An emf of `15 V` is applied in a circuit containing `5 H` inductance 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

Time constant, `tau_(L)=(L)/(R)=(1)/(2)`
We have, `i=i_(0)(1-e^(-t//tau_(L)))`
`therefore" "(i_(0))/(i)=(1)/(1-e^(-t//tau_(L)))`
Substituting, t = 1s and `tau_(L)=(1)/(2)s`, we get
`(i_(0))/(i)=(e^(2))/(e^(2)-1)`