Inductance in Series – Derivation & Formula

When the coils are so connected that the end of coil one is connected to the starting of second coil and so on as to form a shape of a chain, these are connected in series. In this case the flux of first coil does not effect the second coil at all.

Inductance in Series - Derivation & Formula

Let their inductances be L1 and L2 are connected across E volt of f frequency. In that case when same current is flowing in both coils (Fig. 1). And E1 and E2 are voltage across inductors L1 and L2 respectively. Then,

\[E={{E}_{1}}+{{E}_{2}}\]

also

\[E=I\times {{X}_{L}}\]

or

\[{{E}_{1}}=I\times {{X}_{L1}}\]

\[{{E}_{2}}=I\times {{X}_{L2}}\]

Than

\[I\times {{X}_{L}}=I\times {{X}_{L1}}+I\times {{X}_{L2}}\]

\[{{X}_{L}}={{X}_{L1}}+{{X}_{L2}}\]

since,

\[{{X}_{L}}=2\pi fL\]

Thus

\[2\pi fL=2\pi f{{L}_{1}}+2\pi f{{L}_{2}}\]

\[L={{L}_{1}}+{{L}_{2}}\]

It is a very ideal case. But in actual practice the mutual induction takes place and let M be the Coefficient of mutual induction. Than equivalent inductance,

Series-adding

Inductance in Series

Fig. 1: Inductance in Series (Series-aiding)

If both the coils are connection in such a way as to result in a additive mmfs than equivalent inductance (Fig. 1) is

\[L={{L}_{1}}+{{L}_{1}}+2M\]

Series-opposing

Inductance in Series derivation

Fig. 2: Inductance in Series (Series-opposing)

If both the coils are connected in such a way as to result in a subtractive mmfs than equivalent inductance (Fig. 2) is

\[L={{L}_{1}}+{{L}_{1}}-2M\]

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