Inductors in Series and Parallel Formula

Inductors in series and parallel formula are not that different to series resistors or parallel resistors. Just an opening, the equivalent inductance for series inductors is the sum of all the inductances.

The scientific reason behind it can be proven from the explanations and examples below. But since it is in the same group as resistors, we can analyze it using voltage and current.

How to Calculate Equivalent Inductance

Remember that an inductor is also a passive element, we can utilize a series-parallel combination tool. How we calculate equivalent inductance can be done with analyzing voltage and current in the actual electrical circuit.

We will still use our previous formulas where we learnt about the inductance formula.

The induced voltage in a inductor when a current flows through it:

    \begin{align*}v=L\frac{di}{dt}\end{align*}

We will treat this induced voltage like voltage drops in the resistor to make it easier for us to remember.

Without further ado, let’s proceed to the next formulas.

Series Inductance Formula

Observe the circuit below consists of n inductors,

inductors in series and parallel formula 1

Since it is a series circuit, the flowing current will be the same through all inductors, while each of them may have different voltages across them.

The current in the circuit is

    \begin{align*}i=i_1=i_2=i_3=i_n\end{align*}

Using Kirchhoff’s Voltage Law, the total voltage is

    \begin{align*}v=v_1+v_2+v_3+...+v_n\end{align*}

Substituting

    \begin{align*}v_n=L_n\frac{di}{dt}\end{align*}

Results in

    \begin{align*}v&=L_1\frac{di}{dt}+L_2\frac{di}{dt}+L_3\frac{di}{dt}+...+L_n\frac{di}{dt}\\&=(L_1+L_2+L_3+...+L_n)\frac{di}{dt}\\&=(\sum_{m=1}^nL_m)=L_{eq}\frac{di}{dt}\end{align*}

We can redraw the circuit above into

inductors in series and parallel formula 2

Hence, the equivalent inductance in series is

    \begin{align*}L_{eq}=L_1+L_2+L_3+...+L_n\end{align*}

In conclusion,

The equivalent inductance of n inductors in series is the sum of all the individual inductances.

The total inductance of series inductors is always bigger than the biggest individual inductance in the calculation.

As an example, assume that we have three inductors in series as seen below:

inductors in series and parallel formula 3

Utilizing inductors in series formula, we get

    \begin{align*}L_{eq}&=L_1+L_2+L_3\\&=\mbox{10mH+20mH+30mH}\\&=\mbox{60mH}\end{align*}

This way we can remember that both inductors and resistors have the same characteristic in series connection. But, is that also true for parallel inductors? Let’s read the next part.

Parallel Inductance Formula

Observe the circuit below consists of n inductors,

inductors in series and parallel formula 4

Since it is a parallel circuit, the voltage will be the same across all inductors, while each of them may have different current through them.

The voltage across the inductors are

    \begin{align*}v=v_1=v_2=v_3=v_n\end{align*}

The total current in the circuit is

    \begin{align*}i=i_1+i_2+i_3+...+i_n\end{align*}

Substituting

    \begin{align*}i_m=\frac{1}{L_m}\int_{t_0}^{t}v\space dt +i_m(t_0) \end{align*}

Results in

    \begin{align*}i=&\frac{1}{L_1}\int_{t_0}^{t}v\space dt+i_1(t_0)\\&+\frac{1}{L_2}\int_{t_0}^{t}v\space dt+i_2(t_0)\\&+...+\frac{1}{L_n}\int_{t_0}^{t}v\space dt +i_n(t_0)\\=&(\frac{1}{L_1}+\frac{1}{L_2}+...+\frac{1}{L_n})\int_{t_0}^{t}v\space dt+i_1(t_0)+i_2(t_0)+...+i_n(t_0)\\=&(\sum_{m=1}^n\frac{1}{L_m}\int_{t_0}^{t}v\space dt+\sum_{m=1}^ni_m(t_0)\\=&\frac{1}{L_{eq}}\int_{t_0}^{t}v\space dt+i(t_0)\end{align*}

We can redraw the circuit into

inductors in series and parallel formula 5

Hence, equivalent inductance in parallel is

    \begin{align*}\frac{1}{L_{eq}}=\frac{1}{L_1}+\frac{1}{L_2}+...+\frac{1}{L_n}\end{align*}

From Kirchhoff’s Current Law, the initial current i(t0) through Leq (at t=t0) is the sum of all the currents through each inductor, in at t0.

    \begin{align*}i(t_0)=i_1(t_0)+i_2(t_0)+i_3(t_0)+...+i_n(t_0)\end{align*}

If we only have two parallel inductors then we can simply use the equation below.

    \begin{align*}L_{eq}=\frac{L_1L_2}{L_1+L_2}\end{align*}

As an example, assume that we have three inductors in parallel as seen below:

inductors in series and parallel formula 6

Utilizing inductors in parallel formula, we get

    \begin{align*}\frac{1}{L_{eq}}&=\frac{1}{L_1}+\frac{1}{L_2}+\frac{1}{L_3}\\&=\frac{1}{10\mbox{mH}}+\frac{1}{20\mbox{mH}}+\frac{1}{30\mbox{mH}}\\&=\frac{6}{60\mbox{mH}}+\frac{3}{60\mbox{mH}}+\frac{2}{60\mbox{mH}}\\\frac{1}{L_{eq}}&=\frac{11}{60\mbox{mH}}\\L_{eq}&=\frac{60\mbox{mH}}{11}=5.4545\mbox{mH}\end{align*}

In conclusion,

The equivalent inductance of n inductors in parallel is the reciprocal of the sum of all the individual inductances.

The total inductance of parallel inductors is always smaller than the smallest individual inductance in the calculation.

From the inductance formula in series and parallel above, we can conclude that the way we find equivalent inductance is the same as when we find equivalent resistance.

Summary – Inductors in Series and Parallel Formula

After reading and learning the simple and complete explanations above, we can summary them in few lines:

  • The way we calculate equivalent series inductors will be same with how we calculate equivalent series resistors.
  • The way we calculate equivalent parallel inductors will be same with how we calculate equivalent parallel resistors.
  • The equivalent inductance of n inductors in series is the sum of all the individual inductances.
  • The equivalent inductance of n inductors in parallel is the reciprocal of the sum of all the individual inductances.

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