**This post answers the question: “How do you calculate inductors in series and parallel?”.**

Let’s consider series connection of two inductors depicted below.

Current through each conductor in the series connection $i\left(t\right)=\frac{{\lambda}_{1}\left(t\right)}{{L}_{1}}=\frac{{\lambda}_{2}\left(t\right)}{{L}_{2}}$, the total flux here $\lambda \left(t\right)={\lambda}_{1}\left(t\right)+{\lambda}_{2}\left(t\right)$. It means that effective inductance $L=\frac{\lambda \left(t\right)}{i\left(t\right)}=\frac{{\lambda}_{1}\left(t\right)+{\lambda}_{2}\left(t\right)}{{\displaystyle \frac{{\lambda}_{1}\left(t\right)}{{L}_{1}}}}={L}_{1}+{L}_{1}\frac{{\lambda}_{2}\left(t\right)}{{\lambda}_{1}\left(t\right)}={L}_{1}+{L}_{2}$.

Let’s now consider parallel combination of two inductors depicted below.

As soon as these inductors have equal voltage we can write that $\lambda \left(t\right)={L}_{1}{i}_{1}\left(t\right)={L}_{2}{i}_{2}\left(t\right)$. At the same time $i\left(t\right)={i}_{1}\left(t\right)+{i}_{1}\left(t\right)$. From here we can obtain $\frac{1}{L}=\frac{{i}_{1}\left(t\right)+{i}_{2}\left(t\right)}{{L}_{1}{i}_{1}\left(t\right)}=\frac{1}{{L}_{1}}+\frac{1}{{L}_{1}}\frac{{i}_{2}\left(t\right)}{{i}_{1}\left(t\right)}=\frac{1}{{L}_{1}}+\frac{1}{{L}_{2}}=\frac{{L}_{1}{L}_{2}}{{L}_{1}+{L}_{2}}$.

More educational content can be found at Reddit community **r/ElectronicsEasy.**