Question

In the circuit of Fig.27-65, $\mathbf{\epsilon }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{\text{kV}}$, $\mathbf{C}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\mu F}$, ${\mathbf{R}}_{\mathbf{1}}\mathbf{=}{\mathbf{R}}_{\mathbf{2}}\mathbf{=}{\mathbf{R}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{73}\mathbf{}\mathbf{M\Omega }$. With C completely uncharged, switch S is suddenly closed (at$\mathbf{t}\mathbf{=}\mathbf{0}$). At $\mathbf{t}\mathbf{=}\mathbf{0}$, what are (a) current ${\mathit{i}}_{\mathbf{1}}$in resistor 1, (b) current${\mathit{i}}_{\mathbf{2}}$in resistor 2, and (c) current${\mathit{i}}_{\mathbf{3}}$in resistor 3? At $\mathbf{t}\mathbf{=}\mathbf{\infty }$(that is, after many time constants), what are (d) ${\mathit{i}}_{\mathbf{1}}$, (e)${\mathit{i}}_{\mathbf{2}}$, and (f) ${\mathit{i}}_{\mathbf{3}}$? What is the potential difference${\mathit{V}}_{\mathbf{2}}$across resistor 2 at (g) $\mathit{t}\mathbf{=}\mathbf{0}$and (h) $\mathit{t}\mathbf{=}\mathbf{\infty }$? (i) Sketch${\mathit{V}}_{\mathbf{2}}$versus$\mathit{t}$between these two extreme times.

Short answer

a) The current $i_1$in resistor 1 is $1.1mA$.

b) The current $i_2$in resistor 2 is $5.5\times {10}^{-4}A$.

C) The current $i_3$in resistor 3 is$5.5\times {10}^{-4}A$.

d) The current $i_1$at$t=\infty$is$8.2\times {10}^{-4}A$.

e) The current $i_2$at $t=\infty$is $8.2\times {10}^{-4}A$.

f) The current $i_3$at $t=\infty$is$0A$.

g) The potential difference $V_2$across resistor 2 at$t=0$is$4\times {10}^{2}V$.

h) The potential difference $V_2$across resistor 2 at $t=\infty$is $6\times {10}^{2}V$.

i) The graph$V_2$versus t between these two extreme times is sketched.

Step-by-step solution

The given data

Emf of the battery in the circuit,$\epsilon =1.2\text{kV}$

Resistances of the resistors in the circuit,$R_1=R_2=R_3=0.73M\Omega$

The capacitor is uncharged at$\text{t}=0$ when the switch is closed.

Understanding the concept of time constant

The law states that at any circuit junction, the sum of the currents flowing into and out of that junction is equal. Now, from Kirchhoff's voltage law, the net emf of a closed loop can be equated to zero. Thus, using these concepts, we get the equations required to get the values of current and voltage of the resistors.

Formulae:

The voltage equation using Ohm’s law, $\text{V}=\text{IR}$ (1)

Kirchhoff’s voltage law, $\sum _{closedloop}V=0$ (2)

Kirchhoff’s junction rule, ${\text{I}}_{\text{in}}={\text{I}}_{\text{out}}$ (3)

The charge within the plates of a capacitor, $\text{q}=\text{CV}$ (4)

a) Calculation of the current in resistor 1

Let,$i_1$be the current in$R_1$and taken to be positive if it is to the right. Let,$i_2$be the current in$R_2$and taken to be positive if it is downward. Let,$i_3$be the current in$R_3$and taken to be positive if it is downward.

