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

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 ,

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

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

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