# Overture to Magnetic Circuits | Self and mutual inductance

Hello everyone. I hope that you all are doing good in your lives. However, the topic for today is Magnetic Circuits. Two circuits are name coupled circuits when energy transfer takes place from one circuit to the other without having any electrical connection between them. Such circuits are frequently used in network analysis. For example transformer, gyrator, etc. Future we are also going to study mutual and self-inductance.

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### Self Inductance in Magnetic circuits

Consider a coil of N turns carrying current i as appearing below;

Whenever the current flows through the coil, a flux Φ is produced in the coil. The flux in the coil links itself with the coil. However, if the current flowing through the coil changes, thus the flux linking the coil also changes. Hence, an emf is induced in the coil. This is known as self-induced emf. the direction of this emf is obtained by Lenz’s law.

However we know that,

Φ ∝ i

Φ = k i

Hence, rate of change of flux = k * rate of change of current

dΦ/dt = k * di/dt

Certainly according to Faraday’s laws of electromagnetic induction, a self induce emf can be express as;

v = -N * dΦ/dt = -Nk * di/dt = -N (Φ/i) (di/dt) = -L * di/dt

Where L = NΦ/i and is coefficient of self – inductance.

Thus the property of a coil that resists any change in the current flowing through it is self-inductance or inductance of that particular coil. If the current in the coil is increasing, the self induces electromotive force is set up in such a direction so that it opposes the rising current. Similarly, if the current in the coil decreases, the self-induced emf will be in the same direction as that of the voltage applied.

However self-inductance does not prevent the current from changing, it serves only to delay the change.

### Mutual Inductance in a Magnetic circuits

If the flux produce by the coil links with the other coil, present near the first coil. thus the emf induced in the second coil is due to change the flux in the first coin. This is name as mutually induced emf.

However, consider two coils 1 and 2 places adjacent to each other as shown in the figure below;

Let coil 1 have N_{1} turns while the other coil has N_{2} turns. however now assume that a current i_{1} flows in coil 1, flux is produced and a part of this flux links itself to coil 2. Thus the Induce emf in coil 2 is mutually induced emf.

However, we know that;

Φ_{2} ∝ i_{1}

Φ_{2} = k * i_{1}

Hence, rate of modification or change of flux = k * rate of change of current i_{1}

dΦ_{2}/dt = k * di_{1}/dt

Certainly according to faraday’s law of electromagnetic induction, the induce emf is express as;

v_{2} = -N_{2} dΦ_{2}/dt = -N_{2}k di_{1}/dt = -N_{2} (Φ_{2} / dt) (di_{1} / dt) = -M di_{1}/dt

where M = N_{2}Φ_{2} / I_{1} and is coefficient of mutual inductance.

### Coefficient of Coupling (k) in magnetic circuits

Thus the coefficient of coupling (k) between coils is defined as a fraction of magnetic flux produce by the current in the coil that links the other.

So start by assuming two coils having a number of turns N1 and N2 respectively. Whenever a current i1 is flowing in coil 1 and is changing, an emf is induced in coil 2.

M = N_{2}Φ_{2} / i_{1}

Besides let,

K1 = Φ_{2} / Φ_{1}

M = N_{2} k_{1} Φ_{1} / i_{1}____________(1)

However, if the current i_{2} is flowing in coil 2 and is changing, an emf is induce in the coil 1

M = N_{1} Φ_{1} / i_{2}

Similarly let,

K2 = Φ_{2} / Φ_{1}

M = N_{1} k_{2} Φ_{2} / i_{2}_______________(2)

Thus by multiplying equation 1 and 2 we get;

M^{2} = K^{2} L_{1} L_{2}

M = k **√** L_{1}L_{2}

So, k = **√** k_{1}k_{2}

### Inductance in Series

#### Cumulative Coupling

Whenever two coils are in series in such a way that the current through the two coils is in the same direction in order to produce flux in the same direction. Thus such connections are named Cumulative coupling.

Firstly let;

Thus total induce emf is;

L = L_{1} + L_{2} + 2M

#### Differential Coupling

Similarly when these Coils are also connected in series but the direction of current in coil 2 is now opposite to that in coil 1. thus such a connection of two coils is known as differential coupling.

L = L1 + L2 – 2M

### Inductance in Parallel

#### Cumulative Coupling

Whenever coil 1 and 2 are connected in parallel in such a way that fluxes produced by the coil act in the same direction. then such a connection is referred to be as cumulative coupling.

Thus the total inductance, in this case, is given by;

#### Differential Coupling

Similarly when both the coils i.e coil 1 and 2 are connected in parallel but in such a way that the flux produce in the coil act in the opposite direction. thus such connections of two coil is named as differential coupling.

Hence the total impedance is;

### Example problem on self and mutual inductance

#### Example No. 1

However, the combination of the inductance of two coils connected non-parallel to each is 0.6 H or 0.1 H on the basis of relative directions of current in the two coils. If any one of the coils features a self-inductance of 0.2 H. Hence find (a) mutual inductance and (b) coefficient of coupling of a magnetic circuit.

Given :- L_{1} = 0.2 H, L_{diff} = 0.1 H, L_{cum} = 0.6 H

Solution :- (a) Firstly Mutual inductance;

Thus adding both the equations above;

Certainly by Substracting the equations of L_{cum} and L_{diff}

Hence the value of mutual inductance is 0.125 H.

(b)And lastly Coefficient of coupling

Thus the value of the coefficient of coupling is 0.72

#### Example No. 2

Two coils with a coefficient of coupling of 0.6 among them are in series so as to magnetize in (a) the same direction and (b) the opposite direction. The total inductance in the same direction is 1.5 H and in the opposite direction is 0.5 H. Thus find the self-inductance of the coil i.e. magnetic circuits.

Given :- k = 0.6, L_{diff} = 0.5 H, L_{cum} = 1.5 H

Solution;

However, Subtracting both equations above;

lastly Adding the equations of L_{diff} and L_{cum}

Thus by solving the equation we get

#### Example 3

Two coils are connected parallel to each other having 4 mH and 7 mH self-inductance respectively. Then mutual inductance between them is 5 mH, Thus find the equivalent inductance of the magnetic circuits.

Given :- L1 = 4mH, L2 = 7mH, M = 5mH

Solution; Firstly for cumulative coupling,

And lastly differential coupling;

#### Example No. 4

Two inductors are connected in parallel. Thus their equivalent inductance when the mutual inductance aids the self-inductance is 6mH and it is 2 mH when the mutual inductance opposes the self-inductance. If the ratio of the self-inductance is 1:3 and the mutual inductance between the coils is 4mH, so find the self-inductance.

Given :- L_{cum} = 6 mH, L_{diff }= 2 mH, L_{1}/L_{2} = 1.3, M = 4mH

Solution; Firstly cumulative coupling

Secondly differential coupling;

Thus solving equations (1) and (2);

However, we know that L_{1}/L_{2} = 1.3

### Conclusion

However here we are at the last part of the blog. I hope that you liked the blog and are happy with the content provided then please make sure to share it with others and also mention the part which you like the most. Besides, feel free to ask any doubt if present. Also would be happy to know the topic which you would like to read next on.

Have a nice day 🙂

Regards.

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