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 localid="1664179721041" ${\mathit{V}}_{\mathbf{2}}$across resistor 2, (e) the potential difference localid="1664179725185" ${\mathit{V}}_{\mathbf{L}}$across the inductor, and (f) the rate of change localid="1661859249391" $\mathit{d}{\mathit{I}}_{\mathbf{2}}\mathbf{/}\mathit{d}\mathit{t}$? A long time later, what are (g)${\mathit{I}}_{\mathbf{1}}$, (h)${\mathit{I}}_{\mathbf{2}}$, (i) localid="1664179712268" ${\mathit{I}}_{\mathit{s}}$, (j) localid="1664180045587" ${\mathit{V}}_{\mathbf{2}}$, (k) localid="1664180042725" ${\mathit{V}}_{\mathbf{L}}$, and (l)localid="1664180040049" $\mathit{d}{\mathit{I}}_{\mathbf{2}}\mathbf{/}\mathit{d}\mathit{t}$?

Short 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-by-step solution

Given

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

Understanding the concept

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

Formula:

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

(a) Calculate current /1    

When the switch is closed,

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

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

(b) Calculate current  I2

As the switch is closed,

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

From this we can say that,

$I_2=0.0A$

(c) Calculate current through the switch, Is

We apply junction rule, we get,

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

(d) Calculate potential difference across resistor 2, V2

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

So the potential difference across it will be,

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

(e) Calculate potential difference across inductor VL

As the switch is closed,

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

(f) Calculate rate of change of current dI2/dt

We have,

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

(g) Calculate a long time later value for I1

After certain time we still have,

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

According to ohm’s law,

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

(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,

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

(i) Calculate a long time later value for Is

We apply junction rule, we get,

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

(j) Calculate a long time later value for V2

After certain time,

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

(k) Calculate a long time later value for VL

After certain time,

$V_L=0.0V$

(l) Calculate a long time later value for dI2/dt

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

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