unrelated
d) Much greater than
Answer: b
Explanation: The formula for frequency in an inductive circuit is:
XL=2*π*f*L.
Therefore: f is inversely proportional to L.


2. In case of Inductive circuit, Frequency is ______________ to the current.
a) Directly proportional
b) Inversely proportional
c) Unrelated
d) Much greater than
Answer: b
Explanation: The formula for frequency in an inductive circuit is:
XL=2*π*f*L => i=V/(2*π*f*L)
Therefore: f is inversely proportional to i.


3. In an inductive circuit, when the XL value increases, the circuit power factor?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero
Answer: b
Explanation: tan ϕ = XL/R and Power factor=cos ϕ
As XL increases, tan ϕ increases, ϕ increases, cos ϕ decreases and hence power factor decreases.


4. If the current and voltage are 90 degrees out of phase, the power factor will be?
a) 0
b) Infinity
c) 1
d) Insufficient information provided
Answer: a
Explanation: The power factor is the cosine of the angle between the voltage and the current. If the angle between the voltage and the current is 90, then cos90=0. Hence, the power factor is zero.


5. In a pure inductive circuit, the power factor is __________
a) Maximum
b) Minimum
c) 0
d) Infinity
Answer: c
Explanation: In a pure inductive circuit, the current is lagging by 90 degrees from the voltage. The power factor is the cosine of the angle between the voltage and the current. If the angle between the voltage and current is 90, then cos90=0. Hence, the power factor is zero.


6. If the power factor is 1/10 and the value of impedance is 20 ohm, calculate the resistance in the circuit.
a) 1 ohm
b) 2 ohm
c) 3 ohm
d) 4 ohm
Answer: b
Explanation: We know that:
cos(ϕ)=R/Z
R=Z cos(ϕ) = 20/10 = 2 ohm.


7. If the resistance in a circuit is 2 ohm and the impedance is 20 ohm, calculate the power factor.
a) 1/10
b) 1/20
c) 1/30
d) 1/40
Answer: a
Explanation: We know that:
cos(ϕ)=R/Z = 2/20 = 1/10 ohm.


8. If tan ϕ = 10 and the resistance is 2 ohm, calculate the inductive reactance.
a) 10 ohm
b) 20 ohm
c) 30 ohm
d) 40 ohm
Answer: b
Explanation: We know that:
tan(ϕ)=XL/R
From the given question, we find that the inductive reactance is 20 ohm.


9. What is the unit for inductive reactance?
a) Henry
b) Ohm
c) Farad
d) Volts
Answer: b
Explanation: Inductive reactance is nothing but the impedance. Impedance is the AC equivalent of resistance, hence the unit for inductive reactance is ohm.


10. An induced emf is said to be ________
a) Inductive
b) Capacitive
c) Resistive
d) Cannot be determined
Answer: a
Explanation: Any circuit in which a change of current is accompanied by a change of flux, and therefore by an induced emf, is said to be inductive.

Inductance in Terms of Flux Linkages Per Ampere

1. Among the following, which is the right formula for inductance?
a) L=emf*t/I
b) L=emf/t*I
c) L=emf*I/t
d) L=emf*t*I
Answer: a
Explanation: The average emf induced is proportional to the current per unit time, the constant of proportionality being L. Hence emf=LI/t. Making L the subject of the formula, we get L=emf*t/I.


2. Among the following, which is the right formula for inductance of N turns?
a) L=et/Ni
b) L=N*i *e*t
c) L=Ni/et
d) L=N/iet
Answer: a
Explanation: We know that:
emf=NLi/t
Inductance = L = et/N.


3. For a coil having a magnetic circuit of constant reluctance, the flux is ___________ to the current.
a) Directly proportional
b) Inversely proportional
c) Not related
d) Very large compared to
Answer: a
Explanation: For a coil having a magnetic circuit of constant reluctance, the flux is directly proportional to the current.


4. For a coil having a magnetic circuit of constant reluctance, if the flux increases, what happens to the current?
a) Increases
b) Decreases
c) Remains constant
d) Becomes zero
Answer: a
Explanation: For a coil having a magnetic circuit of constant reluctance, the flux is directly proportional to the current. Hence as the flux increases, the current also increases.


