An emf of `15 V` is applied in a circuit containing `5 H` indutance and `10 Omega` resistance. The ratio of the currents at time `t = oo` and `t = 1 s` is

Options

A. `(e^(1//2))/(e^(1//2)-1)`
B. `(e^(2))/(e^(2)-1)`
C. `1 - e^(-1)`
D. `e^(-1)`

Answer

B

Explanation

`I_(t = oo) = I_(0)`
`I_(t = Is)= I_(0) (1 - e^(-(10)/(5) xx 1))`
`= I_(0) (I - e^(-2)) = I_(0) (1 - (1)/(e^(2)))`
`:. =(I_(t = oo))/(I_(t = Is)) = (I_(0)e^(2))/(I_(0)(e^(2)-1)) = (e^(2))/(e^(2) - 1)`