Consider the circuit shown in Fig. P30.71. Switch $S$ is closed at time $t=0,$ causing a current $i_{1}$ through the inductive branch and a current $i_{2}$ through the capacitive branch. The initial charge on the capacitor is zero, and the charge at time $t$ is $q_{2}$.
(a) Derive expressions for $i_{1}, i_{2},$ and $q_{2}$ as functions of time. Express your answers in terms of $\mathcal{E}, L, C, R_{1}, R_{2},$ and $t$. For the remainder of the problem let the circuit elements have the following values: $\mathcal{E}=48 \mathrm{~V}, L=8.0 \mathrm{H}, C=20 \mu \mathrm{F}, R_{1}=25 \Omega,$ and $R_{2}=5000 \Omega$
(b) What is the initial current through the inductive branch? What is the initial current through the capacitive branch? (c) What are the currents through the inductive and capacitive branches a long time after the switch has been closed? How long is a "long time"? Explain. (d) At what time $t_{1}$ (accurate to two significant figures) will the currents $i_{1}$ and $i_{2}$ be equal? (Hint: You might consider using series expansions for the exponentials.) (e) For the conditions given in part (d), determine $i_{1}$.
(f) The total current through the battery is $i=i_{1}+i_{2}$. At what time $t_{2}$ (accurate to two significant figures) will $i$ equal one-half of its final value? (Hint: The numerical work is greatly simplified if one makes suitable approximations. A sketch of $i_{1}$ and $i_{2}$ versus $t$ may help you decide what approximations are valid.)