In the circuit shown in figure `L=10H, R=5Omega, E=15V.` The switch `S` is closed at `t=0`. At `t=2s` the current in the circuit is

Options

A. `3(1-1/e)A`
B. `3(1-1/e^2)A`
C. `3(1/e)A`
D. `3(1/e^2)A`

Answer

A

Explanation

`tau_L=L/R=2s`
`i_0=E/R=3A, t=2s`
`i=i_0(1-e^(-t//tau_L))`
Substituting the given values, we can find i.