In the figure below, the switches S _ 1 and S _ 2 are closed...

The current in the middle, $I$, as shown in the figure above has a maximum value,
{i}_\max = {\left({i}_{2} - {i}_{1}\right)}_\max

Using the equation of current decay in an $L - R$ circuit, for the given values in the figure above,
$i = \left({i}_{2} - {i}_{1}\right) = \dfrac{V}{R}\left[1 - {e}^{- \left(\dfrac{R}{2L}\right)t}\right] - \dfrac{V}{R}\left[1 - {e}^{\left(- \dfrac{R}{L}\right)t}\right]$
$\dfrac{V}{R}\left[{e}^{\left(- \dfrac{R}{L}\right)t} - {e}^{\left(- \dfrac{R}{2L}\right)t}\right]$
For {\left(i\right)}_\max, \dfrac{d\left(i\right)}{dt} = 0
$⇒ \dfrac{V}{R}\left[- \dfrac{R}{L}{e}^{- \left(\dfrac{R}{L}\right)t} - \left(- \dfrac{R}{2L}\right){e}^{- \left(\dfrac{R}{2L}\right)t}\right] = 0$
$⇒ {e}^{- \left(\dfrac{R}{L}\right)t} = \dfrac{1}{2}{e}^{- \left(\dfrac{R}{2L}\right)t}$
$⇒ {e}^{- \left(\dfrac{R}{2L}\right)t} = \dfrac{1}{2}$
$⇒ \left(\dfrac{R}{2L}\right)t = \ln 2$
$t = \dfrac{2L}{R}\ln 2 \to$ time when I is maximum
{i}_\max = \dfrac{V}{R}\left[{e}^{- \dfrac{R}{L}\left(\dfrac{2L}{R}\ln 2\right)} - {e}^{- \left(\dfrac{R}{2L}\right)\left(\dfrac{2L}{R}\ln 2\right)}\right]
⇒ \left|{i}_\max\right| = \dfrac{V}{R}\left|\left[\dfrac{1}{4} - \dfrac{1}{2}\right]\right| = \dfrac{1}{4}\dfrac{V}{R}