2.1f.3i Inductor Series Equivalence

Series Equivalence

Two inductors in series are equivalent to

\usepackage{circuitikz}
\begin{document}
\begin{circuitikz}[american]
 \draw (3,3) to [inductor, v=$V_1(t)$, i>^=$I_{test}(t)$, L=$L_1$] (3,0)
 to [inductor, v=$V_2(t)$, L=$L_2$] (3, -3)
 to [short] (0, -3)
 to [I, l=$I_{test}$] (0, 3)
 to [short] (3, 3);
 \draw [->] (4, 0)--(6, 0) node[below=5pt, xshift=-28]{$L_{eq}=L_1+L_2$};
 \draw (10,2) to [inductor, v=$V_{eq}(t)$, i>^=$I_{test}(t)$, L=$L_{eq}$] (10,-2)
 to [short] (8, -2)
 to [I, l=$I_{test}$] (8, 2)
 to [short] (10, 2);

\end{circuitikz}
\end{document}

Proof

Using $V_{test}=V_{eq}$ and KVL, we deduce that
$V_{eq}=V_1+V_2.$
We then plug-in the equations of the inductors,
$V_{eq}=L_1\frac{d{I}_{test}}{dt}+L_2\frac{d{I}_{test}}{dt}$
Which leads to
$V_{eq}=(L_1+L_2)\frac{d{I}_{test}}{dt}.$