5. The unit for inductance is ___________
a) Ohm
b) Henry
c) A/m
d) A/s
Answer: b
Explanation: The unit of induction is named after a famous scientist Joseph Henry who independently discovered electromagnetic induction.


6. If either the inductance or the rate of change of current is doubled, the induced e.m.f?
a) Remains constant
b) Becomes zero
c) Doubles
d) Becomes half
Answer: c
Explanation: If either the inductance or the rate of change of current is doubled, the induced e.m.f. becomes double because of emf=LI/t.


7. If the current changes from 5A to 3A in 2 seconds and the inductance is 10H, calculate the emf.
a) 5V
b) 10V
c) 15V
d) 20V
Answer: b
Explanation: We know that:
emf=L(i2-i1)/t
Substituting the values from the question, we get emf=10V.


8. If the current changes from 5A to 3A in x sec and inductance is 10H. The emf is 10V, calculate the value of x.
a) 2s
b) 3s
c) 4s
d) 5s
Answer: a
Explanation: We know that:
emf=L(i2-i1)/t
Substituting the values from the question, we get x=2s.


9. If the current changes from 3A to 5A in 2s and the emf is 10V. Calculate the inductance.
a) 10H
b) 20H
c) 30H
d) 40H
Answer: a
Explanation: If the current changes from 5A to 3A in 2s and the emf is 10V. Calculate the inductance.

Factors Determining the Inductance of a Coil

1. As the number of turns in the coil increases, what happens to the inductance of the coil?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero
Answer: a
Explanation: Inductance is directly proportional to the square of the number of turns in the coil, hence as the number of turns increases, inductance also increases.


2. What happens to the inductance when the magnetic field strength decreases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero
Answer: b
Explanation: Inductance is directly proportional to the magnetic field strength in the coil, hence as the magnetic field strength decreases, inductance decreases.


3. What happens to the inductance when the current in the coil becomes double its original value?
a) Becomes half
b) Becomes four times
c) Becomes double
d) Remains same
Answer: d
Explanation: ϕ is directly proportional to i.
Φ=Li where L is constant of proportionality
So, when current get double ϕ also becomes double keeping L same.


4. When the coil is wrapped around a ferromagnetic core, why is it difficult to determine the inductance?
a) The variation of flux is no longer proportional to the variation of current
b) Current does not exist in the coil
c) Flux does not exist in the coil
d) The value of current is too large to measure
Answer: a
Explanation: When a coil is wrapped around a ferromagnetic core, it is difficult to determine the inductance because the variation of flux is no longer proportional to the variation of the current.


5. What happens to the inductance as the area of the cross section of the coil increases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero
Answer: a
Explanation: L=µ0*N2*A/l, hence as the area of cross section A increases, the inductance also increases.


6. What happens to the inductance as the length of the magnetic circuit increases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero
Answer: b
Explanation: L=µ0*N2*A/l, hence as the length of the magnetic circuit l increases, the inductance decreases.


7. If the current changes from 20A to 10A in 5 seconds and the value of inductance is 1H, calculate the emf induced.
a) 8V
b) 6V
c) 4V
d) 2V
Answer: d
Explanation: We know that:
emf=L(i2-i1)/t
Substituting the values from the question we get emf=2V.

Ferromagnetic Cored Inductor in a DC Circuit

1. When a ferromagnetic core is inserted into an inductor, what happens to the flux linkage?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero
Answer: a
Explanation: When a ferromagnetic core is introduced into an inductor, its flux increases because the number of magnetic field lines increases due to the introduction of magnetic field within the coil.


2. What happens to the current when a ferromagnetic material is introduced within an inductor?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero
Answer: c
Explanation: When a ferromagnetic is introduced within an inductor, the current remains fairly constant. This is because the current does not depend on the magnetic field.


