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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‼
Network Theory - Revision Test 1
Parameters of Periodic Waves
Control Systems - Revision Test 1
Routh Hurwitz Stability Criteria
Special Cases in Routh Stability Criteria
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