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:

Magnetic flux is equal to inductance times current

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

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:

Voltage is proportional to the negative time change of the current

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

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:

Sum of the individual voltages in a series circuit

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

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

Sum of the individual voltages via induction law

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

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:

Sum of the individual voltages via induction law and equal current

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

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

Total inductance for two coils

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

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:

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

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:

Total current through a parallel circuit

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

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

Time derivative of the total current is the sum of the time derivatives of the individual currents

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

Ratio of voltage to inductance

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

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:

Total voltage per total inductance

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

We can cancel out the voltage \(U\) in Eq. 11 on both sides:

Reciprocal of the total inductance for a parallel circuit

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

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:

Formula for total inductance of two coils connected in parallel

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

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:

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