Consider the circuit shown below. When switch `S_(1)` is closed, let I be the current at time t, then applying Kirchhoff's law
`E-iR-L (di)/(dt) = 0 or int_(0)^(i) (di)/(E-iR) = 1/L int_(0)^(1) dt`
`i=E/R (1-e^(-R/l *t))`
`L/R` = time constant of circuit
When current reaches its steady value `(=i_(0)`, open `S_(1)` and close `S_(2)`, the current does reach to zero finally but decays expnentially. The decay equation is given as `i=i_(0)e^(-R/L *T)`).

When a coil carrying a steady current is short circuited, the current decreases in it `eta` times in time `t_(0)`. The time constant of the circuit is

Options

A. `t_(0)1n eta`
B. `(t_(0))/(1n eta)`
C. `(t_(0))/(eta)`
D. `(t_(0))/(eta)-1`

Answer

B

Explanation

`I=I_(0)e^(-t//tau)`
Given , `I-cI_(0) at t=t_(0)`
`implies eta = e^(-t_(0)/tau)`
`implies 1n eta = -(t_0)/(tau)`
`implies tau = (t_0)/(1n eta)`.