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Magnetic Coupled Circuits - GATE Study

Material in PDF

Earlier we learnt all about Transient Analysis. The next chapter in Network Theory is

Magnetic Coupling Circuits. These free GATE 2018 Study Notes will deal

with the chapter of Analysis of Magnetic Coupled Circuits.

These GATE Study Material are designed to help you ace your GATE EE, GATE

EC, IES, BARC, BSNL, DRDO and other PSU and Placement exams. You can get

Magnetic Coupled Circuits downloaded in PDF for that your GATE Preparation is

made easy.

Before you start reading AC Transients though, you need to understand the basics on

which this topic is built, using the articles listed below.

Recommended Reading –

Laplace Transforms

Types of Matrices

Properties of Matrices

Rank of a Matrix

Basic Network Theory Concepts

Kirchhoff’s Laws, Node & Mesh Analysis

i. These are the circuits in the presence of mutual inductance. This mutual

inductance is due to the mutual flux between the coils.

ii. The mutual flux may aid (or) oppose the self-flux based on the dot convention

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iii. If the current enters (or) leaves the dots in both the coils simultaneously, then

mutual flux will be added to self-flux otherwise it will oppose the self-flux.

iv. Mutual flux in one coil is due to the current flowing through the other coil.

Coupled Inductors in Series Connection

Case i: Series Aiding

∴ Leq = L1 + L2 + 2M

M = K√L1L2

0 ≤ K ≤ 1

Where K = coefficient of coupling

Note:

i) K =Useful flux

Total flux

ii) For ideal circuits K = 1 (Maximum Coupling)

iii) For isolated circuit K = 0

iv) For practical circuit it is always greater than 1

Example 1:

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Find the equivalent inductance of the given circuit and also find coefficient of

coupling

Solution:

Here in the both the coils current is entered simultaneously. Hence mutual flux is

added to self-flux

∴ Leq = L1 + L2 + 2M

M = K√L1L2

2 = K√2 × 8

∴ K =2

4= 0.5

∴ Leq = 2 + 8 + 2 (2) = 14 H

Case ii: Series Opposing

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∴ Leq = L1 + L2 − 2M

M = K√L1L2

0 ≤ K ≤ 1

K = coefficient of coupling

Note:

i) Leakage Factor =Total flux

Useful flux=

1

K

Example 2:

Find the equivalent inductance of the given circuit. Also find coefficient of coupling

and leakage factor.

Solution:

Here in the first coil current is entering and in the second coil current is leaving. So

here mutual flux is opposing the self-flux.

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∴ Leq = L1 + L2 – 2M

= 4 + 9 – 2(3) = 13 – 6 = 7H

M = K√L1L2

3 = K√4 × 9 = 6K

∴ K = 0.5

∴ Leakage factor = 1

K=

1

0.5= 2

Case iii: Parallel Aiding

∴ Leq =L1L2−M2

L1+L2−2M>

L1L2

L1+L2(M = 0)

Case iv: Parallel Opposing

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∴ Leq =L1L2−M2

L1+L2+2M<

L1L2

L1+L2(M = 0)

Example 3:

Find the equivalent inductance of the given circuit assume k = 0.5

Solution:

In this case mutual flux is opposing the self-flux

∴ Leq =L1L2−M2

L1+L2+2M

M = K√L1L2 = 0.5√16 × 4 = 0.5 × 8 = 4H

∴ Leq =16×4−16

16+4+2(4)=

64−16

20+8=

48

28= 1.71 H

Transformer Coupling

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Transformer coupling is used when load is small

Apply KVL at input then we get

−V1(t) + i1(t)R1 + L1di1(t)

dt+ M.

di2(t)

dt= 0

∴ V1(t) = i1(t). R1 + L1.di1(t)

dt+ M.

di2(t)

dt

Apply KVL at output, then we get

L2.di2(t)

dt+ i2(t). R2 + M.

di1(t)

dt= 0

A transformer is replaced by its T – equivalent i.e.

Case i: For Magnetic Aiding

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Case ii: For Magnetic Opposing

Example 4:

Determine the steady state currents i1 and i2 in the given circuit

Solution:

Here i1 is entering into the dot whereas i2 is leaving the dot hence mutual flux is

opposing the self – flux.

Represent the given network in phasor domain then we get

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5∠0° = I1 + j8I1 − j4I2 ____________ (1)

0 = I2 + j4I2 − j4 I1

j4I1 = (1 + j4)I2

I1 = (1

j4+ 1) I2___________(2)

From (1) and (2) we get

5∠0° = (1 + j8) (1

j4+ 1) I2 − j4I2

5∠0° = (1

j4+ 1 + 2 + j8 − j4) I2

20∠90° = (12j − 16)I2

20∠90° = 20∠143.13I2

∴ I2 = 1∠ − 53.13°

From (2) we get

I1 = (1 – j 0.25)I2

∴ I1 = 1.03 ∠ − 67.16°

Example 5:

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Consider the following circuit

The value of equivalent inductance between the terminals a and b is 4H with the

terminals c and d open and it is 3 H with the shorted terminals c and d. Then

determine the value of Coefficient of Coupling K.

Solution:

The equivalent circuit for the given circuit is given as

Given that,

(L1 + M) + (−M) = 4

and (L1 + M) +(L2+M)(−M)

L2+M−M= 3

∴ (4 + M) +(2+M)(−M)

2= 3

8 + 2M – 2M – M2 = 3

M2 = 5

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But we know M = K√L1L2

∴ 5 = K2(L1L2) = K2(4 × 2)

∴ K = √5

8= 0.79 ≃ 0.8

Did you enjoy reading this article on Magnetic Couple Circuits? Let us know in the

comments. You may also like some more articles in our series to help you ace your

exam and have concepts made easy. Best of luck for GATE 2018‼

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