3. What is the relation between the flux and the magnetizing current when a ferromagnetic core is introduced within the inductor?
a) Directly proportional
b) Inversely proportional
c) Not proportional
d) Current is double of flux
Answer: c
Explanation: When a ferromagnetic core is introduced within an inductor the flux changes rapidly, whereas the current changes at the same pace. Hence the two are not proportional.


4. What happens to the effective inductance when a ferromagnetic core is introduced?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero
Answer: a
Explanation: The effective inductance increases when a ferromagnetic core is introduced within an inductor because of the flux increases. Inductance varies directly with the flux hence it increases.

 

Capacitance And Capacitors MCQs




5. A laminated steel ring is wound with 200 turns. When the magnetizing current varies between 5 and 7 A, the magnetic flux varies between 760 and 800 Wb. Calculate the inductance of the coil.
a) 40H
b) 4 H
c) 4000H
d) 0.004 H
Answer: c
Explanation: From the formula of incremental inductance, we know that:
L=(Change in flux/Change in current)*Number of turns
Substituting the values from the given question, we get L = 4000 H.


6. Calculate the number of turns in an inductor with a ferromagnetic core when the inductance is 4000 H, the current changes from 5A to 7A and the flux changes from 760 to 800 Wb.
a) 100
b) 200
c) 300
d) 400
Answer: b
Explanation: From the formula of incremental inductance, we know that:
L=(Change in flux/Change in current)*Number of turns
Substituting the values from the given question, we get N=200.


7. Calculate the change in current in an inductor having inductance 4000H, number of turns is 200 and the flux changes from 760 to 800 Wb.
a) 2A
b) 4A
c) 6A
d) 8A
Answer: a
Explanation: From the formula of incremental inductance, we know that:
L=(Change in flux/Change in current)*Number of turns
Substituting the values from the given question, we get a change in current = 2A.


8. Calculate the initial current in an inductor having inductance 4000 H, number of turns is 200 and the flux changes from 760 to 800 Wb. Current changes to 7A.
a) 10A
b) 2A
c) 3A
d) 5A
Answer: d
Explanation: From the formula of incremental inductance, we know that:
L=(Change in flux/Change in current)*Number of turns
Substituting the values from the given question, we get change in current = 2A.
Change in current = final current- initial current.
2=7-initial current.
Initial current = 5A.


9. Calculate the change in flux of an inductor having inductance 4000 H, number of turns is 200 and the current changes from 5A to 7A.
a) 20 Wb
b) 40 Wb
c) 60 Wb
d) 80 Wb
Answer: b
Explanation: From the formula of incremental inductance, we know that:
L=(Change in flux/Change in current)*Number of turns
Substituting the values from the given question, we get change in flux = 40 Wb.


10. Calculate the final flux in an inductor having inductance 4000 H, number of turns is 200 and the current changes from 5A to 7A. The initial flux is 760 Wb.
a) 200 Wb
b) 400 Wb
c) 600 Wb
d) 800 Wb
Answer: d
Explanation: From the formula of incremental inductance, we know that:
L=(Change in flux/Change in current)*Number of turns
Substituting the values from the given question, we get a change in flux = 40 Wb.
Change in flux = final flux- initial flux.
Thus final flux = 800 Wb.

Growth in an Inductive Circuit

1. In a pure inductive circuit, the power factor is?
a) Maximum
b) Minimum
c) 0
d) Infinity
Answer: c
Explanation: In a pure inductive circuit, the current is lagging by 90 degrees from the voltage. The power factor is the cosine of the angle between the voltage and the current. If the angle between the voltage and the current is 90, then cos90=0. Hence, the power factor is zero.


2. Among the following, which is the right formula for growth in an inductive circuit?
a) VL=V(1-e-tR/L)
b) VL=(e-tR/L)
c) VL=(1-e-tR/L)
d) VL=V(e-tR/L)
Answer: d
Explanation: The correct formula for growth in an inductive circuit is VL=V(e-tR/L). As the time increases, voltage decreases.


3. The charging time constant of a circuit consisting of an inductor is the time taken for the voltage in the inductor to become __________ % of the initial voltage.
a) 33
b) 63
c) 37
d) 36
Answer: c
Explanation: We know that: V=V0(1-e-tR/L).
When time constant=t, we have: V=V0(1-e-1) = 0.37*V0.
Hence the time constant is the time taken for the charge in an inductive circuit to become 0.37 times its initial charge.


