An inductor of inductance L = 400 mH and res | vedclass

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(D) When the switch $S$ is closed at $t = 0$,the branch containing $R_1$ is in parallel with the branch containing $L$ and $R_2$. The potential differ

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

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(D) When the switch $S$ is closed at $t = 0$,the branch containing $R_1$ is in parallel with the branch containing $L$ and $R_2$. The potential difference across the $L-R_2$ branch is equal to the emf of the battery,$E = 12 \ V$.The current $i$ in the $L-R_2$ branch grows according to the equation: $i = \frac{E}{R_2}(1 - e^{-R_2 t / L})$.The potential drop across the inductor $L$ is given by $V_L = L \frac{di}{dt}$.Differentiating the current expression with respect to time $t$:$\frac{di}{dt} = \frac{E}{R_2} \cdot \frac{R_2}{L} e^{-R_2 t / L} = \frac{E}{L} e^{-R_2 t / L}$.Substituting this into the expression for $V_L$:$V_L = L \left( \frac{E}{L} e^{-R_2 t / L} ight) = E e^{-R_2 t / L}$.Given $E = 12 \ V$,$R_2 = 2 \ \Omega$,and $L = 400 \ mH = 0.4 \ H$,the time constant $	au = \frac{L}{R_2} = \frac{0.4}{2} = 0.2 \ s$.Thus,$V_L = 12 e^{-t / 0.2} = 12 e^{-5t} \ V$.

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