Question

In Fig. 25-34 the battery has potential difference $\mathbf{V}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{V}\mathbf{,}\mathbf{}{\mathbf{C}}_{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\mu F}\mathbf{,}{\mathbf{C}}_{\mathbf{4}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\mu F}$and all the capacitors are initially uncharged. When switch S is closed, a total charge of$\mathbf{12}\mathbf{}\mathbf{\mu C}$passes through point aand a total charge of$\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\mu C}$passes through point b. What are (a)${\mathbf{C}}_{\mathbf{1}}$and (b)${\mathbf{C}}_{\mathbf{3}}$?

Short answer

1. The value of $C_1$is$4.0\mathrm{\mu F}$
2. The value of ${\mathrm{C}}_3$is${\mathrm{C}}_3=2.0\mathrm{\mu F}$

Step-by-step solution

Step 1: Given data

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

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

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

Determining the concept

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${\mathrm{C}}_{\mathrm{eq}}=\sum _{\mathrm{j}=1}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{j}}$

For series combination$\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\sum _{\mathrm{j}=1}^{\mathrm{n}}\frac{1}{{\mathrm{C}}_{\mathrm{j}}}$

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

(a) Determining the value of C1

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

${\mathrm{q}}_1={\mathrm{q}}_2=12\mathrm{\mu F}\mathrm{and}{\mathrm{q}}_4=8\mathrm{\mu C}$

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

$$\begin{gathered} {\mathrm{q}}_1=12\mathrm{\mu C}-8\mathrm{\mu C} \\ =4\mathrm{\mu C} \end{gathered}$$

From the equation 25 - 1,

Since q = CV, therefore,

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

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

$$\begin{gathered} {\mathrm{V}}_4=\frac{{\mathrm{q}}_4}{{\mathrm{C}}_4} \\ =\frac{8\mathrm{\mu C}}{4\mathrm{\mu F}} \\ =2\mathrm{V} \end{gathered}$$

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

$$\begin{gathered} {\mathrm{V}}_3=2\mathrm{V} \\ \mathrm{Thus}, \\ {\mathrm{C}}_3=\frac{{\mathrm{q}}_3}{{\mathrm{V}}_3} \\ =\frac{4\mathrm{\mu C}}{2\mathrm{V}} \\ =2\mathrm{\mu F} \end{gathered}$$

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

$$\begin{gathered} {\mathrm{C}}_{\mathrm{eq}}=\sum _{\mathrm{j}=1}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{j}} \\ {\mathrm{C}}_{34}={\mathrm{C}}_3+{\mathrm{C}}_4 \\ =2\mathrm{\mu F}+4\mathrm{\mu F} \\ =6\mathrm{\mu F} \end{gathered}$$

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

$$\begin{gathered} \frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\sum _{\mathrm{j}=1}^{\mathrm{n}}\frac{1}{{\mathrm{C}}_{\mathrm{j}}} \\ \frac{1}{{\mathrm{C}}_{234}}=\frac{1}{{\mathrm{C}}_2}+\frac{1}{{\mathrm{C}}_{34}} \\ =\frac{1}{3\mathrm{\mu F}}+\frac{1}{6\mathrm{\mu F}} \\ =\frac{1}{2\mathrm{\mu F}} \\ {\mathrm{C}}_{234}=2\mathrm{\mu F} \end{gathered}$$

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

Therefore, the equivalent capacitance can be given by,

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

$\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\frac{1}{2\mathrm{\mu F}}+\frac{1}{{\mathrm{C}}_1}$………………….(1)

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

$$\begin{gathered} {\mathrm{C}}_{\mathrm{eq}}=\frac{12\mathrm{\mu C}}{{\mathrm{V}}_{\mathrm{battery}}} \\ =\frac{12\mathrm{\mu C}}{9\mathrm{V}} \\ =\frac{4}{3}\mathrm{\mu F} \end{gathered}$$

Substituting this value in the equationwe get

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

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

(b) Determining the value of C3

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

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

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.