#### 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) Since, $L\dfrac{di}{dt}+Ri=0$
Also, $i=(\dfrac{V}{R})(1-e^{-(\frac{R}{L})t})$
$\implies i=(\dfrac{V}{R}) (1-e^{-3})\approx (0.9502) \dfrac{V}{R}$
Hence, 95% of the steady state value $\dfrac{V}{R}$
(b)Consider $L\dfrac{di}{dt}+Ri=0$
Also, $i=(\dfrac{V}{R})(1-e^{-(\frac{R}{L})t})$
Suppose $t=\dfrac{2L}{R}$, then we have $i=(\dfrac{V}{R})(1-e^{-(2)})$
$\implies i=(\dfrac{V}{R}) (1-e^{-3})\approx (0.8647) (\dfrac{V}{R})$
Hence, 86% of the steady state value $\dfrac{V}{R}$