My name is Alexander FufaeV and here I will explain the following topic:

Total (Equivalent) Inductance of a Series and Parallel Circuit of Coils

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In the following we want to derive the total inductance $L$ (also called equivalent inductance) of two coils connected in parallel and of two coils connected in series. One coil has the inductance $L_1$ and the other $L_2$.

Since the magnetic field generated by the coil is proportional to the current $ \class{red}{I} $ flowing through the coil, the magnetic field and hence the magnetic flux $\Phi_{\text m}$ can be written as follows:

$$ \begin{align} \Phi_{\text m} ~=~ L \, \class{red}{I} \end{align} $$

The inductance $L$ of the coil is the proportionality constant here. The applied AC voltage $U$ and the resulting current $I$ are related via the Faraday's law of induction:

$$ \begin{align} U ~=~ -L \, \frac{\text{d}\class{red}{I}}{\text{d}t} \end{align} $$

Total inductance of a series circuit of coils

Consider two coils connected in series, to which an alternating voltage $U$ is applied. Thus, the time-dependent total voltage $U$ is between the both coils. The voltages $U_1$ and $U_2$ are between the ends of the individual coils:

Two coils connected in series to which an AC voltage is applied.

$$ \begin{align} U ~=~ U_1 + U_2 \end{align} $$

For all three voltages $U$, $U_1$ and $U_2$ we insert the induction law 2 (the minus sign cancels out on both sides):

$$ \begin{align} L \, \frac{\text{d} \class{red}{I}}{\text{d}t} ~=~ L_1 \, \frac{\text{d}\class{red}{I_1}}{\text{d}t} + L_2 \, \frac{\text{d}\class{red}{I_2}}{\text{d}t} \end{align} $$

The total current $ \class{red}{I} $ passes through the two coils according to the junction rule (1st Kirchhoff rule). So the currents are all equal: $ \class{red}{I} = \class{red}{I_1} = \class{red}{I_2}$. Insert the current $ \class{red}{I} $ into Eq. 4:

$$ \begin{align} L \, \frac{\text{d} \class{red}{I}}{\text{d}t} ~=~ L_1 \, \frac{\text{d} \class{red}{I}}{\text{d}t} + L_2 \, \frac{\text{d} \class{red}{I}}{\text{d}t} \end{align} $$

The time derivative of the current occurs on both sides of the equation, so it can be canceled out:

$$ \begin{align} L ~=~ L_1 + L_2 \end{align} $$

The inductance in a series circuit add up to a total inductance. If we had $n$ instead of two coils connected in series, then we analogously sum up the individual inductances:

$$ \begin{align} L ~=~ L_1 ~+~ L_2 ~+~ ... ~+~ L_n \end{align} $$

Total inductance of a parallel circuit of coils

Let us now consider two parallel connected coils. If an AC voltage $U$ is applied to the parallel circuit, an AC current $ I $ flows. This current splits at the junction into the currents $I_1$ and $I_2$, which flow to the first and second coils, respectively. According to Kirchhoff's junction rule, the total current is given by the sum of the individual currents:

Two coils connected in parallel through which an alternating current is passed.

$$ \begin{align} \class{red}{I} ~=~ \class{red}{I_1} ~+~ \class{red}{I_2} \end{align} $$

Differentiate both sides of Eq. 8 with respect to time:

$$ \begin{align} \frac{\text{d} \class{red}{I}}{\text{d}t} ~=~ \frac{\text{d} \class{red}{I_1}}{\text{d}t} + \frac{\text{d} \class{red}{I_2}}{\text{d}t} \end{align} $$

This way you can insert the induction law 2 into Eq. 9 for the time derivatives:

$$ \begin{align} \frac{U}{L} ~=~ \frac{U_1}{L_1} + \frac{U_2}{L_2} \end{align} $$

According to the mesh rule (2nd Kirchhoff rule), the total voltage $U$ also drops at the individual coils. This means: $U = U_1 = U_2$. Substitute this voltage into Eq. 10:

$$ \begin{align} \frac{U}{L} ~=~ \frac{U}{L_1} + \frac{U}{L_2} \end{align} $$

We can cancel out the voltage $U$ in Eq. 11 on both sides:

$$ \begin{align} \frac{1}{L} ~=~ \frac{1}{L_1} + \frac{1}{L_2} \end{align} $$

As you can see: the total inductance of a parallel circuit of two coils is not equal to the sum of the individual inductances. We can also rearrange the equation with respect to the total inductance:

$$ \begin{align} L ~=~ \frac{ L_1 \, L_2 }{ L_1 ~+~ L_2 } \end{align} $$

If we have not 2 but $n$ coils connected in parallel, then analogously we sum the reciprocals of the individual inductances to get the reciprocal of the total inductance:

$$ \begin{align} \frac{1}{L} ~=~ \frac{1}{L_1} ~+~ \frac{1}{L_2} ~+~ ... ~+~ \frac{1}{L_n} \end{align} $$