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

Options

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

Answer

D

Explanation

According to the question, The circuit as shown below ,
The current in the circuit,
`I(t) = I(infty)- [I(infty) - I(0) ]e^(-t//tau) " ". . . (i)`
`:.` Time constant , `tau = (L)/(R) = (5)/(10) = (1)/(2) sec`
at `t to infty` , inductor behaves as short circit hence,circuit will be
So, `I(infty) = (15)/(10) = 1.5A" ". . . (ii)`
at `t to 0` , inductor behaves as open circuit
so, no current flow in the circuit , I(0) = 0
So from Eq. (i) ,
`rArr I(t) = 1.5 - (1.5 - 0)e^((-1)/((1//2)))=1 . 5 (1 - e^(-2t))`
at `t = 1 sec rArr I(1) = 1.50 (1-e^(-2)) " ". . . (iii)`
From Eq. (ii) and (iii) , we get
`rArr (I(infty))/(I(1)) = (1.5)/(1.5 (1 - e^(-2)))=(e^(2))/(e^(2)-1)`