An ideal coil of `10` henry is joined in series with a resistance of `5` ohm and a battery of `5` volt. `2` second after joining, the current flowing in ampere in the circuit will be

Options

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

Answer

B

Explanation

Growth of current in the circuit is given by
`i=i_(0)(1-e^(-Rt//L))`
where, `i_(0)` is peck value of current and
`i_(0)=(5)/(5)=1A`
`therefore" "i=1(1-e^(-5xx2//10))=(1-e^(-1))A`