Magnetic Coupling Circuits MCQ Quiz - Objective Question with Answer for Magnetic Coupling Circuits - Download Free PDF

Last updated on Jun 10, 2025

Latest Magnetic Coupling Circuits MCQ Objective Questions

Magnetic Coupling Circuits Question 1:

The total inductance of two coils, A and B, when connected in series is 0.5 H or 0.2 H, depending upon the relative directions of the current in the coils. Coil A, when isolated from coil B, has a self-inductance of 0.2 H. The mutual inductance between the two coils is

  1. 0.25 H
  2. 0.05 H
  3. 0.15 H
  4. 0.075 H

Answer (Detailed Solution Below)

Option 4 : 0.075 H

Magnetic Coupling Circuits Question 1 Detailed Solution

Explanation:

Mutual Inductance Between Two Coils

Definition: Mutual inductance is a measure of the interaction between two coils, where the magnetic field generated by the current in one coil induces a voltage in the other coil. It depends on the geometry of the coils, the number of turns in each coil, and the relative positioning or orientation of the coils.

Problem Statement: The total inductance of two coils, A and B, when connected in series is given as either 0.5 H or 0.2 H, depending upon the relative directions of the current in the coils. The self-inductance of coil A is 0.2 H, and we need to determine the mutual inductance (M) between the two coils.

Solution:

When two coils are connected in series, the equivalent inductance (Ltotal) can be calculated using the following formula:

Case 1: Currents in the same direction

The total inductance is given by:

Ltotal = LA + LB + 2M

Where:

  • LA = Self-inductance of coil A
  • LB = Self-inductance of coil B
  • M = Mutual inductance between the two coils

In this case, Ltotal = 0.5 H (as given in the problem).

Substituting the values:

0.5 = 0.2 + LB + 2M

Case 2: Currents in opposite directions

The total inductance is given by:

Ltotal = LA + LB - 2M

In this case, Ltotal = 0.2 H (as given in the problem).

Substituting the values:

0.2 = 0.2 + LB - 2M

Step-by-Step Calculation:

From Case 2:

0.2 = 0.2 + LB - 2M

Rearranging:

LB - 2M = 0

LB = 2M

From Case 1:

0.5 = 0.2 + LB + 2M

Substitute LB = 2M:

0.5 = 0.2 + 2M + 2M

0.5 = 0.2 + 4M

Rearranging:

4M = 0.5 - 0.2

4M = 0.3

M = 0.3 ÷ 4

M = 0.075 H

Conclusion:

The mutual inductance between the two coils is 0.075 H, which corresponds to Option 4.

Important Information

To further analyze the other options:

  • Option 1: 0.25 H - This value is incorrect. If mutual inductance were 0.25 H, the calculated total inductance values would not match the given problem statement.
  • Option 2: 0.05 H - This value is too low and does not satisfy the equations for total inductance in both cases (same direction and opposite direction currents).
  • Option 3: 0.15 H - This value is incorrect. Substituting M = 0.15 H into the equations for total inductance leads to inconsistencies with the given values of 0.5 H and 0.2 H.
  • Option 4: 0.075 H - This is the correct answer, as demonstrated above.

Magnetic Coupling Circuits Question 2:

Two magnetic paths BE and BCDE are in parallel and form a parallel magnetic circuit, as shown in the given figure. The Ampere Turn required for this parallel circuit is equal to: 

qImage680f58b1066652bd53e12c0d

  1. inverse of Ampere Turn required for BE
  2. ampere Turn required for any one of the paths
  3. square of Ampere Turn required for any one of the paths 
  4. sum of Ampere Turn required for BE and BCDE

Answer (Detailed Solution Below)

Option 2 : ampere Turn required for any one of the paths

Magnetic Coupling Circuits Question 2 Detailed Solution

Concept

The reluctance in a magnetic circuit is given by:

\(R={MMF\over Flux}\)

\(R={l\over μ A}\)

 

Explanation

In a magnetic circuit, reluctance is analogous to resistance in an electrical circuit.  In the given question and diagram, two magnetic paths, BE and BCDE, are in parallel, forming a parallel magnetic circuit.

In a parallel magnetic circuit, the magnetomotive force (MMF) or Ampere-Turns (AT) across parallel branches is the same, just like voltage in an electric parallel circuit.

Hence, the correct answer is option 2.

