1

CHAPTER 7:CHAPTER 7:MAGNETICALLY MAGNETICALLY

COUPLED CIRCUITCOUPLED CIRCUIT

2

OBJECTIVESOBJECTIVES• To understand the basic concept of self

inductance and mutual inductance.• To understand the concept of coupling coefficient

and dot determination in circuit analysis.

3

SUB - TOPICSSUB - TOPICS

7-1 SELF AND MUTUAL INDUCTANCE.7-2 COUPLING COEFFICIENT (K)7-3 DOT DETERMINATION

4

7-1 SELF AND MUTUAL 7-1 SELF AND MUTUAL INDUCTANCEINDUCTANCE

• When two loops with or without contacts between them affect each other through the magnetic field generated by one of them, it called magnetically coupled.

• Example: transformer An electrical device designed on the basis of the

concept of magnetic coupling. Used magnetically coupled coils to transfer energy

from one circuit to another.

5

a) Self Inductancea) Self Inductance

• It called self inductance because it relates the voltage induced in a coil by a time varying current in the same coil.

• Consider a single inductor with N number of turns when current, i flows through the coil, a magnetic flux, Φ is produces around it.

i(t)Φ

+

V

_

Fig. 1

6

• According to Faraday’s Law, the voltage, v induced in the coil is proportional to N number of turns and rate of change of the magnetic flux, Φ;

• But a change in the flux Φ is caused by a change in current, i.Hence;

• Thus, (2) into (1) yields;

• From equation (3) and (4) the self inductance L is define as;

)1.......(dt

dN v

)2.......(dt

di

di

d

dt

d

)3.......(dt

di

di

dNv

)4.......(

dt

diLv or

)5........(H di

dNL

The unit is in Henrys (H)

7

b) Mutual Inductanceb) Mutual Inductance• When two inductors or coils are in close proximity to

each other, magnetic flux caused by current in one coil links with the other coil, therefore producing the induced voltage.

• Mutual inductance is the ability of one inductor to induce a voltage across a neighboring inductor.

8

Consider the following two cases:• Case 1:

two coil with self – inductance L1 and L2 which are in close proximity which each other (Fig. 2). Coil 1 has N1 turns, while coil 2 has N2 turns.

• Magnetic flux Φ1 from coil 1 has two components;

* Φ11 links only coil 1.

* Φ12 links both coils.

Hence; Φ1 = Φ11 + Φ12 ……. (6)

i1(t)

Φ12+

V1

_

+

V2

_Φ11

L2L1

N1 turns N2 turns

Fig. 2

9

)7.......(11

1

1

1111 dt

diL

dt

di

di

dNv

Thus, voltage induces in coil 1:

Voltage induces in coil 2

)8.......(121

1

1

1222 dt

diM

dt

di

di

dNv

Subscript 21 in M21 means the mutual inductance on

coil 2 due to coil 1

10

• Case 2:Same circuit but let current i2 flow in coil 2.(Fig. 3)

• The magnetic flux Φ2 from coil 2 has two components:

* Φ22 links only coil 2.

* Φ21 links both coils.

Hence; Φ2 = Φ21 + Φ22 ……. (9)

i2(t)

Φ21+

V1

_

+

V2

_Φ22

L2L1

N1 turns N2 turns

Fig. 3

11

Thus; voltage induced in coil 2

Voltage induced in coil 1

)10.......(22

2

2

2222 dt

diL

dt

di

di

dNv

)11.......(212

2

2

2111 dt

diM

dt

di

di

dNv

Subscript 12 in M12 means the

Mutual Inductance on

coil 1 due to coil 2

Since the two circuits and two current are the same:

MMM 1221

Mutual inductance M is measured in Henrys (H)

12

7-2 COUPLING 7-2 COUPLING COEFFICIENT, (k)COEFFICIENT, (k)

• It is measure of the magnetic coupling between two coils.

• Range of k : 0 ≤ k ≤ 1 k = 0 means the two coils are NOT COUPLED. k = 1 means the two coils are PERFECTLY

COUPLED. k < 0.5 means the two coils are LOOSELY

COUPLED. k > 0.5 means the two coils are TIGHTLY COUPLED.

• k depends on the closeness of two coils, their core, their orientation and their winding.

• The coefficient of coupling, k is given by;21LL

Mk

21LLkM or

13

7-3 DOT DETERMINATION7-3 DOT DETERMINATION• Required to determine polarity of “mutual” induced

voltage.• A dot is placed in the circuit at one end of each of the

two magnetically coupled coils to indicate the direction of the magnetic flux if current enters that dotted terminal of the coil.

Fig. 4

14

• Dot convention is stated as follows:if a current ENTERS the dotted terminal of one coil, the reference polarity of the mutual voltage in the second coil is POSITIVE at the dotted terminal of the second coil.

• Alternatively;if a current LEAVES the dotted terminal of one coil, the reference polarity of the mutual voltage in the second coil is NEGATIVE at the dotted terminal of the second coil.