4. What is the time constant of an inductive circuit?
a) LR
b) R/L
c) 1/LR
d) L/R
Answer: d
Explanation: The time constant in an inductive circuit is the time taken for the voltage across the inductor to become 63 percent of its initial value. It is given by Time constant = L/R.


5. Calculate the time constant of an inductive circuit having resistance 5 ohm and inductance 10H.
a) 2s
b) 4s
c) 5s
d)10s
Answer: a
Explanation: We know that: Time constant = L/R
Substituting the values from the given question, we get a time constant = 2s.


6. Calculate the resistance in an inductive circuit whose time constant is 2s and the inductance is 10H.
a) 7ohm
b) 10ohm
c) 2ohm
d) 5ohm
Answer: d
Explanation: We know that: Time constant = L/R
Substituting the values from the given question, we get R=5ohm.


7. Calculate the inductance in an inductive circuit whose time constant is 2s and the resistance is 5 ohm.
a) 10H
b) 20H
c) 5H
d) 15H
Answer: a
Explanation: We know that: Time constant = L/R
Substituting the values from the given question, we get L=10H.


8. The charging time constant of a circuit consisting of an inductor is the time taken for the current in the inductor to become __________% of the initial current.
a) 33
b) 63
c) 37
d) 36
Answer: b
Explanation: We know that: i=i0(1-e-tR/L).
When t=L/R, we have: i=i0(1-e-1) = 0.63*i0.
Hence the time constant is the time taken for the current in an inductive circuit to become 0.63 times its initial current.

Analysis of Growth & Decay

1. What is the total applied voltage in an inductive circuit?
a) V=Ri+Ldi/dt
b) V=Ri+di/dt
c) V=i+Ldi/dt
d) V=R+Ldi/dt
Answer: a
Explanation: The total voltage in an inductive circuit is the sum of the voltage due to the resistor which is Ri and the voltage due to the inductor which is Ldi/dt. Hence V=Ri+Ldi/dt.


2. What is Helmholtz equation?
a) i=I(eRt/L)
b) i=I(1-e-Rt/L)
c) i=I(1+e-Rt/L)
d) i=I(e-Rt/L)
Answer: b
Explanation: Helmholtz equation is an equation which gives the formula for the growth in an inductive circuit. Hence the Helmholtz formula is: i=I(1-e-Rt/L).


3. A coil has a resistance of 4 ohms and an inductance of 2H. It is connected to a 20V dc supply. Calculate the initial value of the current in the circuit.
a) 5A
b) 10A
c) 0 A
d) 20A
Answer: c
Explanation: Initially, inductor behave as open circuit for dc current so, i=0.


4. A coil has a resistance of 4 ohms and an inductance of 2H. It is connected to a 20V dc supply. Calculate the final value of the current in the circuit.
a) 5A
b) 10A
c) 15A
d) 20A
Answer: a
Explanation: The final value of the current in the circuit is:
I=V/R = 5A.


5. A coil has a resistance of 4 ohms and an inductance of 2H. It is connected to a 20V dc supply. Calculate the value of current 1s after the switch is closed.
a) 5.44A
b) 4.32A
c) 6.56A
d) 2.34A
Answer: b
Explanation: We know that:
i=I(1-eRt/L)
I=V/R=5A
Substituting the remaining values from the given question, we get i=4.32A.


6. A coil has a resistance of 4 ohms and an inductance of 2H. It is connected to a 20V dc supply. Calculate the value of voltage 1s after the switch is closed.
a) 5.4V
b) 10.8V
c) 0 V
d) 2.7V
Answer: d
Explanation: V=V0e-Rt/L
V=20e-2=2.7V.


7. Among the following, which is the right formula for decay in an inductive circuit?
a) i=I(1-e-t/time constant)
b) i=I(1-et /time constant)
c) i=(1-e-t /time constant)
d) i=I(e-t /time constant)
Answer: d
Explanation: The correct formula for decay in an inductive circuit is i=I(e-t /time constant). As the time increases, the current in the inductor decreases, the voltage also increases.