According to the junction rule from equation (3), the current equation can be given as follows:

$i_1=i_2+i_3................................\left(5\right)$

Now, using the loop rule from equations (1) and (2) for the left-hand loop, the voltage equation can be given as:

$\epsilon -i_1{R}_1-i_2{R}_2=0...................\left(6\right)$

Now, using equation (1) in equation (2) for the right-hand loop, the voltage equation can be given as:

$i_2{R}_2-i_3{R}_3=0.........................\left(7\right)$

Since, all the resistance values are the same, thus we can say that

$R_1=R_2=R_3\left(=R\right)=0.73M\Omega$

So from equation (7), we get that

$$\begin{gathered} i_{2}R=i_{3}R \\ {i}_2=i_3............................\left(8\right) \end{gathered}$$

Now, substituting the above value in equation (5), we get that

$$\begin{gathered} i_1=2i_2 \\ {i}_2=\frac{i_1}{2}........................\left(9\right) \end{gathered}$$

Substituting the above value in equation (6), we can get the current value $i_1$in resistor 1 as follows:

$$\begin{gathered} \epsilon -i_{1}R-\frac{i_1}{2}R=0 \\ \epsilon =\frac{3i_{1}R}{2} \\ {i}_1=\frac{2\epsilon }{3R} \\ {i}_1=\frac{2\left(1.2\times {10}^{3}V\right)}{3\left(0.73\times {10}^6\Omega \right)} \end{gathered}$$

$$\begin{gathered} i_1=1.1\times {10}^{-3}\text{A}\left[\frac{1mA}{{10}^{-3}A}\right] \\ {}_1=1.1mA \end{gathered}$$

Hence, the value of the current is$1.1mA$ .

b) Calculation of the current in resistor 2

The current value $i_2$in resistor 2 can be given using equation (9) as follows:

$$\begin{gathered} i_2=\frac{i_1}{2} \\ =\frac{1.1\times {10}^{-3}\text{A}}{2} \\ =5.5\times {10}^{-4}\text{A} \end{gathered}$$

Hence, the value of the current is $5.5\times {10}^{-4}\text{A}$.

c) Calculation of the current in resistor 3

The current value ${\text{i}}_{\text{3}}$in resistor 3 can be given using equation (8) as follows:

$$\begin{gathered} i_3=i_2 \\ =5.5\times {10}^{-4}\text{A} \end{gathered}$$

Hence, the value of the current is $5.5\times {10}^{-4}\text{A}$.

d) Calculation of the current in resistor 1 at t = infinity

At,$t=\infty$the capacitor is fully charged and the current in the capacitor branch is 0. Thus, the current equation (5) becomes:

(The current through resistor 3,$i_{3=}0$as it is the capacitor branch.)

$i_1=i_2$

Thus, using the above value in equation (6), the current value $i_1$at $t=\infty$can be given as follows:

$$\begin{gathered} \epsilon -i_{1}R-i_{1}R=0 \\ \epsilon =2i_{1}R \\ {i}_1=\frac{\epsilon }{2R} \\ {i}_1=\frac{\left(1.2\times {10}^{3}V\right)}{2\left(0.73\times {10}^6\Omega \right)} \\ {i}_1=8.2\times {10}^{-4}\text{A} \end{gathered}$$

Hence, the current value is $8.2\times {10}^{-4}\text{A}$.

e) Calculation of the current in resistor 2 at t = infinity

Now, from the calculations of part (d), the current value $i_2$at $t=\infty$can be given as follows:

$$\begin{gathered} i_2=i_1 \\ =8.2\times {10}^{-4}\text{A} \end{gathered}$$

Hence, the current value is$8.2\times {10}^{-4}\text{A}$ .

f) Calculation of the current in resistor 3 at t = infinity

No, as per the calculations and the given condition, the current value$i_2$ at$t=\infty$ is found to be $0A$.

g) Calculation of the potential difference across the resistor 2 at t = 0

The upper plate of the capacitor is taken to be positive.