Magnetic Coupling Circuits Question 3:

Coefficient of coupling between two coils is given as ______, where M = Mutual Inductance, L1 = Self Inductance of coil 1 and L2 = Self Inductance of coil 2 

  1. \(\rm \frac{M}{\sqrt{L_1L_2}}\)
  2. \(\rm \frac{\sqrt{L_1L_2}}{M}\)
  3. \(\rm M\sqrt{L_1L_2}\)
  4. L1L2.M

Answer (Detailed Solution Below)

Option 1 : \(\rm \frac{M}{\sqrt{L_1L_2}}\)

Magnetic Coupling Circuits Question 3 Detailed Solution

Concept:

The coefficient of coupling (k) between two coils indicates how effectively energy is transferred from one coil to another through mutual inductance. It is a dimensionless quantity ranging from 0 (no coupling) to 1 (perfect coupling).

Calculation

The formula for coefficient of coupling is:

\( k = \frac{M}{\sqrt{L_1 \cdot L_2}} \)

This formula expresses how mutual inductance relates to the geometric mean of the self-inductances of the two coils.

Hence, the correct option is the one showing: M / √(L₁L₂)

Magnetic Coupling Circuits Question 4:

The differential coupling of two coils in series connection has self-inductance of 2 mH & 4 mH & a mutual inductance of 0.15 mH. The equivalent inductance of the combination is:  

  1. 5.7 mH
  2. 5.85 mH
  3. 6 mH 
  4. 6.15 mH

Answer (Detailed Solution Below)

Option 1 : 5.7 mH

Magnetic Coupling Circuits Question 4 Detailed Solution

Explanation:

Equivalent Inductance in Series Connection of Coils

Problem Statement: The differential coupling of two coils in series has self-inductances of 2 mH and 4 mH, with a mutual inductance of 0.15 mH. The task is to find the equivalent inductance of this combination.

Understanding the Concept:

When two inductors are connected in series, the equivalent inductance depends on their individual inductances and the mutual inductance between them. The equivalent inductance, denoted as Leq, is given by:

Formula:

Leq = L1 + L2 ± 2M

Where:

  • L1 = Self-inductance of the first coil (2 mH in this case)
  • L2 = Self-inductance of the second coil (4 mH in this case)
  • M = Mutual inductance between the two coils (0.15 mH in this case)
  • The sign of 2M depends on whether the mutual coupling is aiding or opposing. The positive sign is used if the coupling is aiding, and the negative sign is used if the coupling is opposing.

Calculation:

Assume the coupling is aiding (positive sign for 2M).

Substitute the given values into the formula:

Leq = L1 + L2 + 2M

Leq = 2 + 4 + 2 × 0.15

Leq = 2 + 4 + 0.3

Leq = 6.3 mH

However, the actual correct answer provided in the problem is 5.7 mH. This indicates that the coupling is opposing (negative sign for 2M). Let us re-calculate with the opposing case:

Substitute with the negative sign for 2M:

Leq = L1 + L2 - 2M

Leq = 2 + 4 - 2 × 0.15

Leq = 2 + 4 - 0.3

Leq = 5.7 mH

Correct Option Analysis:

The correct answer is Option 1: 5.7 mH. This is the equivalent inductance when the coupling is opposing.

Important Information:

To further understand the analysis, let’s evaluate why other values do not match:

  • Option 2: 5.85 mH: This value could arise from an incorrect calculation where the mutual inductance is not properly considered.
  • Option 3: 6 mH: This would be the result of ignoring the mutual inductance entirely, i.e., simply adding L1 and L2 without accounting for the coupling.
  • Option 4: 6.15 mH: This might be a miscalculation involving an incorrect positive sign for 2M or an incorrect value for M.

Conclusion:

Understanding the role of mutual inductance is critical in calculating the equivalent inductance of coupled coils. The sign of the mutual inductance term depends on whether the coupling is aiding or opposing. In this case, the coupling is opposing, resulting in an equivalent inductance of 5.7 mH.

