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Q&A

Coupling of inductors

+1
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Which will be the total series inductance?

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How coupling effects the series inductance is a good question, +1. However, it's ambiguous to talk a... (3 comments)

2 answers

+1
−1

Which will be the total inductance of this circuit?

Quite simply, if both the individual 1 henry inductors are perfectly magnetically coupled then, the net inductance will be 4 henries. If both 1 henry inductors are totally uncoupled, then the net inductance is 2 henries.

This can be realized by considering two electrically identical parallel windings that are 100% coupled on the same magnetic core. Two parallel windings produce no-more or no-less inductance than one winding. However, each winding must take 50% of the overall current hence each winding must be regarded as having twice the inductance of the single winding scenario.

Therefore, each winding is 2 henries because, in parallel, they produce an inductance of 1 henry.

If you then disconnected the ends of the two windings and rearranged the connections so that the two windings became in series, you would find that the overall inductance has become 4 henries.


The more important and interesting scenario is when you have two inductors of different value that are perfectly coupled. Let's say you have a 3 henry inductor in series with a 1 henry inductor: -

Image alt text

Starting with the inductors totally uncoupled (a net inductance of 4 henry): -

Image alt text

The rate of change of current is 0.25 amps per second and that is exactly what we'll get with 1 volt connected across 4 henries. However, if I added 100% coupling like this: -

Image alt text

The time response becomes this: -

Image alt text

Now, we have a rate of change of current of approximately 0.133975 amps per second (values taken from the simulator). This is an inductance value equal to the reciprocal of that value i.e. 7.4641 henries.

Given that the mutual inductance equals $k\cdot\sqrt{L_1 L_2} = \sqrt{3}$ (when k=1), the net inductance is: -

$$L_1 + L_2 + 2M = 1 + 3 + 2\sqrt3 = 7.4641 \text{ henries}$$


But, I'll tell you what, I'm not a big fan of mutual inductance. I mean, what does it bring to the table other than yet another formula to remember. Consider this and optionally remember it; if the two separate inductors (3 henry and 1 henry) were wound on separate but identical cores, the turns on the 3 henry inductor would be $\sqrt3$ times more than the turns on the 1 henry inductor.

Now, combine the turns onto one common core and you get $1+\sqrt3$ turns. To calculate inductance on the common core, it's just $(1+\sqrt3)^2 =$ 7.4641 henry. Job done without bothering to wake up mister mutual inductance.

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+0
−3

When 2 inductors share the same magnetic field they become coupled and besides their self inductance they have a mutual inductance as well. For this circuit:

Now current entering one coil marked with a dot, produces a current exiting the couple coil at its dotting end. In L1 current enters from the dotted end so this creates a current which exits from the dotted end of L2 and in L2 current enters from the dotted end so this again creates a current which exits from the dotted end of L1 so we get 2 voltage drops from self inductance of each coil and 2 voltage drops from the mutual inductance of the coupled coils.

I

so that is translated to this equation:

Now obviously :

The total inductance of the circuit becomes

the value of the mutual inductance is: (with k=1)

$L_{T}=4 \text{H}$

and the equation of current of this circuit becomes

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*Now current entering one coil marked with a dot, produces a current exiting the couple coil at its d... (3 comments)

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