Q79P - Fundamentals Of Physics

Question

In Fig. 30-71, the battery is ideal and ${\mathbf{\epsilon }}_{\mathbf{battery}}\mathbf{=}\mathbf{10}\mathbf{ }\mathbf{V}\mathbf{ }\mathbf{,}\mathbf{ }{\mathbf{R}}_{\mathbf{1}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\Omega }\mathbf{ }\mathbf{,}{\mathbf{R}}_{\mathbf{2}}\mathbf{=}\mathbf{10}\mathbf{\Omega }\mathbf{,}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{L}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{H}$ . Switch S is closed at time t = 0. Just afterwards, what are (a),${\mathit{I}}_{\mathbf{1}}$(b) ${\mathit{I}}_{\mathbf{2}}$ (c) the current is through the switch, (d) the potential difference ${\mathit{V}}_{\mathbf{2}}$ across resistor 2, (e) the potential difference ${\mathit{V}}_{\mathbf{L}}$ across the inductor, and (f) the rate of change $\mathit{d}{\mathit{I}}_{\mathbf{2}}\mathbf{/}\mathit{d}\mathit{t}$? A long time later, what are (g)${\mathit{I}}_{\mathbf{1}}$, (h)${\mathit{I}}_{\mathbf{2}}$, (i) ${\mathit{I}}_{\mathit{s}}$, (j) ${\mathit{V}}_{\mathbf{2}}$, (k) ${\mathit{V}}_{\mathbf{L}}$, and (l)$\mathit{d}{\mathit{I}}_{\mathbf{2}}\mathbf{/}\mathit{d}\mathit{t}$ ?

Step-by-Step Solution

Answer

a)   Current $I_1$will be 2.0 A

b)   Current$I_2$will be 0.0 A

c)   Current through the switch $I_s$ will be 2.0 A 

d)   Potential difference across resistor 2 $V_2$will be 0.0 V

e)   Potential difference across inductor $V_L$ will be 10.0 V

f)   Rate of change of current $d{I}_2/dt$ will be 2.0 A/s

g)  A long time later value for $I_2$ will be 2.0 A

h)  A long time later value for $I_2$ will be 1.0 A

i)   A long time later value for $I_s$ will be 3.0 A 

j)   A long time later value for $V_2$ will be 10.0 V

k)  A long time later value for $V_L$ will be 0.0 V

l)   A long time later value for $d{I}_2/dt$ will be 0.0 A/s

Step 1: Given

$$\begin{gathered} {\mathrm{\epsilon }}_{\mathrm{battery}}=10 \mathrm{V} \\ {\mathrm{R}}_1=5.0\mathrm{\Omega } \\ {\mathrm{R}}_2=10\mathrm{\Omega } \\ \mathrm{L}=5.0\mathrm{H} \end{gathered}$$

Step 2: Understanding the concept

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

Formula:

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

Step 3: (a) Calculate current / 1        

When the switch is closed,

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

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

Step 4: (b) Calculate current   I 2

As the switch is closed,

${\epsilon }_{battery}=V_L$

From this we can say that,

$I_2=0.0 A$

Step 5: (c) Calculate current through the switch, I s

We apply junction rule, we get,

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

Step 6: (d) Calculate potential difference across resistor 2, V 2

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

So the potential difference across it will be,

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

Step 7: (e) Calculate potential difference across inductor V L

As the switch is closed, 

$$\begin{gathered} {\epsilon }_{battery}=V_L \\ V_L=10 V \end{gathered}$$

Step 8: (f) Calculate rate of change of current d I 2 / d t

We have,

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

Step 9: (g) Calculate a long time later value for I 1

After certain time we still have,

$$\begin{gathered} V_1={\epsilon }_{battery} \\ {V}_1=10 V \end{gathered}$$

According to ohm’s law,

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

Step 10: (h) Calculate a long time later value for

After certain time,

$$\begin{gathered} V_L=0 \mathrm{and} V_2={\epsilon }_{battery} \\ {V}_2-10V \end{gathered}$$

According to ohm’s law,

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

Step 11: (i) Calculate a long time later value for I s

We apply junction rule, we get, 

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

Step 12: (j) Calculate a long time later value for V 2

After certain time,

$$\begin{gathered} V_2={\epsilon }_{battery} \\ {V}_2=10.0 V \end{gathered}$$

Step 13: (k) Calculate a long time later value for V L

After certain time,

$V_L=0.0 V$

Step 14: (l) Calculate a long time later value for d I 2 / d t

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

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