91Ó°ÊÓ

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

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

In the parallel combination, the total capacitance C is:

$C=C_1+C_2+C_3+……$

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 ${\mathrm{C}}_1={\mathrm{C}}_2=4\mathrm{μ¹ó},{\mathrm{C}}_4=8\mathrm{μ¹ó}$and the stored energy $\mathrm{U}=2.90×{10}^{−3}\mathrm{J}$

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 $V_{ab}=48\mathrm{V}\text{and}\mathrm{U}=2.90×{10}^{−3}\mathrm{J}$

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

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

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

So $\mathrm{°äÏ•}$is given by,

$C\mathrm{Ï•}=C_{12}+C_3$

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

$C\mathrm{Ï•}:=2+C_3$

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

So $C_t$is given by,

$\frac{1}{C_t}=\frac{1}{C_{\mathrm{Ï•}}}+\frac{1}{C_4}$

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

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

Hence the capacitor $C_3$must have capacitance $1.68\mathrm{μ}F$in the network.