91Ó°ÊÓ

1. Capacitance value set by the horizontal scale, ${\mathrm{C}}_{3\mathrm{s}}=12\mathrm{μ¹ó}$
2. Electric potential approaches an asymptote,${\mathrm{V}}_1=10\mathrm{V}$

We need to find the equivalent capacitance for the series and parallel capacitor first. Now, using this relation and the information from the graph, we will find the voltage across the battery and the capacitance.

Formulae:

The equivalent capacitance of a series connection of capacitors,

$\frac{1}{{\mathrm{C}}_{\mathrm{equivalent}}}=\underset{}{∑}\frac{1}{{\mathrm{C}}_{\mathrm{i}}}$ ...(i).

The equivalent capacitance of a parallel connection of capacitors,

${\mathrm{C}}_{\mathrm{equivalent}}\underset{}{=∑}{\mathrm{C}}_{\mathrm{i}}$ ...(ii)

The charge stored between the plates of the capacitor, q = CV ...(iii)

In the given circuit diagram, capacitor and are parallel. So, the equivalent capacitance of the two capacitors is given using equation (ii) by:

$C_{23}=C_2+C_3$

Now, we have the equivalent capacitorwhich is in series with the capacitor.

So, the equivalent capacitance ofandis given by using equation (i) as:

$$\begin{gathered} \frac{1}{C_{123}}=\frac{1}{C_1}+\frac{1}{C_2+C_3} \\ =\frac{C_1+C_2+C_3}{C_1\left(C_2+C_3\right)} \end{gathered}$$ ...(iv)

Now using equation (iii), we get that$q=C_{123}V$and

$$\begin{gathered} q=q_1 \\ =C_1{V}_1 \end{gathered}$$

Thus, using the above values, we get the equation of potential as:

$$\begin{gathered} V_1=\frac{q_1}{C_1} \\ =\frac{q}{C_1} \\ =\frac{C_{123}}{C_1}V \end{gathered}$$

Substituting equation (a) in the above value of potential, we get that

$$\begin{gathered} V_1=\frac{C_1+\left(C_2+C_3\right)}{C_1\left(C_1+C_2+C_3\right)}V \\ =\frac{C_2+C_3}{C_1+C_2+C_3}V \end{gathered}$$ ...(v)

Substituting the value of $C_3→∞$in equation (v), we get that ( $V_1$approach 10 V in this limit as$C_3→∞$)

$$\begin{gathered} V_1=V \\ V=10V \end{gathered}$$

Hence, the value of the potential is 10 V.

From the above graph at $C_3=0$, the graph shows

${\mathrm{V}}_1=2.0\mathrm{V}$

Substituting these values in equation (v), we get the following equation as:

$$\begin{gathered} 2.0V=\frac{C_2+0}{C_1+C_2+0}10V \\ 2C_1+2C_2=10C_2 \\ 2C_1=8C_2 \\ {C}_1=4C_2 \end{gathered}$$ ...(vi)

From the graph we see that when $C_3=6\mathrm{μ}F$the voltage across $C_1$ is exactly half the battery voltage, that is

$$\begin{gathered} {\mathrm{V}}_1=\frac{10}{2} \\ =5\mathrm{V} \end{gathered}$$

Substituting these values in equation (v), we get the following equation as:

$$\begin{gathered} 5\mathrm{V}=\frac{{\mathrm{C}}_2+6\mathrm{μ¹ó}}{{\mathrm{C}}_1+{\mathrm{C}}_2+6\mathrm{μ¹ó}}10\mathrm{V} \\ {\mathrm{C}}_1+{\mathrm{C}}_2+6\mathrm{μ¹ó}=\left({\mathrm{C}}_2+6\mathrm{μ¹ó}\right)×2 \\ {\mathrm{C}}_1+\frac{{\mathrm{C}}_1}{4}+6\mathrm{μ¹ó}=2\frac{{\mathrm{C}}_1}{4}+12\mathrm{μ¹ó}(\mathrm{from}\mathrm{equation}(\mathrm{vi}\left)\right) \end{gathered}$$

Hence, the value of the required capacitance is 8.0 $\mathrm{μ¹ó}$.

Using the value of capacitance ${\mathrm{C}}_1$in equation (vi), we can get the value of capacitance as follows:

$$\begin{gathered} {\mathrm{C}}_2=\frac{8\mathrm{μ¹ó}}{4} \\ =2.0\mathrm{μ¹ó} \end{gathered}$$

Hence, the value of the capacitance is 2.0 $\mathrm{μ¹ó}$.