Now, according to the junction rule from equation (5), the current equation can be given as follows:

$i_1=i_2+i_3................................\left(5\right)$

Now, in the left-hand loop, the voltage equation can be given using equation (6):

$\epsilon -i_{1}R-i_{2}R=0...................\left(6\right)$

The right-hand loop voltage equation can be given using equations (1), (2), and (4) as follows:

$-\frac{q}{C}+i_{2}R-i_{3}R=0.........................\left(10\right)$

Solving equation (6) by using equation (5), we get the current value for resistor 2 as follows:

$$\begin{gathered} \epsilon -i_{2}R-i_{3}R-i_{2}R=0 \\ {i}_2=\frac{\epsilon -i_{3}R}{2R}......................\left(11\right) \end{gathered}$$

Substituting the above value in equation (10), we get the following equation:

$$\begin{gathered} -\frac{q}{C}+\left(\frac{\epsilon -i_{3}R}{2R}\right)R-i_{3}R=0 \\ -\frac{q}{C}+\frac{\epsilon }{2}-\frac{3i_{3}R}{2}=0 \\ -\frac{q}{C}+\frac{\epsilon }{2}-\frac{3R}{2}\left(\frac{dq}{dt}\right)=0\left(Q{i}_3\left(or\text{i}\right)=\frac{dq}{dt}\right) \\ \frac{3R}{2}\left(\frac{dq}{dt}\right)+\frac{q}{C}=\frac{\epsilon }{2} \end{gathered}$$

The above equation is similar to that of an equation of an RC circuit, where the time constant is equal to $\tau =\frac{\text{3RC}}{\text{2}}$and the impressed potential difference is $\frac{\epsilon }{2}$. Thus, the solution can be given as follows:

$q=\frac{C\epsilon }{2}\left(1-e^{-2t/3RC}\right)$

Thus, the current in the capacitor branch can be given as:

$$\begin{gathered} i_3\left(t\right)=\frac{dq}{dt} \\ =\frac{d}{dt}\left[\frac{C\epsilon }{2}\left(1-e^{-2t/3RC}\right)\right] \\ =\frac{C\epsilon }{2}\times \frac{2}{3RC}{e}^{-2t/3RC} \\ =\frac{\epsilon }{3R}{e}^{-2t/3RC} \end{gathered}$$

Now, the current in the center branch is given using the above value in equation (9) as follows:

$$\begin{gathered} i_2\left(t\right)=\frac{\epsilon }{2R}-\left(\frac{\epsilon }{3R}{e}^{-2t/3RC}\right)\frac{R}{2R} \\ =\frac{\epsilon }{2R}-\frac{\epsilon }{6R}{e}^{-2t/3RC} \\ =\frac{\epsilon }{6R}\left(3-e^{-2t/3RC}\right) \end{gathered}$$

Now, the required potential difference across the resistor$R_2$ branch can be given using equation (1) with the above current value as follows:

$$\begin{gathered} V_2\left(t\right)=\left[\frac{\epsilon }{6R}\left(3-e^{-2t/3RC}\right)\right]R \\ =\frac{\epsilon }{6}\left(3-e^{-2t/3RC}\right) \end{gathered}$$

Thus, the potential difference at$t=0$ can be given as:

$$\begin{gathered} V_2\left(0\right)=\frac{\epsilon }{6}\left(3-e^{-2\times 0/3RC}\right) \\ =\frac{\epsilon }{6}\left(3-1\right) \\ =\frac{\epsilon }{3} \\ =\frac{1.2\times {10}^3\text{V}}{3} \\ =4\times {10}^2\text{V} \end{gathered}$$

Hence, the value of the potential difference at t=0 is $=4\times {10}^2\text{V}$.

h) Calculation of the potential difference across the resistor 2 at t=infinity

Now, the potential difference across the resistor $R_2$branch at $t=\infty$can be given as follows:

$$\begin{gathered} V_2\left(0\right)=\frac{\epsilon }{6}\left(3-e^{-2\times \infty /3RC}\right) \\ =\frac{\epsilon }{6}\left(3-0\right) \\ =\frac{\epsilon }{2} \\ =\frac{1.2\times {10}^3\text{V}}{2} \\ =6\times {10}^2\text{V} \end{gathered}$$

Hence, the value of the potential difference is$=6\times {10}^2\text{V}$ .

i) Calculation of the values for plotting the graph of V2 versus time

The following graph represents the plot of potential difference $V_2\left(t\right)$as a function of time.