A unit step u(t – 5) is applied to the RL network:The

1. 1 – e^{– t}
2. [1 – e^{– (t – 5)}] u(t – 5)
3. (1 – e^{– t}) u(t – 5)
4. 1 – e^{– (t – 5)}

I(∞) = I(∞) [1 – e^–R/Lt]

I(∞) =	V	=	u(t - 5)
	R		1

= u(t – 5) [1 – e^–1/1t]
Alternative Method In frequency domain

e^–5s	=I(s) + I(s) s
s

I(s) =	e^–5s
	(s + 1)s

.............. (i)

I(s) = e^–5s		1	-	20
		s		1 + s

or i(t) = [1 – e–t] u [t – 5]
Hence alternative (C) is the correct choice.