Magnetic Coupling Circuits Question 5:

What is the equivalent inductance of the circuit below, between A and B?

qImage67c97ddb514b82c2f1599776

  1. 16 H
  2. 12 H
  3. 28 H
  4. 13 H

Answer (Detailed Solution Below)

Option 2 : 12 H

Magnetic Coupling Circuits Question 5 Detailed Solution

Concept

When 'n' inductors are connected in series, the equivalent inductance is given by:

\(L_{eq}=L_1+L_2.........L_n\)

If all inductors are equal, then:

\(L_{eq}=nL\)

When 'n' inductors are connected in parallel, the equivalent inductance is given by:

\({1\over L_{eq}}={1\over L_1}+{1\over L_2}.........{1\over L_n}\)

If all inductors are equal, then:

\(L_{eq}={L\over n}\)

Calculation

From the figure, 6Ω inductors are connected in parallel.

\(L_{1}={6\over 3}=2\space H\)

Now 2H, 2H and 8H are connected in series.

 \(L_{AB}=2+2+8=12\space H\)

Top Magnetic Coupling Circuits MCQ Objective Questions

Two identical coils A and B of 1000 turns each lie in parallel plane such that 80% of the flux produced by one coil links with the other. If a current of 5 A flowing in A produces a flux of 0.05 mWb, then the flux linking with coil B is:

  1. 0.4 mWb
  2. 0.04 mWb
  3. 4 mWb
  4. 0.004 mWb

Answer (Detailed Solution Below)

Option 2 : 0.04 mWb

Magnetic Coupling Circuits Question 6 Detailed Solution

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Concept:

Consider two coils having self-inductance L1 and L2 placed very close to each other. Let the number of turns of the two coils be N1 and N2 respectively. Let coil A carries current I1 and coil B carries current I2.

F1 U.B M.P 31.07.19 D 20

Due to current I1, the flux produced is ϕ1 which links with both the coils. Then the mutual inductance between two coils can be written as

\(M = \frac{{{N_1}{ϕ _{12}}}}{{{I_1}}}\)

Here, ϕ12 is the part of the flux ϕ1 linking with the coil 2

Calculation:

Flux produced in coil X (ϕ1) = 0.05 mWb

As we are just required to find the flux linked with the second coil, and we are given that 80% of the flux produced by one coil links with the other. 

∴ Flux linked with Y (ϕ12) = 80% of flux produced in coil 1 

= 0.05 × 0.8 mWb

0.04 mWb

For magnetically coupled circuits, mutual inductance is always _______.

  1. infinite
  2. negative
  3. positive
  4. zero

Answer (Detailed Solution Below)

Option 3 : positive

Magnetic Coupling Circuits Question 7 Detailed Solution

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Concept:

Mutual inductance (M) between two coupled coils is given by

\(M = k\sqrt {{L_1}{L_2}} \)

Where,

L1 and L2 are self-inductance of the two coils

k is the linking coefficient of the two coils

Explanation:

For two coupled coils, K ≤ 1

If K = 1, it is said to be tight coupling i.e, ideal.

If K < 1, it is said to be loosely coupled

Therefore, for magnetically coupled circuits mutual inductance is always positive.

A coil of 360 turns is linked by a flux of 200 μ Wb. If the flux is reversed in 0.01 second, then find the EMF induced in the coil.

  1. 7.2 V
  2. 0.72 V
  3. 14.4 V
  4. 144

Answer (Detailed Solution Below)

Option 3 : 14.4 V

Magnetic Coupling Circuits Question 8 Detailed Solution

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Concept:

Average induced emf is given by \(E = N\frac{{d\phi }}{{dt}}\)

Where N is the number of turns

dϕ is changing in flux

dt is changing in time

Calculation:

Given that, number of turns (N) = 360

Change in time (dt) = 0.01 s

Magnetic flux (ϕ) = 200 μWb

Since the flux is reversed, it changes from 200 μWb to -200 μWb, which is a change of 200 – (-200), i.e. 400 μWb

Change in magnetic flux (dϕ) = 400 μWb

\(E = 360 \times \frac{{400 \times {{10}^{ - 6}}}}{{0.01}} = 14.4\;V\)

Find voltage across 16 Ω that is V0 in the circuit shown

 F1 Shubham.B 10-04-21 Savita D1

  1. 100 V
  2. 74 V
  3. 96 V
  4. 48 V

Answer (Detailed Solution Below)

Option 4 : 48 V

Magnetic Coupling Circuits Question 9 Detailed Solution

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Concept:

Turns Ratio:

The turns ratio for the transformer is given by

\(\frac{{{V_p}}}{{{V_s}}} = \frac{{{I_s}}}{{{I_p}}} = \frac{{{N_p}}}{{{N_s}}}\)

Kirchhoff’s Voltage Law (KVL):

It states that the algebraic sum of all voltage around a close path or loop is zero.

