The battery shown in the figure is ideal.The values are `epsilon=10 V,R=5Omega,L=2H` Intially the current in the inductor is zero.The current through the battery at `t=2s` is

Options

A. `12 A`
B. `7 A`
C. `3 A`
D. None of these

Answer

A

Explanation

`I=I_(1)+I_(2)`
`I_(1)=E//R`
`L=(dI)/(dt)=E I_(2)=(Et)/L rArr I=E//R+(Et)/L rArr I=12A`