Switch S is in position (1) for a long time at t = 0 comes

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Switch S is in position (1) for a long time at t = 0 comes to position (2). Find i _L (t)—

1. i (t) = (1.25 – 0.25) e ^{– t / 5 × 10³} + .25
2. i (t) = (1.25 – 0.25) e ^{– t / 5 × 10^–3} + .25
3. i (t) = (.25 – 1.25) e ^{– t / 0.5 × 10^–3} + .5
4. i (t) = (.25 – .25) e ^{– t / 0.5 × 10^–3} + .5

Correct Option: B

It may be noted that in this problem differential approach becomes complex. So we will apply here general approach.
We know that
i (t)=[i (0) – i (∞)] e ^{–t / τ} + i (∞) .....(i)

here i (0^–) = i (0⁺) =	50	= 1.25 amp.
	40

i (∞) =	50	= .25
	40

τ =		L	=	20 × 10^–3	= .5 × 10^–3
		R		40

so on putting these values in equation (i), we get
i _L (t) = (1.25 – 0.25) e ^{–1 5 × 10^–3} + .25