Answer
a) 95% of the steady state value $\dfrac{V}{R}$
b) 86% of the steady state value $\dfrac{V}{R}$
Work Step by Step
a) Here, we have
$L\dfrac{di}{dt}+Ri=0$
As we know that
$i=\dfrac{V}{R}(1-e^{-(\frac{R}{L})t})$
Then, we have $i=\dfrac{V}{R}(1-e^{-3})\approx (0.9502) \dfrac{V}{R}$
b) Here, we have
$L\dfrac{di}{dt}+Ri=0$
As we know that
$i=\dfrac{V}{R}(1-e^{-(\frac{R}{L})t})$
Plug $t=\dfrac{2L}{R}$
we get $i=\dfrac{V}{R}(1-e^{-2})$
$i=\dfrac{V}{R}(1-e^{-3})\approx (0.8647) \dfrac{V}{R}$
Hence, a) 95% of the steady state value $\dfrac{V}{R}$
b) 86% of the steady state value $\dfrac{V}{R}$
