An inductor of inductance $\mathrm{L}=400 \mathrm{mH}$ and resistors of resistances $\mathrm{R}_1$ $=2…
An inductor of inductance $\mathrm{L}=400 \mathrm{mH}$ and resistors of resistances $\mathrm{R}_1$ $=2 \Omega$ and $R_2=2 \Omega$ are connected to a battery of emf $12 \mathrm{~V}$ 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
$6 e^{-5 t} V$
$\frac{12}{t} e^{-3 t} V$
$6\left(1-e^{-t / 0.2}\right) V$
$12 e^{-5 t} V$
Solution
$
\begin{aligned}
& \mathrm{I}_1=\frac{\mathrm{F}}{\mathrm{R}_1}=\frac{12}{2}=6 \mathrm{~A} \\
& \mathrm{E}=\mathrm{L} \frac{\mathrm{dl_{2 }}}{\mathrm{dt}}+\mathrm{R}_2 \times \mathrm{I}_2 \\
& \mathrm{I}_2=\mathrm{I}_0\left(1-\mathrm{e}^{-\mathrm{t} / \mathrm{t}_{\mathrm{c}}}\right) \Rightarrow \mathrm{I}_{\mathrm{o}}=\frac{\mathrm{E}}{\mathrm{R}_2}=\frac{12}{2}=6 \mathrm{~A} \\
& \mathrm{t}_{\mathrm{c}}=\frac{\mathrm{L}}{\mathrm{R}}=\frac{400 \times 10^{-3}}{2}=0.2 \\
& \mathrm{I}_2=6\left(1-\mathrm{e}^{-\mathrm{t} / 0.2}\right)
\end{aligned}
$
Potential drop across $L=E-R_2 l_2=12-2 \times 6\left(1-e^{-b t}\right)=12 e^{-5 t}$
Directions: Question numbers 28,29 and 30 are based on the following paragraph. Two moles of helium gas are taken over the cycle $A B C D A$, as shown in the $P-T$ diagram.