• The following dot rule may be used:i. when the assumed currents both entered or both

leaves a pair of couple coils by the dotted terminals, the signs on the L – terms.

ii. if one current enters by a dotted terminals while the other leaves by a dotted terminal, the sign on the M – terms will be opposite to the signs on the L – terms.

• Once the polarity of the mutual voltage is already known, the circuit can be analyzed using mesh method.

15

Application of the dot convention is illustrated in the four pairs of mutual coupled coils. (Fig. a,b,c,d)

The sign of the mutual voltage v2 is determined by the reference polarity for v2 and the direction of i1. Since i1enters the dotted terminal of coil 1 and v2 is positive at thedotted terminal of coil 2, the mutual voltage is +M di1/dt. (Fig. a)

Current i1 enters the dotted terminal of coil 1 and v2 is negative at the dotted terminal of coil 2. The mutualvoltage is –M di1/dt. (Fig. b)

16

Same reasoning with Fig. a and fig. b are applies to the coil in Fig. c and Fig. d.

17

Dot convention for coils in Dot convention for coils in seriesseries

MLLL 221

i

L2L1

M

i

(+)

i

L2L1

M

i

(-)

MLLL 221

Series – aiding

connection

Series – opposing

connection

18

Below are examples of the sets of Below are examples of the sets of equations derived from basic equations derived from basic

configurations involving mutual configurations involving mutual inductanceinductance

• Circuit 1

+

M

ja

R2R3

R1

jb

Vs I1 I2

)2.......(0)(:I KVL

)1.......()(:I KVL

1232122

21211

MIIjbRRIR

VMIIjaRR s

Solution:

19

• Circuit 2

Solution:

+

M

ja R2

-jc

R1

jbVs I1 I2

)2.......(0)(:I KVL

)1.......()()(:I KVL

12212

1212111

MIIjcjbRjbI

VMIIIMjbIIjbjaR s

20

• Circuit 3

Solution:

R2+

M

ja

R1 jb

Vs I1 I2

)2.......(0)()(:I KVL

)1.......()(:I KVL

1222212

22111

IIMMIIjbjaRjaI

VMIjaIIjaR s

21

• Circuit 4

Solution:

+

M

ja

R2

-jcR1

jbVs I1

I2

)2.......(02)(:I KVL

)1.......()(:I KVL

222122

221211

MIIjbjaRIR

VIRIjcRR s

22

Circuit 5

Solution:

+

M2

ja

R2

jc

R1

-jdVs

I1I2

M1

M3

jb

)2.......(0)()()(:I KVL

)1.......()()()(:I KVL

12213221122122

221121123221211

IIMIMIMIMIjdjbjcRIjcR

VIMIMIIMIMIjcRIjcjaRR s

23

Example Example 11

+

4Ω

100VI1 I2

-j3Ωj8Ω

j2Ω

5Ωj6Ω

Calculate the mesh currents in the circuit shown below.

Solution:Solution:

)1.......(1008)34(

10026)34(:I KVL

21

2211

IjIj

IjIjIj

)2.......(0)185(8

0)(22)145(6:I KVL

21

122212

IjIj

IIjIjIjIj

24

0

100

1858

834

2

1

I

I

jj

jjIn matrix form;

The determinants are:

80008

10034

18005001850

8100

87301858

834

2

1

jj

j

jj

j

jjj

jj

AI

AI

197.8

5.33.20

22

11

Hence:

25

Example Example 22

+

5Ω

10VI1 I2

j3Ω

j2Ω

-j4Ω

+

Vo

_

j6Ω

Determine the voltage Vo in the circuit shown below.

Solution: Solution:

)1.......(104)55(

102)(26)95(:I KVL

21

121211

IjIj

IjIIjIjIj

)2.......(024

0226:I KVL

21

1212

IjIj

IjIjIj

26

0

10

24

455

2

1

I

I

jj

jjIn matrix form:

Answers:

76.1104.7

hence,

4

or 2)(6

or 2)(6

76.194.2

88.047.1

2

112

1210

2

1

jV

IjV

IjIIjV

IjIIjV

jI

jI

o

o

o

27

Example 3Example 3Calculate the phasor currents I1 and I2 in the circuit below.

j6Ωj5Ω

j3Ω

+

I1 I2

-j4Ω

12ΩV012

Solution:Solution:

For coil 1, KVL gives;

-12 + (-j4+j5)I1 – j3I2 = 0Or jI1 – j3I2 = 12 1

28

2

For coil 2, KVL gives;

-j3I1 + (12 + j6)I2 = 0Or I1 = (12 + j6)I2 = (2 – j4)I2

j3

Substituting into :

2 1

(j2 + 4 – j3)I2 = (4 – j)I2 = 12 Or A14.04 2.91 j-4

12I

2 3

From eqn. and :2 3

A 49.39- 13.01

)14.04(2.91 )63.43- (4.472 j4)I-(2 I 21