8. The discharging time constant of a circuit consisting of an inductor is the time taken for the voltage in the inductor to become __________ % of the initial voltage.
a) 33
b) 63
c) 37
d) 36
Answer: c
Explanation: We know that: V=V0(e-tR/L).
When t=L/R, we have: V=V0(e-1) = 0.37*Vsub>0.
Hence the time constant is the time taken for the voltage in an inductive circuit to become 0.37 times its initial voltage.


9. A coil has a resistance of 4 ohms and an inductance of 2H. It is connected to a 20V dc supply. Calculate the initial value of the voltage across the inductor.
a) 5V
b) 10V
c) 0 V
d) 20V
Answer: d
Explanation: Initially, inductor behave as open circuit for dc current so, V = V0 = 20V i.e. same as voltage source.


10. A coil has a resistance of 4 ohms and an inductance of 2H. It is connected to a 20V dc supply. Calculate the final value of the voltage across the inductor.
a) 5V
b) 10V
c) 0 V
d) 20V
Answer: c
Explanation: At steady state, inductor behaves as a short circuit for dc current so, V=0

Transients in LR Networks

1. An RL network is one which consists of ____________
a) Resistor and capacitor in parallel
b) Resistor and capacitor in series
c) Resistor and inductor in parallel
d) Resistor and inductor in series
Answer: d
Explanation: An R-L network is a network which consists of a resistor which is connected in series to an inductor.


2. If the switch is opened at t=0, what is the current in the circuit?
basic-electrical-engineering-questions-answers-transients-lr-networks-q4
a) 0A
b) 1A
c) 2A
d) 3A
Answer: c
Explanation: Initially when switch was closed,current in the inductor was 60/30=2A.
Current in inductor doesn’t change suddenly so when switch is opened, current in inductor remains same i.e. 2A.


3. In an RL series circuit, when the switch is closed and the circuit is complete, what is the response?
a) Response does not vary with time
b) Decays with time
c) Increases with time
d) First increases, then decrease
Answer: b
Explanation: In an RL series circuit, the response decays with time because according to the equation, there is an exponential decrease in the response.


4. If the switch is closed at t=0, what is the current in the circuit?
basic-electrical-engineering-questions-answers-transients-lr-networks-q4
a) 0A
b) 10A
c) 20A
d) 30A
Answer: a
Explanation: Initially, when the switch is open, the current in the circuit is 0. As soon as the switch is closed at t=0+, the inductor acts as an open circuit, hence the current in the circuit is zero. Since the current in the circuit is zero, there is no voltage drop across the resistor and hence voltage across the inductor is equal to the supply voltage, i.e. 60V.


5. What is the voltage across the inductor at t=0?
basic-electrical-engineering-questions-answers-transients-lr-networks-q4
a) 0V
b) 20V
c) 60V
d) 58V
Answer: c
Explanation: Initially, when the switch is open, the current in the circuit is 0. As soon as the switch is closes at t=0+, the inductor acts as an open circuit, hence the current in the circuit is zero. Since the current in the circuit is zero, there is no voltage drop across the resistor and the voltage across the inductor is equal to the supply voltage, which is equal to 60V.


6. What is the expression for current in the given circuit?
basic-electrical-engineering-questions-answers-transients-lr-networks-q4
a) i=2(e-2t)A
b) i=2(1-e-2t)A
c) i=2(e2t)A
d) i=2(1+e-2t)A
Answer: b
Explanation: Applying KVL in above circuit, we get
60-30i-15di/dt =0
i=2(1-e-2t)A


7. What is the expression for voltage in the given circuit?
basic-electrical-engineering-questions-answers-transients-lr-networks-q4
a) V=60e-0.5t
b) V=30e-0.5t
c) V=60e-2t
d) V=30e-2t
Answer: c
Explanation: Applying KVL in above circuit, we get
60-30i-15di/dt = 0
i=2(1-e-2t)A
di/dt = 4e-2t
V=Ldi/dt=15*4e--2t=60e-2t.


