Demystify Inductance: Understanding Self & Mutual Inductance

Demonstrative Video


Self Inductance

\[\begin{aligned} L &=\frac{e}{d I / d t}\left(\text { since } e=L \frac{d I}{d t}\right) \\ &=\frac{N \phi}{I}\left(\text { since } e=N \frac{d \phi}{d t}=L \frac{d I}{d t}\right)\\ &=\frac{N^{2}}{l / a \mu_{0} \mu_{r}}\left(\text { since } \phi=\frac{N I}{l / a \mu_{0} \mu_{r}}\right) \end{aligned}\]


Mutual Inductance

\[\begin{aligned} M &=\frac{e_{m}}{d I_{1} / d t}\left(\text { since } e_{m}=M \frac{d I_{1}}{d t}\right) \\ &=\frac{N_{2} \phi_{12}}{I_{1}}\left(\text { since } e_{m}=N_{2} \frac{d \phi_{12}}{d t}=M \frac{d I_{1}}{d t}\right) \\ &=\frac{N_{1} N_{2}}{l / a \mu_{0} \mu_{r}}\left(\text { since } \phi_{12}=\frac{N_{1} I_{1}}{l / a \mu_{0} \mu_{r}}\right) \end{aligned}\]


Co-efficient of Coupling

  • When current flows through one coil, it produces flux (\(\phi_1\)).

  • The whole \(\phi\) may not be linking with the other coil coupled to it.

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\[L_{1}=\frac{N_{1} \phi_{1}}{I_{1}} \text { and } M=\frac{N_{2} \phi_{12}}{I_{1}}=\frac{N_{2} k \phi_{1}}{I_{1}} \quad \ldots(i)\left(\because \phi_{12}=k \phi_{1}\right)\] Now considering coil- 2 carrying current \(I_{2}\) \[L_{2}=\frac{N_{2} \phi_{2}}{I_{2}} \text { and } M=\frac{N_{1} \phi_{21}}{I_{2}}=\frac{N_{1} k \phi_{2}}{I_{2}} \quad \ldots(i i)\left(\because \phi_{21}=k \phi_{2}\right)\] Multiplying equation (i) and (ii), we get, Or \[\begin{aligned} M \times M &=\frac{N_{2} k \phi_{1}}{I_{1}} \times \frac{N_{1} k \phi_{2}}{I_{2}} \\ M^{2} &=k^{2} \frac{N_{1} \phi_{1}}{I_{1}} \times \frac{N_{2} \phi_{2}}{I_{2}}=k^{2} L_{1} L_{2} \end{aligned}\] Or \[M=k \sqrt{L_{1} L_{2}}\] \[\begin{aligned} & \boxed{k=\frac{M}{\sqrt{L_{1} L_{2}}}} \end{aligned}\]