Question

For the capacitor network shown in Fig., a potential difference across abis 12.0 V. Find

(a) the total energy stored in this network?

(b) the energy stored in the $\mathbf{4}\mathbf{.}\mathbf{80}\mathit{\mu }\mathit{F}$ capacitor?

Short answer

1. The total energy stored in this network is $1.58\mu J$.
2. The energy stored in the $4.80\mu F$capacitor is$72.0\mu J$ .

Given:

$C_1=6.2\mu F,C_2=11.8\mu F,C_3=3.5\mu F,C_4=4.8\mu F,C_5=8.6\mu F,V_{ab}=12.0V$

Required:

$U_t,U_4$

Step-by-step solution

Circuit diagram.

Calculate the total energy stored in this network.

The total energy stored in that network is given by

$U_t=\frac{1}{2}{C}_t{V_{ab}}^2$ (1)

So let's get the equivalent capacitance.

The capacitance$C_{12}$ resulted from $C_1,\&C_2,$is given by

$\frac{1}{C_{12}}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{6.2}+\frac{1}{11.8}=\frac{450}{1829}$

The capacitance C' resulted from $C_{12},\&C_3$,is given by

$C\text{'}=C_3+C_{12}=4.06+3.5=7.56\mu F$

The capacitance resulting from $C\text{'}\&C_4\&C_5$,is given by

$$\begin{gathered} \frac{1}{C_t}=\frac{1}{C_4}+\frac{1}{C_5}+\frac{1}{C\text{'}}=\frac{1}{4.8}+\frac{1}{7.56}+\frac{1}{8.6}=0.457 \\ {C}_t=2.19\mu F \end{gathered}$$

So Substitution in (1) gives

$U_t=\frac{1}{2}\times 2.19\times {10}^{-6}\times {\left(12\right)}^2=1.58\times {10}^{-4}J$

Calculate the energy stored in the C4 capacitor.

To get the energy stored in$C_4$we need to get the charge or voltage across it so getting charge is easier so the total charge is given by

$Q=C_t{V}_{ab}=2.19\times {10}^{-6}\times 12=2.63\times {10}^{-5}C$

Since the charge is the same in series capacitors then get the energy to be

$U_4=\frac{Q^2}{2C_4}=\frac{\left(2.63\times {10}^{-5}\right)}{2\times 4.8\times {10}^{-6}}=72.0\mu J$