8. At steady state, the current in the inductor is?
a) Maximum
b) Minimum
c) Zero
d) Infinity
Answer: a
Explanation: At steady state maximum current flows in the inductor because it acts as an open circuit.


9. Initially, when the switch in a series RL circuit is closed, the inductor acts as?
a) Open circuit
b) Short circuit
c) Resistor
d) Capacitor
Answer: a
Explanation: Before switch is closed, current in inductor is zero. When the switch in a series RL circuit is closed, current in the inductor remains zero since current in inductor doesn’t change suddenly. So, the inductor acts as an open circuit.


10. Initially, when the switch in a series RL circuit is closed, the current in the inductor is?
a) Maximum
b) Minimum
c) Zero
d) Infinity
Answer: c
Explanation: Initially, when the switch in a series RL circuit is closed, the inductor acts as an open circuit. Current in an open circuit is zero, hence the inductor current is zero.

Energy Stored in an Inductor

1. If the current in a coil having a constant inductance of L henrys grows at a uniform rate, what is the value of the average current?
a) I
b) I/2
c) I/4
d) 0
Answer: b
Explanation: The average current is the average of the current which flows in the inductor. Hence it is I/2.


2. What is the power in the magnetic field if the current in a coil has a constant inductance of L henrys grows at a uniform rate?
a) LI/2t
b) LI2/2t
c) L/2It
d) L/2I2t
Answer: b
Explanation: EMF induced in the coil is Ldi/dt.
Energy=ei*dt=Lidi/dt*dt=Li2/2.
Power=Energy/Time = Li2/2t.


3. What is the energy stored in the magnetic field if the current in a coil has a constant inductance of L henrys grows at a uniform rate?
a) LI/2
b) LI2/2
c) L/2I
d) L/2I2
Answer: b
Explanation: EMF induced in the coil is Ldi/dt.
Energy=ei*dt=Lidi/dt*dt=Li2/2.


4. Find the average current in an inductor if the total current in the inductor is 26A.
a) 10A
b) 26A
c) 13A
d) 5A
Answer: c
Explanation: Average current = I/2.
Substituting the value of I from the equation, average current = 13A.

 

Electromagnetism MCQs




5. Calculate the power in an inductive circuit if the inductance is 10H, the current flowing in the inductor is 2A in 4s.
a) 50W
b) 4W
c) 5W
d) 10W
Answer: c
Explanation: The expression for power in an inductive circuit is:
P = LI2/2t
Substituting the values from the given question, we get P=5W.


6. Calculate the value of stored energy in an inductor if the value of inductance is 20H and 4A of current flows through it.
a) 220J
b) 150J
c) 190J
d) 160J
Answer: d
Explanation: The expression for energy in an inductor is
E = LI2/2. Substituting the values from the given question, we get E = 160J.


7. Calculate the emf induced in an inductor if the inductance is 10H and the current is 2A in 4s.
a) 2.5V
b) 1.5V
c) 3.5V
d) 5V
Answer: d
Explanation: The expression for emf in an inductive circuit is:
emf = LdI/dt
Substituting the values from the given question, we get emf = 5V.


8. Calculate the value of emf in an inductor if the value of inductance is 15H and an average current of 5A flows through it in 10s.
a) 15V
b) 7.5V
c) 10V
d) 5.5V
Answer: b
Explanation: The expression for emf in an inductive circuit is:
emf = LdI/dt
Substituting the values from the given question, we get emf = 7.5V.


9. Calculate the current in an inductor if the energy stored is 160J and the inductance is 20H.
a) 1A
b) 2A
c) 3A
d) 4A
Answer: d
Explanation: The expression for energy in an inductor is:
E = LI2/2
Substituting the values from the given question, we get I=4A.


10. Find the time taken for the current in an inductor to change to 2A from 0A if the power in the inductor is 5W. The value of inductance is 10H.
a) 1s
b) 2s
c) 3s
d) 4s
Answer: d
Explanation: The expression for power in an inductive circuit is:
P = LI2/2t
Substituting the values from the given question, we get t=4s.

