Self Inductance and Mutual Inductance


Consider a circuit consisting of n-coils that are magnetically coupled with each other. Under linear conditions, the total flux linking any one coil is the sum of self flux and mutual flux due to (n-1) coils.
To find

  • the self flux of a coil, excite that coil alone.
  • the mutual flux in a coil due to other coils, excite all other coils except that coil.

Now apply superposition to find total flux due self and mutual flux in the coil.

Consider the case where k^{th} coil alone is excited. The current i_k flowing through the k^{th} coil produces a total flux of \Phi_{kk}. Out of the flux \Phi_{kk}, a part of this flux links with all other coils called magnetizing flux(\Phi_{mk}) and rest of flux that links to itself is called leakage flux(\Phi_{lk}).

(1)   \begin{equation*} \Phi_{kk} = \Phi_{lk}+\Phi_{mk} \end{equation*}

Consider the case where all the coils except k^{th} coil are excited. The flux produced by all other coils, links with k^{th} coil produces a total mutual flux of \Phi_{Mk}

(2)   \begin{equation*} \Phi_{Mk} = \sum\limits_{\substack{j=1\\ j\neq k}}^{j=n}\Phi_{kj} \end{equation*}

where \Phi_{kj} is the flux produced by j^{th}-coil linking with k^{th}-coil.

By superposition theorem, the total flux \Phi_k linking with k^{th} coil is,

(3)   \begin{equation*} \Phi_{k} = \Phi_{kk} + \Phi_{Mk} = \Phi_{lk}+\Phi_{mk} + \sum\limits_{\substack{j=1 \\ j\neq k}}^{j=n}\Phi_{kj} \end{equation*}

If the k^{th}-coil has N_k turns, then flux linkages is given by,

(4)   \begin{equation*} \Psi_{k} = N_k \Phi_k = \Psi_{lk}+\Psi_{mk}+\sum\limits_{\substack{j=1 \\ j\neq k}}^{j=n}\Psi_{lk} \end{equation*}

Inductance is defined as rate of change of flux linkages(N\Phi) with current.

(5)   \begin{equation*} L = {N\Phi \over i} \end{equation*}

If \mathcal{R} is the reluctance offered to the flux(\Phi), then it is related to magneto-motive force(\mathcal{F}) as, \mathcal{F}=\Phi \mathcal{R}. The reciprocal of reluctance is permeance(\matcal{P}). Therefore

(6)   \begin{equation*} \Phi = \mathcal{F P} \end{equation*}

For an N-turn k^{th}-coil, by Ampere’s law,

(7)   \begin{equation*} \underbrace{\oint H.dl}_{MMF} = N_k i_k \end{equation*}

Using Eq-(5) to Eq-(7), for k^{th}-coil we can define

  • Leakage inductance,

    (8)   \begin{equation*} \boxed{L_{lk}= {N_k\Phi_{lk} \over i_k} = N_k^2 \mathcal{P}_{lk}} \end{equation*}

  • Magnetizing inductance,

    (9)   \begin{equation*} \boxed{L_{mk}= {N_k\Phi_{mk} \over i_k} = N_k^2 \mathcal{P}_{mk}} \end{equation*}

  • Self-inductance,

    (10)   \begin{equation*} \boxed{L_{k}={N_k\Phi_{kk} \over i_k} = N_k^2 \mathcal{P}_{kk} =N_k^2 ( \mathcal{P}_{lk} + \mathcal{P}_{mk})} \end{equation*}

Mutual inductance bewteen j^{th} and k^{th} coil is

(11)   \begin{equation*} M_{jk}= {N_j\Phi_{jk} \over i_k} = N_j N_k \mathcal{P}_{jk} \end{equation*}

and

(12)   \begin{equation*} M_{kj}= {N_k\Phi_{kj} \over i_j} = N_k N_j \mathcal{P}_{kj} \end{equation*}

but, \mathcal{P}_{kj} = \mathcal{P}_{jk}.
Therefore,

    \[M_{jk}= M_{kj}\]

From EQ-11 and EQ-12,

(13)   \begin{eqnarray*} M_{jk}^2 &=& {N_jN_k\Phi_{jk}\Phi_{kj} \over i_j i_k } \\ &=& L_j L_k {\Phi_{jk} \over \Phi_{jj}}{\Phi_{kj} \over \Phi_{kk}}\\ &=& k_j k_k L_j L_k \end{eqnarray*}

    \[\boxed{M_{jk} = \sqrt{k_jk_k}\sqrt{L_jL_k}=k_{jk}\sqrt{L_jL_k}}\]

Leakage coefficient is

    \[\boxed{\sigma_{jk} = 1-k_{jk}^2 = 1 - {M_{jk}^2 \over L_j L_k}}\]

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