91Ó°ÊÓ

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:

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

Given

$$\begin{gathered} R_1=20k\mathrm{Î\copyright } \\ {R}_2=20\mathrm{Î\copyright } \\ L=50mH=50×{10}^{-3}H \\ \mathrm{Î\mu }=40V \end{gathered}$$

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

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

At$t=0$the current is

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

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

${\mathrm{Î\mu }}_L=-L\frac{di}{dt}$

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

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

First find theequivalentresistance of the circuit,

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

$$\begin{gathered} R_{eq}=\frac{R_1{R}_2}{R_1+R_2} \\ {R}_{eq}=\frac{20000×20}{20000+20} \\ {R}_{eq}=\frac{400000}{20020} \\ {R}_{eq}≈20\mathrm{Î\copyright } \end{gathered}$$

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

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

The rate of change of the current is

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

So we have to first find the ${\mathrm{Î\mu }}_L$as

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

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

By substituting the value

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

As$t→∞$, the circuit reaches steady state so that the equation becomes

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

By substituting the value, where R is the equivalent resistance

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

As$t→∞$the circuit is in steady state condition so,

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