The growth of current in two series LR circuits A and B is shown in graph C. Let L₁, R₁ and L₂, R₂ be the corresponding inductance, resistance values in the two circuits. Then

Since the peak value (steady state) current for both 1 and 2 is the same. So, we have

 $\frac{\mathrm{V}}{{\mathrm{R}}_1}=\frac{\mathrm{V}}{{\mathrm{R}}_2}$

$\Rightarrow {\mathrm{R}}_1={\mathrm{R}}_2$

Further 2 grows faster than 1, so we conclude that

$\begin{array}{l}{\mathrm{\tau }}_{{\mathrm{L}}_2}<{\mathrm{\tau }}_{{\mathrm{L}}_1}\\ \Rightarrow {\mathrm{L}}_2<{\mathrm{L}}_1 \left\{\because {\mathrm{\tau }}_{\mathrm{L}}=\frac{\mathrm{L}}{\mathrm{R}}\right\}\end{array}$