# GATE 2015 EE – SET 2 – Question 18

Solution :

Current in an RL circuit ,

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

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

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

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

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

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

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

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

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