A potential difference ${\mathit{V}}_{\mathbf{a}\mathbf{b}}\mathbf{=}\mathbf{48}\mathbf{.0}\mathbf{V}$is applied across the capacitor network of Fig. E24.17.${\mathbf{C}}_{\mathbf{1}}\mathbf{=}{\mathbf{C}}_{\mathbf{2}}\mathbf{=}\mathbf{4.00}\mathit{\mu }\mathbf{F}\mathbf{\text{and}}{\mathbf{C}}_{\mathbf{4}}\mathbf{=}\mathbf{8.00}\mathit{\mu }\mathbf{F}\mathbf{\text{,}}$what must the capacitance ${\mathit{c}}_{\mathbf{3}}$be if the network is to store $\mathbf{2}\mathbf{.90}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{J}$of electrical energy?

Capacitance is the ratio of the amount of electric charge stored on a conductor to a difference in electric potential.

The energy U stored in a capacitor of capacitance with voltage V is

$\mathrm{U}=\frac{1}{2}C{\mathrm{C}}^2$

For capacitors in the series combination, the total capacitance C is given by

$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\dots$

In the parallel combination, the total capacitance C is:

$C=C_1+C_2+C_3+\dots \dots$

Where $C_1,C_2,C_3$... are individual capacitances.

Across the capacitor network the value of the potential difference $V_{ab}$and capacitance are ${\mathrm{C}}_1={\mathrm{C}}_2=4\mathrm{\mu F},{\mathrm{C}}_4=8\mathrm{\mu F}$and the stored energy $\mathrm{U}=2.90\times {10}^{-3}\mathrm{J}$

And the equivalent of Fig E24.17 is,

To find the capacitance $C_3$.

The total capacitance $C_t$is given by,

$\mathrm{U}=\frac{1}{2}{\mathrm{C}}_t{{\mathrm{V}}_{ab}}^2$

Substitute the value of $V_{ab}=48\mathrm{V}\text{and}\mathrm{U}=2.90\times {10}^{-3}\mathrm{J}$

$\begin{array}{r}{C}_t=\frac{2U}{V_{ab}^2}\\ =\frac{2\times 2.90\times {10}^{-3}}{(48{)}^2}\\ =2.52\times {10}^{-6}\mathrm{F}\\ =2.52\mu \mathrm{F}\end{array}$

Since the capacitance $C_1\text{and}{C}_2$are in the series combination.

So $C_{12}$is given by,

$\frac{1}{C_{12}}=\frac{1}{C_1}+\frac{1}{C_2}$

Substitute the value of ${\mathrm{C}}_1={\mathrm{C}}_2=4\mu \mathrm{F}$localid="1668325565024" $\begin{array}{r}\Rightarrow \frac{1}{{\mathrm{C}}_{12}}=\frac{1}{4}+\frac{1}{4}\\ \Rightarrow \frac{1}{{\mathrm{C}}_{12}}=\frac{1}{2}\\ \Rightarrow C_{12}=2\mu \mathrm{F}\end{array}$

Since the capacitance $C_{12}\text{and}{C}_3$in the parallel combination.

So $\mathrm{C\varphi }$is given by,

$C\varphi =C_{12}+C_3$

Substitute the value of $C_{12}=2\mu \mathrm{F}$.

$C\varphi :=2+C_3$

Since the capacitance$\text{Cϕand}{\mathrm{C}}_4$in the series combination.

So $C_t$is given by,

$\frac{1}{C_t}=\frac{1}{C_{\varphi }}+\frac{1}{C_4}$

Substitute the value of $C_t=2.52\mu \mathrm{F}\text{and}C\mathrm{\%}=2+C_3\text{and}{C}_4=8\mu \mathrm{F}\text{.}$

$\begin{array}{r}\frac{1}{2.52}=\frac{1}{2+C_3}+\frac{1}{8}\\ \text{⇒}\frac{1}{2+C_3}=\frac{1}{2.52}-\frac{1}{8}\\ \text{⇒}\frac{1}{2+C_3}=\frac{137}{504}\\ \text{⇒}2+C_3=\frac{504}{137}\\ \text{⇒}{C}_3=\frac{504}{137}-2\\ \text{⇒}{C}_3=1.68\mu \mathrm{F}\end{array}$

Hence the capacitor $C_3$must have capacitance $1.68\mu F$in the network.