Question

In Fig. 30-77,${\mathbf{R}}_{\mathbf{1}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{\Omega }\mathbf{}\mathbf{,}{\mathbf{R}}_{\mathbf{2}}\mathbf{=}\mathbf{10}\mathbf{\Omega }\mathbf{}\mathbf{,}\mathbf{}{\mathbf{L}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{30}\mathbf{H}\mathbf{}\mathbf{,}\mathbf{}{\mathbf{L}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{H}$and the ideal battery has$\mathbf{\epsilon }\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{V}$. (a) Just after switch S is closed, at what rate is the current in inductor 1 changing? (b) When the circuit is in the steady state, what is the current in inductor 1?

Short answer

a) The rate of change of current in inductor 1 is,

$\frac{d{i}_{bat}}{dt}=20\mathrm{A}/\mathrm{s}$

b) The current in the inductor 1 is $i=0.75\mathrm{A}$

Step-by-step solution

Given

$$\begin{gathered} R_1=8.0\Omega \\ {R}_2=10\Omega \\ {L}_2=0.20\mathrm{H} \\ \mathrm{\epsilon }=6.0\mathrm{V} \end{gathered}$$

Understanding the concept

By using the equation 30-35, we can find the rate of current. After that by using the KVL

We can find the current in the inductor 1.

Formula:

$$\begin{gathered} i_{bat}=\frac{\epsilon }{R}\left(1-e^{\frac{Rt}{L}}\right) \\ \frac{d{i}_{bat}}{dt}=\frac{\left|\epsilon \right|}{L_1} \end{gathered}$$

(a) Calculate the rate of change of current in inductor 1.

We have, equation for self-induced emf as,

$L_1\frac{d{i}_{bat}}{dt}=\left|\mathrm{\epsilon }\right|$

So that the rate of current is,

$\frac{d{i}_{bat}}{dt}=\frac{\left|\mathrm{\epsilon }\right|}{L_1}$

By substituting the value

$$\begin{gathered} \frac{d{i}_{bat}}{dt}=\frac{6.0}{0.30} \\ \frac{d{i}_{bat}}{dt}=20\mathrm{A}/\mathrm{s} \end{gathered}$$

(b) Calculate the current in the inductor 1 at steady state.

Now we have to find the current in inductor 1 so that by Appling the KVL for outer loop we can writer as

$\mathrm{\epsilon }-{\mathrm{iR}}_1-\left|{\mathrm{\epsilon }}_{{\mathrm{L}}_1}\right|-\left|{\mathrm{\epsilon }}_{{\mathrm{L}}_2}\right|=0$

But as $t\to \infty emf$in the coil vanishes

$$\begin{gathered} \mathrm{\epsilon }-i{R}_1=0 \\ \mathrm{\epsilon }-{\mathrm{iR}}_1 \\ \mathrm{i}=\frac{\mathrm{\epsilon }}{{\mathrm{R}}_1} \\ \mathrm{i}=\frac{6.0}{8.0} \\ \mathrm{i}=0.75\mathrm{A} \end{gathered}$$