Mathematically, KVL implies that

\(\mathop \sum \limits_{m = 1}^M {v_m} = 0 \)

Where M is the number of voltages in a loop or number of branches in a loop

And, vm is mth voltage.

Consider a circuit shown below in which R1 and R2 are two resistance, v1, v2 are two voltage source which causes current (I) flow in the loop.

F12 Jai Prakash 2-2-2021 Swati D37

The sign of voltage drop across the passive element is such a way that the current entering from the positive terminal.

Applying KVL on this circuit

V1 + V4 – IR1 – IR2 = 0

Or, V1 + V4 = IR1 + IR2

Hence, the Sum of Voltage drops = Sum of Voltage rises

Note: KVL deals with the conservation of energy.

Calculation:

The direction of current and polarity of voltage from each branch in the circuit can be represent as,

F1 Shubham.B 10-04-21 Savita D2

From above concept,

V2 = 2V1 …. (1)

I1 = 2I2 …. (2)

Applying KVL in loop,

120 = 8 (I1 + I0) + 16 I0 + 16 (I0 – I2)

120 = 40 I0 + 8 I1 – 16 I2

From equation (2)

120 = 40 I0 + 8 I1 - 8 I1

40 I0 = 120

I0 = 3 A

Voltage across 16 Ω will be,

V0 = 16 I= 16 × 3 = 48 Volt

For the three coupled coils in the figure shown below, calculate the total inductance:

F3 Vilas Engineering 8.12.2022 D16

  1. 20 H
  2. 48 H
  3. 70 H
  4. 100 H

Answer (Detailed Solution Below)

Option 1 : 20 H

Magnetic Coupling Circuits Question 10 Detailed Solution

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Concept

The total inductance for series connection is given by:

\(L_{eq}=L_1+L_2+L_3+2(\pm M_1 \space \pm M_2\space \pm M_3)\)

where, L = Self inductance

M = Mutual inductance

+ Sign is used in mutual inductance when current either enters or leaves from the dot in both the coils

- Sign is used in mutual inductance when current enters from a dot in one coil and leaves without a dot from another coil

Calculation

Given, L1 = 12 H, L2 = 16 H and L3 = 20 H

M1 = - 8 H, M2 = -10 and M3 = 4 H

\(L_{eq}=12+16+20+2(-8 \space-10\space + 4)\)

Leq = 20 H

Additional Information The total inductance for parallel connection is given by:

\(L_{eq}={L_1L_2\space -\space M^2\over L_1\space + \space L_2 \space \mp \space 2M}\)

- Sign is used in mutual inductance when current either enters or leaves from the dot in both the coils

+ Sign is used in mutual inductance when current enters from a dot in one coil and leaves without a dot from another coil

A linear magnetic circuit has a flux linkage of 2 wb-turn when a current of 20 A flows through its coil. What is the energy stored in the magnetic field of the coil?

  1. 80 J
  2. 40 J
  3. 10 J 
  4. 20 J

Answer (Detailed Solution Below)

Option 4 : 20 J

Magnetic Coupling Circuits Question 11 Detailed Solution

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Concept:

The energy in a coil is given by:

\(E = {1 \over 2}LI^2\) .............(i)

where, E = Energy 

L = Inductance

I = Current

The flux linkage in a coil is given by:

\(ϕ = LI\)

\(L = {ϕ\over I}\) .............(ii)

Putting the value of equation (i) in (ii), we get:

\(E = {1 \over 2}({ϕ \over I})I^2\)

\(E = {1 \over 2}ϕ I\)

Calculation:

Given, I = 20 A

ϕ = 2 AT

\(E = {1 \over 2}\times 2\times 20\)

E = 20 J

Two coupled coils with L1 = L2 = 0.5 H have a coupling coefficient of K = 0.75. The turn ratio N1/N2 =?

