2.1f.3ii Inductor Parallel Equivalence

Parallel Equivalence

Two inductors in parallel are equivalent to

\usepackage{circuitikz}
\begin{document}
\begin{circuitikz}[american]
\draw (-4.5, 6) to [short] (0, 6)
to [short, i=$I_{test}(t)$] (0, 5)
to [short] (1.5, 5)
to [inductor, l=$L_1$, v=$V_1(t)$, i>^=$I_1(t)$] (1.5, 1)
to [short] (0,1)
(0, 5) to [short] (-1.5, 5)
to [inductor, l=$L_2$, v=$V_2(t)$, i>^=$I_2(t)$] (-1.5, 1)
to [short] (0, 1)
to [short] (0, 0)
to [short] (-4.5, 0)
to [I, l=$I_{test}$, v=$V_{test}(t)$] (-4.5, 6);
\draw [->] (3, 3)--(5, 3) node[below=5pt, xshift=-28]{$L_{eq}=\left(\frac{1}{L_1}+\frac{1}{L_2}\right)^{-1}$};
\draw (9,6) to [inductor, v=$V_{eq}(t)$, i>^=$I_{test}(t)$, L=$L_{eq}$] (9,0)
 to [short] (7, 0)
 to [I, l=$I_{test}$] (7, 6)
 to [short] (9, 6);

\end{circuitikz}
\end{document}

Intuition

Using NVA,

$$\begin{array}{r}{V}_{test}=V_1=V_2\\ L_{eq}\frac{d{I}_{test}}{dt}=L_1\frac{d{I}_1}{dt}+L_2\frac{d{I}_2}{dt}\end{array}$$

Since by KCL $I_{eq}=I_1+I_2$,
$\frac{d{I}_{test}}{dt}=\frac{d{I}_1}{dt}+\frac{d{I}_2}{dt}.$
From the first and second equations,
$\frac{d{I}_{test}}{dt}=\frac{L_{eq}}{L_1}\frac{d{I}_{test}}{dt}+\frac{L_{eq}}{L_2}\frac{d{I}_{test}}{dt}.$
Simplifying, we get
$L_{eq}={\left(\frac{1}{L_1}+\frac{1}{L_2}\right)}^{-1}.$