GATE 2015 EE - SET 2 - Question 18

Solution :

Current in an RL circuit ,

i(t)= \frac{V}{R}(1-e^\frac{-t}{\tau }) where \tau = \frac{L}{R}

\frac{di}{dt} = \frac{V}{R}*\frac{1}{\tau }e^{\frac{-t}{\tau }}

\Rightarrow \frac{V}{R}*\frac{R}{L} .e ^{\frac{-t}{\tau }}

\frac{di}{dt} = \frac{V}{L} e^{\frac{-t}{\tau }}

\frac{di^{2}}{dt^{2}} = \frac{V}{L} (\frac{-1}{\tau })e^{\frac{-t}{\tau }}

\Rightarrow \frac{-V}{L\tau } e^{\frac{-t}{\tau }}

\Rightarrow \frac{-V}{L(\frac{L}{R}) } e^{\frac{-t}{\tau }}

\frac{di^{2}}{dt^{2}} = \frac{-VR}{L^{2}} e^{\frac{-t}{\tau }}

\frac{di^{2}}{dt^{2}} = \frac{-VR}{L^{2}} e^{0} = \frac{-VR}{L^{2}}