An idal coil of 10 is connected in series with a resitance of `5Omega` and a battery of 5V. After 2s, after the connection is made, the current flowing ( in ampere) in the circuit is

Options

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

Answer

D

Explanation

Rise of current in L-R circuit is given by
`I=I_(0)(1-e^(-t//tau))`
where, `I_(0) = E/R=5/5=1A`
Now, `tau=L/R = 10/5=2s`
After 2s, i.e. at t=2s
Rise of current, `I=(1-e^(-1))`A