Q. An LR circuit is connected to a battery at time t = 0. The energy stored in the inductor reaches half its maximum value at time

(a) $ \displaystyle \frac{R}{L} ln \frac{\sqrt2}{\sqrt2 -1}$

(b) $ \displaystyle \frac{L}{R} ln \frac{\sqrt2-1}{\sqrt2 }$

(c) $ \displaystyle \frac{L}{R} ln \frac{\sqrt2}{\sqrt2 -1 }$

(d) $ \displaystyle \frac{R}{L} ln \frac{\sqrt2-1}{\sqrt2 }$

Ans: (c)

Sol: $ \displaystyle U = \frac{1}{2}U_{max} $

$ \displaystyle \frac{1}{2}Li^2 = \frac{1}{2}(\frac{1}{2}L i_0^2) $

$ \displaystyle i = \frac{i_0}{\sqrt{2}} $

$\displaystyle i_0(1-e^{-Rt/L}) = \frac{i_0}{\sqrt{2}} $

$\displaystyle (1-e^{-Rt/L}) = \frac{1}{\sqrt{2}} $

$ \displaystyle 1- \frac{1}{\sqrt{2}} = e^{-Rt/L} $

$ \displaystyle \frac{\sqrt{2}-1}{\sqrt{2}} = \frac{1}{e^{Rt/L} } $

$ \displaystyle e^{Rt/L} = \frac{\sqrt{2}}{\sqrt{2}-1} $

Taking log ,

$ \displaystyle ln (e^{Rt/L} ) = ln(\frac{\sqrt{2}}{\sqrt{2}-1} ) $

$ \displaystyle \frac{Rt}{L} = ln(\frac{\sqrt{2}}{\sqrt{2}-1} ) $

$ \displaystyle t = \frac{L}{R} ln(\frac{\sqrt{2}}{\sqrt{2}-1} ) $