In the circuit of Fig. 30-76, ${\mathit{R}}_{\mathbf{1}}\mathbf{=}\mathbf{20}\mathit{k}\mathit{\Omega }\mathbf{}\mathbf{,}\mathbf{}{\mathit{R}}_{\mathbf{2}}\mathbf{=}\mathbf{20}\mathit{\Omega }\mathbf{}\mathbf{,}\mathbf{}\mathit{L}\mathbf{=}\mathbf{50}\mathit{m}\mathbf{}\mathit{H}$and the ideal battery has $\mathit{\epsilon }\mathbf{=}\mathbf{40}\mathbf{}\mathit{V}$. Switch S has been open for a long time when it is closed at time $\mathit{t}\mathbf{=}\mathbf{0}$. Just after the switch is closed, what are (a) the current ${\mathit{i}}_{\mathbf{b}\mathbf{a}\mathbf{t}}$through the battery and (b) the rate $\frac{\mathbf{d}{\mathbf{i}}_{\mathbf{b}\mathbf{a}\mathbf{t}}}{\mathbf{d}\mathbf{t}}$? At $\mathit{t}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathit{\mu }\mathit{s}$, what are (c) ${\mathit{i}}_{\mathbf{b}\mathbf{a}\mathbf{t}}$and (d) $\frac{\mathbf{d}{\mathbf{i}}_{\mathbf{bat}}}{\mathbf{d}\mathbf{t}}$? A long time later, what are (e) ${\mathit{i}}_{\mathbf{b}\mathbf{a}\mathbf{t}}$and (f) $\frac{{\mathbf{di}}_{\mathbf{bat}}}{\mathbf{dt}}$?

Answer is missing

By using the equation 30-35 and the equation 30-40 we can find the rate of current as well as theemfin the coil at various times

Formula:

$$\begin{gathered} i=\frac{\epsilon }{R}\left(1-e^{-\frac{Rt}{L}}\right) \\ {\epsilon }_L=-L\frac{di}{dt} \end{gathered}$$

Given

$$\begin{gathered} R_1=20k\Omega \\ {R}_2=20\Omega \\ L=50mH=50\times {10}^{-3}H \\ \epsilon =40V \end{gathered}$$

As from the equation 30.40 the current in the inductance can be written as

$i=\frac{\epsilon }{R}\left(1-e^{-\frac{Rt}{L}}\right)$

At$t=0$the current is

$$\begin{gathered} i=\frac{\epsilon }{R}\left(1-1\right) \\ i=0A \end{gathered}$$

According to equation 30-35 an inducedappearsin any coil in which the current is changing is

${\epsilon }_L=-L\frac{di}{dt}$

At the$t=0emf$ is same as that of battery voltage so that the rate of current through the battery is

$$\begin{gathered} \epsilon =-L\frac{d{i}_{batt}}{dt} \\ \frac{d{i}_{batt}}{dt}=\frac{\left|\epsilon \right|}{L} \\ \frac{d{i}_{batt}}{dt}=\frac{40}{0.050} \\ \frac{d{i}_{batt}}{dt}=800\mathrm{A}/\mathrm{s} \end{gathered}$$

First find theequivalentresistance of the circuit,

$R_1$And$R_2$are in parallel combination so that the equivalence resistance is

$$\begin{gathered} R_{eq}=\frac{R_1{R}_2}{R_1+R_2} \\ {R}_{eq}=\frac{20000\times 20}{20000+20} \\ {R}_{eq}=\frac{400000}{20020} \\ {R}_{eq}\approx 20\Omega \end{gathered}$$

From equation 30-34 we can write, the current through battery is

$$\begin{gathered} i=\frac{\epsilon }{R}\left(1-e^{-\frac{Rt}{L}}\right) \\ i=\frac{\epsilon }{R}\left(1-e^{-\frac{R_{eq}t}{L}}\right) \\ i=\frac{40}{20}\left(1-e^{-\frac{20\times 3\times {10}^{-6}}{0.05}}\right) \\ i=2\times \left(1-e^{-\frac{6}{5}}\right) \\ i=1.39A \end{gathered}$$

The rate of change of the current is

$$\begin{gathered} {\epsilon }_L=-L\frac{d{i}_{bat}}{dt} \\ \frac{d{i}_{bat}}{dt}=\frac{\left|{\epsilon }_L\right|}{L} \end{gathered}$$

So we have to first find the ${\epsilon }_L$as

From equation $i=\frac{\epsilon }{R_{eq}}\left(1-e^{-\frac{Rt}{L}}\right)$from the loop rule

$$\begin{gathered} i{R}_{eq}=\epsilon -\epsilon e^{-\frac{Rt}{L}} \\ i{R}_{eq}=\epsilon -{\epsilon }_L \\ 0=\epsilon -{\epsilon }_L-i{R}_{eq} \\ \left|{\epsilon }_L\right|=\epsilon -i_{bat}{R}_{eq} \end{gathered}$$

By substituting the value

$$\begin{gathered} \left|{\epsilon }_L\right|=40-i_{bat}\times 20 \\ \left|{\epsilon }_L\right|=40-1.39\times 20 \\ \left|{\epsilon }_L\right|=12.2V \\ \therefore 12.2=50\times {10}^{-3}.\frac{d{i}_{bat}}{dt} \\ \frac{d{i}_{bat}}{dt}=\frac{12.2}{50\times {10}^{-3}} \\ \frac{d{i}_{bat}}{dt}=2.44\times {10}^{2}A/s \end{gathered}$$

As$t\to \infty$, the circuit reaches steady state so that the equation becomes

$$\begin{gathered} i_{bat}=\frac{\epsilon }{R}\left(1-e^{-\frac{Rt}{L}}\right) \\ {i}_{bat}=\frac{\epsilon }{R}\left(1-0\right) \\ {i}_{bat}=\frac{\epsilon }{R} \end{gathered}$$

By substituting the value, where R is the equivalent resistance

$$\begin{gathered} i_{bat}=\frac{\epsilon }{R_{eq}} \\ {i}_{bat}=2A \end{gathered}$$

As$t\to \infty$the circuit is in steady state condition so,

$\frac{d{i}_{bat}}{dt}=0$