In adjacent circuit, switch S is closed at t = 0. The time at which current in the circuit becomes half of the steady current is

Options

A. `tau ln 2`
B. `(ln2)/(tau)`
C. `2 tau ln 2`
D. `(tau)/(2)ln2`

Answer

A

Explanation

`i = i_(0)(1-e^(-vtau))`
`(i_(0))/(2) = i_(0)(1 - e^(-vtau))`
`e^(vtau) = 2`
`t = tau In 2`.