Question

What is the coefficient of coupling of the inductor?

Answer 1

Coefficient of coupling gives the interaction between two coils in terms of mutual induction. Mutual inductance is the property of two coils by which each coils opposes any change in the strength of current in the other coil due to induced emf. Mutual inductance between two coils will be the same.

Complete step by step solution:
An inductor is a two-terminal electronic component that stores energy in the magnetic field when electric current flows through it.
The Coefficient of coupling can be defined as the fraction of the magnetic flux produced by the current in one coil that links with the other coil. It is represented by the symbol (k) and the amount of coupling between two inductively coupled coils is expressed in terms of the coefficient of coupling. It can be given as-
$K=\dfrac{M}{\sqrt{{{L}_{1}}{{L}_{2}}}}$
Where,
M $=$ mutual inductance between two coils
${{L}_{1}}=$ self-induction of the first coil
${{L}_{2}}=$self-induction of the second coil
Now we will prove the coefficient of coupling of an inductor which is as follows
There are two magnetically coupled coils having turn ${{N}_{1}}$ and ${{N}_{2}}$ respectively
The current ${{I}_{1}}$ that is flowing in the first coil produces a magnetic flux of ${{\phi }_{1}}$ ,having a turn ${{N}_{1}}$, then their self-induction will be given as-
${{L}_{1}}=\dfrac{{{N}_{1}}{{\phi }_{1}}}{{{I}_{1}}}$ (Equation 1)

Suppose the coefficient of coupling between two coils is k then $k{{\phi }_{1}}$ represents the magnetic flux linked with the second coil then the mutual inductance in the first coil due to the second coil is given by-
${{M}_{12}}=\dfrac{k{{\phi }_{1}}{{N}_{2}}}{{{I}_{1}}}$ (Equation 2)
The current ${{I}_{2}}$ that is flowing in the second coil will produce a magnetic flux of ${{\phi }_{2}}$, having several turns ${{N}_{2}}$, then their self-induction will be given as –
${{L}_{2}}=\dfrac{{{\phi }_{2}}{{N}_{2}}}{{{I}_{2}}}$(Equation 3)
Since the coefficient of coupling between two coils is k then $k{{\phi }_{2}}$ represents the magnetic flux linked with the first coil then the mutual inductance in the second coil due to the first coil is given by-
${{M}_{21}}=\dfrac{k{{\phi }_{2}}{{N}_{1}}}{{{I}_{2}}}$Equation (4)
On multiplying Eq (2) and Eq (4) with each other we get
${{M}_{12}}\times {{M}_{21}}=\left( \dfrac{k{{\phi }_{1}}{{N}_{2}}}{{{I}_{1}}} \right)\times \left( \dfrac{k{{\phi }_{2}}{{N}_{1}}}{{{I}_{2}}} \right)$Equation (5)
As we know that the mutual inductance between two coils will be equal. So,
${{M}_{12}}={{M}_{21}}$
Now Eq (5) will become
${{M}^{2}}={{k}^{2}}\left( \dfrac{{{\phi }_{1}}{{N}_{1}}}{{{I}_{1}}} \right)\times \left( \dfrac{{{\phi }_{2}}{{N}_{2}}}{{{I}_{2}}} \right)$
From Eq (1) and Eq (3), we get
${{M}^{2}}={{k}^{2}}{{L}_{1}}{{L}_{2}}$

& \Rightarrow {{k}^{2}}=\dfrac{{{M}^{2}}}{{{L}_{1}}{{L}_{2}}} \\\ & \Rightarrow k=\dfrac{M}{\sqrt{{{L}_{1}}{{L}_{2}}}} \\\ \end{aligned}$$ **Note:** Coefficient of coupling always comes out to be less than one and its maximum value is one only. If the coefficient of coupling has value one, then it is called as the perfect coupling that is when the entire flux of one coil links the other. If the coefficient of coupling between two inductors is greater than their mutual inductance will also be greater.