  1. 4
  2. 0.5
  3. 2
  4. 1

Answer (Detailed Solution Below)

Option 4 : 1

Magnetic Coupling Circuits Question 12 Detailed Solution

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Concept:

\(L = \frac{{{\mu _0}{N^2}A}}{l}\)

The self-inductance of a coil is given by

Where L is the self-inductance

N is the number of turns

A is the cross-sectional area

l is the length

Calculation:

The self-inductance is proportional to the square of the turns in a coil i.e. L ∝ N2

\(\Rightarrow \frac{{{L_1}}}{{{L_2}}} = \frac{{N_1^2}}{{N_2^2}}\)

Given that L1 = L2 = 0.5 H

⇒ N/ N2 = 1

For magnetically isolated coils, the value of coefficient of coupling is:

  1. 0
  2. 1
  3. 0.5
  4. 0.75

Answer (Detailed Solution Below)

Option 1 : 0

Magnetic Coupling Circuits Question 13 Detailed Solution

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Coefficient of Coupling (k):
The coefficient of coupling (k) between two coils is defined as the fraction of magnetic flux produced by the current in one coil that links the other.

Two coils have self-inductance L1 and L2, then mutual inductance M between them then Coefficient of Coupling (k) is given by 

\(k=\frac{M}{\sqrt {L_1L_2}}\)

Where,

\(M=\frac{\mu_o \mu_rN_1N_2A}{ l}\)

\(L_1=\frac{\mu_o \mu_rN_1^2A}{ l}\)

\(L_2=\frac{\mu_o \mu_rN_2^2A}{ l}\)

N1 and N2 is the number of turns in coil 1 and coil 2 respectively
A is the cross-section area
l is the length

For fully Coupled Coil:

k = 1

For magnetically Isolated Coil:

k = 0

What is the energy stored in 1 second in an inductor while carrying current i = 20 + 10 t where t is the time in seconds if the electromotive force induced in the coil due to self-induction is 40 mV?

  1. 1.8 J
  2. 0.45 J
  3. 0.9 J
  4. 1.2 J

Answer (Detailed Solution Below)

Option 1 : 1.8 J

Magnetic Coupling Circuits Question 14 Detailed Solution

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Concept:

EMF induced in the coil due to self-induction

Self-induced emf is the e.m.f induced in the coil due to the change of flux produced by linking it with its own turns.

From Faradays Law of Electromagnetic Induction.

\(E\propto\frac{d\phi}{dt}\).............(1)

Since the rate of change of flux linking with the coil depends upon the rate of current in the coil.

\(\phi\propto i\)...........(2)

From equations (1) and (2),

\(\large{E\propto\frac{di}{dt}}\)

\(\large{E=L\frac{di}{dt}}\)..............(3)

Where,

E is the EMF induced in the coil

L is the proportionality constant called as 'Self Inductance'.

Calculation:

Given, E = 40 mV

i = 20 + 10 t

\(\large{\frac{di}{dt}=10}\)

From equation (3),

\(\large{L=\frac{E}{\frac{di}{dt}}}=\frac{40\times10^{-3}}{10}=4mH\)

At t = 1 sec, i = 30 A

Energy store in coil (E) is given by,

\(E = \frac{1}{2}Li^2=\frac{1}{2}\times4\times10^{-3}\times{30}^2=1.8\ J \)

Two coils have coefficient of coupling as 0.8. A current of 3 A in coil 1 produces a total flux of 0.4 mWb. If the current in coil 1 is reduced to zero in 3 ms, the voltage induced in coil 2 is 85 V. Determine the inductance of coil 2 if number of turns of coil 1 is 300.

  1. L= 282 mH
  2. L2 = 282 H
  3. L= 85 mH
  4. L2 = 85 H

Answer (Detailed Solution Below)

Option 1 : L= 282 mH

Magnetic Coupling Circuits Question 15 Detailed Solution

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Concept

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The value of the mutual inductance is given by:

\(M=k\sqrt{L_1L_2}\)

where, M = Mutual inductance

L1 = Self-inductance of coil 1

L2 = Self-inductance of coil 2

Calculation

Given, ϕ1 = 0.4 mWb

I1 = 3 A

N1 = 300

The self-inductance of coil 1 is given by:

\(L_1={N_1ϕ_1\over I_1}\)

\(L_1={300\times 0.4\over 3}\)

L1 = 40 mH

The induced voltage in coil 2 due to the current flowing in coil 1 is given by:

\(V_2=M{dI_1\over dt}\)

\(85=M{3\over 3\times 10^{-3}}\)

M = 85 mH

\(85=0.8\sqrt{40\times L_2}\)

L2 = 282 mH

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