The time constant of an inductor is `tau_(1)`. When a pure resistor of `R Omega` is connected in series with it, the time constant is found to decrease to `tau_(2)`. The internal resistance of the inductor is

Options

A. `(R tau_(2))/(tau_(1)-tau_(2))`
B. `(R tau_(1))/(tau_(1)-tau_(2))`
C. `(R(tau_(1)-tau_(2)))/(tau_(1))`
D. `(R(tau_(1)-tau_(2)))/(tau_(2))`

Answer

A

Explanation

`tau_(1) = (L)/(r )," "tau_(2) = (L)/(R+r)` Solve for from the above equations