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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

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

In the parallel combination, the total capacitance C is:

$C=C_1+C_2+C_3+鈥�鈥�$

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 ${\mathrm{C}}_1={\mathrm{C}}_2=4\mathrm{渭贵},{\mathrm{C}}_4=8\mathrm{渭贵}$and the stored energy $\mathrm{U}=2.90脳{10}^{鈭�3}\mathrm{J}$

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 $V_{ab}=48\mathrm{V}\text{and}\mathrm{U}=2.90脳{10}^{鈭�3}\mathrm{J}$

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

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 ${\mathrm{C}}_1={\mathrm{C}}_2=4渭\mathrm{F}$localid="1668325565024" $\begin{array}{r}鈬�\frac{1}{{\mathrm{C}}_{12}}=\frac{1}{4}+\frac{1}{4}\\ 鈬�\frac{1}{{\mathrm{C}}_{12}}=\frac{1}{2}\\ 鈬�C_{12}=2渭\mathrm{F}\end{array}$

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

So $\mathrm{颁蠒}$is given by,

$C蠒=C_{12}+C_3$

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

$C蠒:=2+C_3$

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

So $C_t$is given by,

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

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

$\begin{array}{r}\frac{1}{2.52}=\frac{1}{2+C_3}+\frac{1}{8}\\ \text{鈬�}\frac{1}{2+C_3}=\frac{1}{2.52}鈭�\frac{1}{8}\\ \text{鈬�}\frac{1}{2+C_3}=\frac{137}{504}\\ \text{鈬�}2+C_3=\frac{504}{137}\\ \text{鈬�}{C}_3=\frac{504}{137}鈭�2\\ \text{鈬�}{C}_3=1.68渭\mathrm{F}\end{array}$

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