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} ) $