Mutual Inductance

1. The phenomenon due to which there is an induced current in one coil due to the current in a neighbouring coil is?
a) Electromagnetism
b) Susceptance
c) Mutual inductance
d) Steady current
Answer: c
Explanation: When there is a current 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 current in neighbouring coil is?
a) Zero
b) Infinity
c) Doubles
d) Halves
Answer: a
Explanation: A current is induced when there is changing magnetic flux. Hence the induced current in neighbouring coil 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
Explanation: A current is induced when there is changing magnetic flux. The induced current in neighbouring coil is zero when the current is steady. So, mutual inductance is zero.


4. What is the SI unit of mutual inductance?
a) Ohm
b) Henry
c) Volt
d) Siemens
Answer: b
Explanation: 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
Explanation: 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 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
Explanation: 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 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
Explanation: 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 = 500 turns.


8. The flux linkage in coil 1 is x 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
b) 2Wb
c) 3Wb
d) 4Wb
Answer: c
Explanation: 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.


9. The flux linkage in coil 1 is 3 Wb 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
Explanation: 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
Explanation: 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.


11. The types of inductors are ____________
a) Fixed and variable
b) Only fixed
c) Only variable
d) Neither fixed nor variable
Answer: d
Explanation: e=Ldi/dt
Leq=2+10+12+20=44H
di/dt=44/44 = 1 A/s.
Voltage across 20H inductor = 20*di/dt = 20*1=20V.

Coupling Coefficient

1. What is the coupling coefficient when all the flux of coil 1 links with coil 2?
a) 0
b) 100
c) 1
d) Insufficient information provided
Answer: c
Explanation: When all the flux of coil 1 links with coil 2 it is known as an ideal coupling where the coupling coefficient is 1.


2. What is the coupling coefficient when there is ideal coupling?
a) 0
b) 100
c) 1
d) Insufficient information provided
Answer: c
Explanation: When all the flux of coil 1 links with coil 2 it is known as an ideal coupling where the coupling coefficient is 1.


3. Can the coupling coefficient practically ever be equal to 1?
a) Yes
b) No
c) Depends on the current in coil 1
d) Depends on the current in coil 2
Answer: b
Explanation: All the flux of coil 1 can never link with coil 2. Loss occurs practically due to which coupling coefficient cannot be equal to 1.


4. Mutual inductance between two coupled coils depend on?
a) Amount of flux linkage
b) Rate of change of flux linkage
c) Rate of change of current
d) Flux density
Answer: b
Explanation: Faraday’s law of induction states that the magnitude of the induced EMF is the product of the number of turns of the coil and the rate of change of flux linkage in it. Hence, the mutual inductance depends on the rate of change of flux linkage.


5. Which, among the following, is the correct formula to find the coupling coefficient?
a) k=M/sqrt(L1L2)
b) k=M/sqrt(L12)
c) k=M/sqrt(L22)
d) k=M/(L1L2)
Answer: a
Explanation: The correct formula for the coupling coefficient is k=M/sqrt(L1L2). Where L1 and L2 are the inductance values of the first and second coil respectively and M is the mutual inductance.


6. What happens to the coupling coefficient when the flux linkage of coil 1 and coil 2 increases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero
Answer: a
Explanation: When the flux linkage of coil 1 and coil 2 increases, its mutual inductance increases. The coupling coefficient is directly proportional to the mutual inductance hence as mutual inductance increases, the coupling coefficient increases.


7. What is the SI unit of coupling coefficient?
a) H
b) H-1
c) No unit
d) H2
Answer: c
Explanation: The expression to find mutual inductance is k=M/sqrt(L1L2)= H/sqrt(H*H)= 1. Therefore it does not have any unit.


8. Find the coupling coefficient if the Mutual inductance is 20H, the inductance of coil 1 is 2H and the inductance of coil 2 is 8H.
a) 5
b) 20
c) 2
d) 8
Answer: a
Explanation: we know that:
k=M/sqrt(L1L2)
Substituting the values from the question, we get k=5.


