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In the circuit shown in the given figure, the values of I(0+) and I(∞), will be, respectively
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- zero and 1.5 A
- 1.5 A and 3 A
- 3 A and zero
- 3 A and 1.5 A
Correct Option: C
| I(s) = | |||||
| (s + 1) |  | 1 + |  | ||
| 2s | |||||
| = | ||
| (s + 0.5)(s + 1) |
| = | + | |||
| (s + 0.5) | (s + 1) |
⇒ i (t) = L– 1(I(s)) = 6e– t – 3e– 0.5t
At t = 0, i (0+) = 3A,
and at t = ∞, i (∞) = 0
