91Ó°ÊÓ

$$\begin{gathered} {\mathrm{Î\mu }}_{\mathrm{battery}}=10\mathrm{V} \\ {\mathrm{R}}_1=5.0\mathrm{Î\copyright } \\ {\mathrm{R}}_2=10\mathrm{Î\copyright } \\ \mathrm{L}=5.0\mathrm{H} \end{gathered}$$

We can use the loop rule and junction rule to get the required answers.

Formula:

$$\begin{gathered} \mathrm{i})\mathrm{For}\mathrm{Any}\mathrm{loop}\mathrm{Î\mu }+\underset{}{∑/\mathrm{R}=0} \\ \mathrm{ii})\mathrm{At}\mathrm{any}\mathrm{junction}∑/=0 \\ \mathrm{iii})\mathrm{V}=/\mathrm{R} \\ \mathrm{iv})\frac{\mathrm{d}/}{\mathrm{dt}}=\frac{\mathrm{V}}{\mathrm{L}} \end{gathered}$$

When the switch is closed,

We apply the loop rule to the left loop, we get,

$$\begin{gathered} \mathrm{Î\mu }+\underset{}{∑/\mathrm{R}=0} \\ {\mathrm{Î\mu }}_{\mathrm{battery}}-{/}_1{\mathrm{R}}_1=0 \\ {\mathrm{Î\mu }}_{\mathrm{battery}}-{/}_1{\mathrm{R}}_1 \\ {\mathrm{I}}_1=\frac{{\mathrm{Î\mu }}_{\mathrm{battery}}}{{\mathrm{R}}_1} \\ {\mathrm{I}}_1=\frac{10}{5} \\ {\mathrm{I}}_1=2.0\mathrm{A} \end{gathered}$$

As the switch is closed,

${\mathrm{Î\mu }}_{battery}=V_L$

From this we can say that,

$I_2=0.0A$

We apply junction rule, we get,

$$\begin{gathered} I_s=I_1+I_2 \\ {I}_s=2.0+0.0 \\ {I}_s=2.0A \end{gathered}$$

As the current flowing the$R_2$i.e.$I_2=0$,

So the potential difference across it will be,

$$\begin{gathered} V_2=I_2{R}_2 \\ {V}_2=0×10 \\ {V}_2=0.0V \end{gathered}$$

As the switch is closed,

$$\begin{gathered} {\mathrm{Î\mu }}_{battery}=V_L \\ {V}_L=10V \end{gathered}$$

We have,

$$\begin{gathered} \frac{dI}{dt}=\frac{V}{L} \\ For\frac{d{I}_2}{dt},\mathrm{will}\mathrm{be} \\ \frac{d{I}_2}{dt}=\frac{10}{5} \\ \frac{d{I}_2}{dt}=2.0A/s \end{gathered}$$

After certain time we still have,

$$\begin{gathered} V_1={\mathrm{Î\mu }}_{battery} \\ {V}_1=10V \end{gathered}$$

According to ohm’s law,

$$\begin{gathered} V=IR \\ {V}_1=I_1{R}_1 \\ {I}_1=\frac{V_1}{R_1} \\ {I}_1=\frac{10}{5} \\ {I}_1=2.0A \end{gathered}$$

After certain time,

$$\begin{gathered} V_L=0\mathrm{and}{V}_2={\mathrm{Î\mu }}_{battery} \\ {V}_2-10V \end{gathered}$$

According to ohm’s law,

$$\begin{gathered} V=IR \\ {V}_2=I_2{R}_2 \\ {I}_2=\frac{V_2}{R_2} \\ {I}_2=\frac{10}{10} \\ {I}_2=1.0A \end{gathered}$$

We apply junction rule, we get,

$$\begin{gathered} I_s=I_1+I_2 \\ {I}_s=2.0+1.0 \\ {I}_s=3.0A \end{gathered}$$

After certain time,

$$\begin{gathered} V_2={\mathrm{Î\mu }}_{battery} \\ {V}_2=10.0V \end{gathered}$$

After certain time,

$V_L=0.0V$

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{\mathrm{dI}}{\mathrm{dt}}=\frac{\mathrm{V}}{\mathrm{L}} \end{gathered}$$

$$\begin{gathered} \mathrm{For}\frac{{\mathrm{dI}}_2}{\mathrm{dt}},\mathrm{will}\mathrm{be} \\ \frac{{\mathrm{dI}}_2}{\mathrm{dt}}=\frac{{\mathrm{V}}_{\mathrm{L}}}{\mathrm{L}} \\ \frac{{\mathrm{dI}}_2}{\mathrm{dt}}=\frac{0}{5} \\ \frac{{\mathrm{dI}}_2}{\mathrm{dt}}=0.0\mathrm{A}/\mathrm{s} \end{gathered}$$