An ideal coil of 10H is connected in series with a resistance of `5(Omega)` and a battery of 5V. 2second after the connections is made, the current flowing in ampere in the circuit is

Options

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

Answer

A

Explanation

Here,`L = 10 h, R = 5 Omega, E = 5 V`
Max. current, `I_(0) = (E)/(R ) = (5)/(5) = 1 A`
During growth of current
`I = I_(0) (1 - e^((-R )/(L) t))`
`:. I = 1 (1 - e^((-5)/(10) xx 2)) = (1 - e^(-1))`