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

`I= I_(0)(1 - e^((-R)/(L)t))` (when current is in growth in LR circuit)
`= (E)/(R) (1 - e^((-R)/(L)t)) = (5)/(5) (1 - e ^((-5)/(10) xx 2)) = (1 - e^(-1))`