An inductor of inductance L=400,mH and resist | vedclass

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(C) When the switch $S$ is closed at $t=0$,the inductor $L$ and resistor $R_2$ form a series circuit connected across the battery of emf $E=12\,V$. Th

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$12 e^{-5 t}\,V$

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(C) When the switch $S$ is closed at $t=0$,the inductor $L$ and resistor $R_2$ form a series circuit connected across the battery of emf $E=12\,V$. The resistor $R_1$ is in parallel with this branch and does not affect the current through the inductor.The current $i$ in the $L-R_2$ branch at time $t$ is given by:$i = \frac{E}{R_2} (1 - e^{-R_2 t / L})$The potential drop across the inductor $L$ is given by $V_L = L \frac{di}{dt}$.First,find $\frac{di}{dt}$:$\frac{di}{dt} = \frac{E}{R_2} \cdot \frac{R_2}{L} \cdot e^{-R_2 t / L} = \frac{E}{L} e^{-R_2 t / L}$Now,calculate $V_L$:$V_L = L \left( \frac{E}{L} e^{-R_2 t / L} \right) = E e^{-R_2 t / L}$Given $E = 12\,V$,$R_2 = 2\,\Omega$,and $L = 400\,mH = 0.4\,H$:$\frac{R_2}{L} = \frac{2}{0.4} = 5\,s^{-1}$Therefore,$V_L = 12 e^{-5t}\,V$.

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