9. Find the value of x if the Mutual inductance is x H, the inductance of coil 1 is 2H and the inductance of coil 2 is 8H. The coupling coefficient is 5.
a) 10H
b) 20H
c) 16H
d) 15H
Answer: b
Explanation: we know that:
k=M/sqrt(L1L2)
Substituting the values from the question, we get M=20H.


10. Find the value of x if the Mutual inductance is 20H, the inductance of coil 1 is xH and the inductance of coil 2 is 8H. The coupling coefficient is 5.
a) 2H
b) 4H
c) 6H
d) 8H
Answer: a
Explanation: we know that:
k=M/sqrt(L1L2)
Substituting the values from the question, we get L1=2H.

Coils Connected in Series

1. What is the equivalent inductance when inductors are connected in series?
a) Sum of all the individual inductances
b) Product of all the individual inductances
c) Sum of the reciprocal of all the individual inductances
d) Product of the reciprocal of all the individual inductances
Answer: a
Explanation: When inductances are connected in series, the equivalent inductance is equal to the sum of all the individual inductance values.


2. When inductances are connected in series, the equivalent inductance is ____________ the largest individual inductance.
a) Greater than
b) Less than
c) Equal to
d) Not related to
Answer: a
Explanation: When inductances are connected in series, the equivalent inductance is equal to the sum of all the individual inductance values. Hence the equivalent inductance is greater than the largest individual inductance.


3. Three inductors having inductance values 3H, 4H and 5H are connected in series, calculate the equivalent inductance.
a) 10H
b) 12H
c) 3H
d) 5H
Answer: b
Explanation: When inductances are connected in series, the equivalent inductance is equal to the sum of all the individual inductance values.
Hence Leq= L1+L2+L3= 12H.


4. Calculate the equivalent inductance between A and B.
basic-electrical-engineering-questions-answers-coils-connected-series-q4
a) 30H
b) 54H
c) 44H
d) 60H
Answer: c
Explanation: The 4 inductors are connected in series, hence their equivalent inductance is:
Leq=L1+L2+L3+L4=44H.


5. When inductors are connected in series, the voltage across each inductor is _________
a) Equal
b) Different
c) Zero
d) Infinity
Answer: b
Explanation: In a series circuit, the current across all elements remain the same and the total voltage of the circuit is the sum of the voltages across all the elements. The voltage across each inductor in series is different.


6. In a series circuit, which of the parameters remain constant across all circuit elements such as resistor, capacitor, inductor etc?
a) Voltage
b) Current
c) Both voltage and current
d) Neither voltage nor current
Answer: b
Explanation: In a series circuit, the current across all elements remain the same and the total voltage of the circuit is the sum of the voltages across all the elements.


7. Find voltage across 2H inductor.
basic-electrical-engineering-questions-answers-coils-connected-series-q7
a) 2V
b) 10V
c) 12V
d) 20V
Answer: a
Explanation: e=Ldi/dt
Leq=2+10+12+20=44H
di/dt=44/44 = 1 A/s.
Voltage across 2H inductor = 2*di/dt = 2*1=2V.


8. Find voltage across 10H inductor.
basic-electrical-engineering-questions-answers-coils-connected-series-q7
a) 2V
b) 10V
c) 12V
d) 20V
Answer: b
Explanation: e=Ldi/dt
Leq=2+10+12+20=44H
di/dt=44/44 = 1 A/s.
Voltage across 10H inductor = 10*di/dt = 10*1=10V.


9. Find voltage across 12H inductor.
basic-electrical-engineering-questions-answers-coils-connected-series-q7
a) 2V
b) 10V
c) 12V
d) 20V
Answer: c
Explanation: e=Ldi/dt
Leq=2+10+12+20=44H
di/dt=44/44 = 1 A/s.
Voltage across 12H inductor = 12*di/dt = 12*1=12V.


10. Find voltage across 20H inductor.
basic-electrical-engineering-questions-answers-coils-connected-series-q7
a) 2V
b) 10V
c) 12V
d) 20V
Answer: d
Explanation: e=Ldi/dt
Leq=2+10+12+20=44H
di/dt=44/44 = 1 A/s.
Voltage across 20H inductor = 20*di/dt = 20*1=20V.

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