An inductor of inductance $L=400 \mathrm{mH}$ and resistors of resistances $R_{1}=2 \Omega$ and $R_{

Question

An inductor of inductance $L=400 \mathrm{mH}$ and resistors of resistances $R_{1}=2 \Omega$ and $R_{2}=2 \Omega$ are connected to a battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch $S$ is closed at $t=0$. The potential drop across $L$ as a function of time is\

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(d) : For the given $R$ - $L$ circuit the potential difference across $A D=V_{B C}$ as they are parallel.\ $I_{1}=E / R_{1} ; \quad I_{2}=I_{0}\left(1-e^{-t / 	au}ight)$ where\ $	au=$ mean life or $L / R . \quad 	au=t_{0}$ (given)\ $E($ across $B C)=L \frac{d I_{2}}{d t}+R_{2} I_{2}$\ $I_{2}=I_{0}\left(1-e^{-t / t 0}ight)$\ But $I_{0}=\frac{E}{R_{2}}=\frac{12}{2}=6 \mathrm{~A}$\ $	au=t_{0}=\frac{L}{R}=\frac{400 	imes 10^{-3} \mathrm{H}}{2 \Omega}=0.2 \mathrm{~s}$\ $	herefore \quad I_{2}=6\left(1-e^{-t / 0.2}ight)$\ Potential drop across $L=E-R_{2} I_{2}$\ $=12-2 	imes 6\left(1-e^{-t / 0.2}ight)=12 e^{-t / 0.2}=12 e^{-5 t} \mathrm{~V}$\