91Ó°ÊÓ

$$\begin{gathered} \mathrm{V}=9\mathrm{V} \\ {\mathrm{C}}_2=3.0\mathrm{μ¹ó} \\ {\mathrm{C}}_4=4.0\mathrm{μ¹ó} \end{gathered}$$

When the switch S is closed, the total charge of $12\mathrm{μ°ä}$passes through the point .

When the switch S is closed, the total charge of $8\mathrm{μ°ä}$passes through the point .

If the capacitors are in series, the charge on each capacitor is the same. Using this and the concept of conservation of charge, find the value of all the charges. find the required value of capacitance by using the concept of equivalent capacitance for series and parallel combination and equation 25 - 1.

Formulae are as follows:

q = CV

For parallel combination${\mathrm{C}}_{\mathrm{eq}}=\underset{\mathrm{j}=1}{\overset{\mathrm{n}}{∑}}{\mathrm{C}}_{\mathrm{j}}$

For series combination$\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\underset{\mathrm{j}=1}{\overset{\mathrm{n}}{∑}}\frac{1}{{\mathrm{C}}_{\mathrm{j}}}$

Where C is capacitance, V is the potential difference, and q is the charge on the capacitor.

According to the given condition in the problem, infer that the charge on ${\mathrm{C}}_1\mathrm{and}{\mathrm{C}}_2$is $12\mathrm{μ¹ó}$and the charge on ${\mathrm{C}}_4\mathrm{is}8\mathrm{μ°ä}$, i.e.

${\mathrm{q}}_1={\mathrm{q}}_2=12\mathrm{μ¹ó}\mathrm{and}{\mathrm{q}}_4=8\mathrm{μ°ä}$

From the conservation of charge, the charge on ${\mathrm{C}}_3$is,

$$\begin{gathered} {\mathrm{q}}_1=12\mathrm{μ°ä}-8\mathrm{μ°ä} \\ =4\mathrm{μ°ä} \end{gathered}$$

From the equation 25 - 1,

Since q = CV, therefore,

$\mathrm{V}=\frac{\mathrm{q}}{\mathrm{C}}$

The voltage across${\mathrm{V}}_4$is,

$$\begin{gathered} {\mathrm{V}}_4=\frac{{\mathrm{q}}_4}{{\mathrm{C}}_4} \\ =\frac{8\mathrm{μ°ä}}{4\mathrm{μ¹ó}} \\ =2\mathrm{V} \end{gathered}$$

Consequently, the voltage across${\mathrm{C}}_3$is also 2 V, i.e.

$$\begin{gathered} {\mathrm{V}}_3=2\mathrm{V} \\ \mathrm{Thus}, \\ {\mathrm{C}}_3=\frac{{\mathrm{q}}_3}{{\mathrm{V}}_3} \\ =\frac{4\mathrm{μ°ä}}{2\mathrm{V}} \\ =2\mathrm{μ¹ó} \end{gathered}$$

Since ${\mathrm{C}}_3\mathrm{and}{\mathrm{C}}_4$ are connected in parallel, their equivalent capacitance can be found by the parallel combination,

$$\begin{gathered} {\mathrm{C}}_{\mathrm{eq}}=\underset{\mathrm{j}=1}{\overset{\mathrm{n}}{∑}}{\mathrm{C}}_{\mathrm{j}} \\ {\mathrm{C}}_{34}={\mathrm{C}}_3+{\mathrm{C}}_4 \\ =2\mathrm{μ¹ó}+4\mathrm{μ¹ó} \\ =6\mathrm{μ¹ó} \end{gathered}$$

This is then in series with${\mathrm{C}}_2$, so, the equivalent capacitance for this combination can be found bythe series combination,

$$\begin{gathered} \frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\underset{\mathrm{j}=1}{\overset{\mathrm{n}}{∑}}\frac{1}{{\mathrm{C}}_{\mathrm{j}}} \\ \frac{1}{{\mathrm{C}}_{234}}=\frac{1}{{\mathrm{C}}_2}+\frac{1}{{\mathrm{C}}_{34}} \\ =\frac{1}{3\mathrm{μ¹ó}}+\frac{1}{6\mathrm{μ¹ó}} \\ =\frac{1}{2\mathrm{μ¹ó}} \\ {\mathrm{C}}_{234}=2\mathrm{μ¹ó} \end{gathered}$$

This capacitance is now in series with the unknown capacitance${\mathrm{C}}_1$,

Therefore, the equivalent capacitance can be given by,

$\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\frac{1}{{\mathrm{C}}_{234}}+\frac{1}{{\mathrm{C}}_1}$

$\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\frac{1}{2\mathrm{μ¹ó}}+\frac{1}{{\mathrm{C}}_1}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦.(1)

It is known that the total effective capacitance of the circuit is,

$$\begin{gathered} {\mathrm{C}}_{\mathrm{eq}}=\frac{12\mathrm{μ°ä}}{{\mathrm{V}}_{\mathrm{battery}}} \\ =\frac{12\mathrm{μ°ä}}{9\mathrm{V}} \\ =\frac{4}{3}\mathrm{μ¹ó} \end{gathered}$$

Substituting this value in the equationwe get

$$\begin{gathered} \frac{3}{4\mathrm{μ¹ó}}=\frac{1}{2\mathrm{μ¹ó}}+\frac{1}{{\mathrm{C}}_1} \\ \frac{1}{{\mathrm{C}}_1}=\frac{1}{2\mathrm{μ¹ó}}-\frac{3}{4\mathrm{μ¹ó}} \\ {\mathrm{C}}_1=4\mathrm{μ¹ó} \end{gathered}$$

Thus, the value of ${\mathrm{C}}_1\mathrm{is}4\mathrm{μ¹ó}$.

From equation (1), it can be concluded that the value of ${\mathrm{C}}_3\mathrm{is}2\mathrm{μ¹ó}$.

Hence, the value of ${\mathrm{C}}_3\mathrm{is}{\mathrm{C}}_3=2\mathrm{μ¹ó}$

Therefore, by using the relation between the charge, the capacitance, and the formula for equivalent capacitance for series and parallel combination, find the required capacitance.