In Fig. 25-53,$\mathbf{V}\mathbf{=}\mathbf{12}\mathbf{V}\mathbf{,}{\mathbf{C}}_{\mathbf{1}}\mathbf{=}{\mathbf{C}}_{\mathbf{4}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\mu F}\mathbf{,}{\mathbf{C}}_{\mathbf{2}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\mu F}\mathbf{,}\mathbf{and}\mathbf{}{\mathbf{C}}_{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\mu F}$and. What is the charge on capacitor 4?

The given values are:

1. V = 12 V
2. ${\mathrm{C}}_1={\mathrm{C}}_4=2\mathrm{\mu F}$
   
3. ${\mathrm{C}}_2=4\mathrm{\mu F}$
   
4. ${\mathrm{C}}_3=1\mathrm{\mu F}$
   

If the capacitors are connected in parallel, the equivalent capacitance is given by the formula and if capacitors are connected in series, the equivalent capacitance is given by the formula as below. We can find the charge on the capacitor by using the relation between charge and capacitance.

Formulae:

The equivalent capacitance of a series connection of capacitors,

$\frac{1}{{\mathrm{C}}_{\mathrm{equivalent}}}=\sum _{}\frac{1}{{\mathrm{C}}_{\mathrm{I}}}$ …(i).

The equivalent capacitance of a parallel connection of capacitors,

role="math" localid="1661343206912" ${\mathrm{C}}_{\mathrm{equivalent}}=\sum _{}{\mathrm{C}}_{\mathrm{i}}$ …(ii)

The charge stored within the capacitors, q = CV …(iii)

$16\mathrm{\mu C}$Since $C_1$and$C_2$are in parallel combination, so we can find the equivalent capacitanceby using equation (ii) as follows:

$$\begin{gathered} {\mathrm{C}}_{12}=2\mathrm{\mu F}+4\mathrm{\mu F} \\ =3\mathrm{\mu F} \end{gathered}$$

Similarly, we can calculate the equivalent capacitance forandconnected in parallel. Thus, the equivalent capacitance is given using equation (ii) as:

$$\begin{gathered} {\mathrm{C}}_{34}=1\mathrm{\mu F}+2\mathrm{\mu F} \\ =3\mathrm{\mu F} \end{gathered}$$

These two combinations are now in series.Their equivalent capacitance is given by equation (i) as follows:

$$\begin{gathered} {\mathrm{C}}_{\mathrm{eq}}=\frac{{\mathrm{C}}_{12}\times {\mathrm{C}}_{34}}{{\mathrm{C}}_{12}+{\mathrm{C}}_{34}} \\ =\frac{\left(6\mathrm{\mu F}\times 3\mathrm{\mu F}\right)}{\left(6+3\right)6\mathrm{\mu F}} \\ =2{\mathrm{C}}_{\mathrm{eq}}=\frac{{\mathrm{C}}_{12}\times {\mathrm{C}}_{34}}{{\mathrm{C}}_{12}+{\mathrm{C}}_{34}} \\ =2\mathrm{\mu F} \end{gathered}$$

The charge on this equivalent capacitance is given by using equation (iii) as:

$$\begin{gathered} \mathrm{q}=2\mathrm{\mu F}\times 12\mathrm{V} \\ =24\mathrm{\mu C} \end{gathered}$$

This charge on the capacitor$C_{34}$has voltage, which can be given using equation (iii) as:

$$\begin{gathered} {\mathrm{V}}_{34}=\frac{\mathrm{q}}{{\mathrm{C}}_{34}} \\ =\frac{24\mathrm{\mu C}}{3\mathrm{\mu F}} \\ =8\mathrm{V} \end{gathered}$$

So, the charge on the capacitor 4 can be given using equation (iii) as:

$$\begin{gathered} {\mathrm{q}}_4=2\mathrm{\mu F}\times 8\mathrm{V} \\ =16\mathrm{\mu C} \end{gathered}$$

Hence, the value of the charge is$16\mathrm{\mu C}$ .