Answer
$t=\dfrac{L}{R} \ln (2)$ seconds
Work Step by Step
Here, we have
$\dfrac{di}{dt}+\dfrac{R}{T}i=\dfrac{V}{L}$
As we know that
$i=\dfrac{V}{R}(1-e^{-(\frac{R}{L})t})$
The value of steady current becomes: $i=(\dfrac{1}{2})(\dfrac{V}{R})$
Then, we have $(\dfrac{1}{2})(\dfrac{V}{R})=\dfrac{V}{R}(1-e^{-(\frac{R}{L})t})$
$\implies \ln (1)-\ln (2)=\ln e^{-(\frac{R}{L})t}$
Thus, we get $t=\dfrac{L}{R} \ln (2)$ seconds
