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

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.

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)

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.

 \(L_{AB}=2+2+8=12\space H\)

Concept

The mutual inductance between the coils is given by:

\(M={μ N_1N_2 A\over l}\)

\(M={ N_1N_2 \over l/\mu A}\)

\(M={ N_1N_2 \over S}\)

where, S = Reluctance

N = No. of turns

Calculation

Given, N1 = 500 and N2 = 1000

S = 106 AT/Wb

\(M={500\times 1000\over 10^6}\)

M = 0.5 H

Coefficient of coupling

The coupling coefficient can be defined as a magnetic flux produced between two different coils while managing the flux successfully.

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

where, k = Coefficient of coupling

M = Mutual inductance

L1 and L2 = Self-inductance of two coils

Calculation

Given, L1  5H &  L2 = 20 H

M = 8 H

\(k={8\over \sqrt{5\times 20}}\)

k = 0.8

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.

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

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.

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

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

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.

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,

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 I₀ = 16 × 3 = 48 Volt

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

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

L_eq = 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

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

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

⇒ N₁ / N₂ = 1

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

Concept

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

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,

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