Chapter 26, Problem 26.56AP

For (he system of four capacitors shown in Figure P26.19. find (a) the total energy stored in the system and (b) the energy stored by each capacitor, (c) (Compare the sum of the answers in part (b) with your result to part (a) and explain your observation.

To determine

The energy stored in the system.

Answer to Problem 26.56AP

The energy stored in the system is $1.35\times {10}^{-2}\text{J}$.

Explanation of Solution

Given info: The electric potential in the system is $90.0\text{V}$.

The Figure of the circuit diagram is shown below.

Figure (1)

Since, the capacitors $C_1$ and $C_2$ of first row are in series. So, the equivalent capacitance of first row for series connection is,

$\frac{1}{C_{\text{eq}}}=\frac{\text{1}}{{\text{C}}_{\text{1}}}+\frac{\text{1}}{{\text{C}}_2}$

Here,

$C_{\text{eq}}$ is the equivalent capacitance for series connection.

Substitute $3.00\text{μF}$ for $C_1$ and $6.00\text{μF}$ for $C_2$ in above equation to find $C_{\text{eq}}$.

$\begin{array}{c}\frac{1}{C_{\text{eq}}}=\frac{\text{1}}{3.00\text{μF}}+\frac{\text{1}}{6.00\text{μF}}\\ C_{\text{eq}}=2.00\text{μF}\end{array}$

Thus, the equivalent capacitance of the first row is $2.00\text{μF}$.

Since, the capacitors $C_3$ and $C_4$ of the second row are in series. So, the equivalent capacitance of the second row is,

$\frac{1}{C_{\text{eq}}\text{'}}=\frac{\text{1}}{{\text{C}}_3}+\frac{\text{1}}{{\text{C}}_4}$

Here,

$C_{\text{eq}}\text{'}$ is the equivalent capacitance of the second row.

Substitute $2.00\text{μF}$ for $C_3$ and $4.00\text{μF}$ for $C_4$ in above equation to find $C_{\text{eq}}\text{'}$.

$\begin{array}{c}\frac{1}{C_{\text{eq}}\text{'}}=\frac{\text{1}}{2.00\text{μF}}+\frac{\text{1}}{4.00\text{μF}}\\ C_{\text{eq}}\text{'}=\frac{4}{3}\text{μF}\end{array}$

Thus, the equivalent capacitance of the second row is $\frac{4}{3}\text{μF}$.

Since, the first and second rows are in parallel to each other. So, the equivalent capacitance of the system is,

$C_{\text{eq}}\text{'}\text{'}=C_{\text{eq}}+C_{\text{eq}}\text{'}$

Here,

$C_{\text{eq}}\text{'}\text{'}$ is the equivalent capacitance of the system.

Substitute $\frac{4}{3}\text{μF}$ for $C_{\text{eq}}\text{'}$ and $2.00\text{μF}$ for $C_{\text{eq}}$ in above equation to find $C_{\text{eq}}\text{'}\text{'}$.

$\begin{array}{c}{C}_{\text{eq}}\text{'}\text{'}=2.00\text{μF}+\frac{4}{3}\text{μF}\\ \text{=3}\text{.33}\text{μF}\end{array}$

Thus, the equivalent capacitance of the system is $\text{3}\text{.33}\text{μF}$.

Formula to calculate the energy stored in the system is,

$U=\frac{1}{2}{C}_{\text{eq}}\text{'}\text{'}{\left(\Delta V\right)}^2$

Here,

$U$ is the energy stored in the system.

$\Delta V$ is the potential difference.

Substitute $\text{3}\text{.33}\text{μF}$ for $C_{\text{eq}}\text{'}\text{'}$ and $90.0\text{V}$ for $\Delta V$ in above equation to find $U$.

$\begin{array}{c}U=\frac{1}{2}\left(\text{3}\text{.33}\text{μF}\right){\left(90.0\text{V}\right)}^2\\ =1.348\times {10}^{-2}\text{J}\\ \approx 1.35\times {10}^{-2}\text{J}\end{array}$

Conclusion:

Therefore, the energy stored in the system is $1.35\times {10}^{-2}\text{J}$.

To determine

The energy stored by each capacitor.

Answer to Problem 26.56AP

The energy stored in capacitors $C_1$, $C_2$, $C_3$ and $C_4$ are $5.4\times {10}^{-3}\text{J}$, $2.7\times {10}^{-3}\text{J}$, $3.6\times {10}^{-3}\text{J}$ and $1.8\times {10}^{-3}\text{J}$ respectively.

Explanation of Solution

Given info: The electric potential in the system is $90.0\text{V}$.

Since, the electric potential through each capacitor is same.

Formula to calculate the charge on the first row of capacitor is,

$Q_1=C_{\text{eq}}V$

Here,

$Q_1$ is the charge on the first row of capacitor.

Substitute $2.00\text{μF}$ for $C_{\text{eq}}$ and $90.0\text{V}$ for $V$ in above equation to find $Q_1$.

$\begin{array}{c}{Q}_1=\left(2.00\text{μF}\right)\left(90.0\text{V}\right)\\ =180\text{μC}\end{array}$

Thus, the charge on the first row of capacitor is $180\text{μC}$.

Formula to calculate the energy stored is,

$U=\frac{Q^2}{2C}$

Here,

$C$ is the capacitance of the capacitor.

$Q$ is the charge through first row of capacitor.

The energy stored in capacitor $C_1$ is,

$U_1=\frac{Q_1{}^2}{2C_1}$

Here,

$U_1$ is the energy stored in capacitor $C_1$.

Substitute $180\text{μC}$ for $Q_1$ and $3.00\text{μF}$ for $C_1$ in above equation to find $U_1$.

$\begin{array}{c}{U}_1=\frac{{\left(180\text{μC}\left(\frac{{10}^{-6}\text{C}}{\text{1}\text{μC}}\right)\right)}^2}{2\left(3.00\text{μF}\left(\frac{{10}^{-6}\text{F}}{\text{1}\text{μF}}\right)\right)}\\ =5.4\times {10}^{-3}\text{J}\end{array}$

Thus, the energy stored in capacitor $C_1$ is $5.4\times {10}^{-3}\text{J}$.

The energy stored in capacitor $C_2$ is,

$U_2=\frac{Q_1{}^2}{2C_2}$

Here,

$U_2$ is the energy stored in capacitor $C_2$.

Substitute $180\text{μC}$ for $Q_1$ and $3.00\text{μF}$ for $C_2$ in above equation to find $U_2$.

$\begin{array}{c}{U}_2=\frac{{\left(180\text{μC}\left(\frac{{10}^{-6}\text{C}}{\text{1}\text{μC}}\right)\right)}^2}{2\left(6.00\text{μF}\left(\frac{{10}^{-6}\text{F}}{\text{1}\text{μF}}\right)\right)}\\ =2.7\times {10}^{-3}\text{J}\end{array}$

Thus, the energy stored in capacitor $C_2$ is $2.7\times {10}^{-3}\text{J}$.

Formula to calculate the charge on the first row of capacitor is,

$Q_2=C_{\text{eq}}\text{'}V$

Here,

$Q_2$ is the charge on the second row of capacitor.

Substitute $\frac{4}{3}\text{μF}$ for $C_{\text{eq}}\text{'}$ and $90.0\text{V}$ for $V$ in above equation to find $Q_2$.

$\begin{array}{c}{Q}_1=\frac{4}{3}\text{μF}\left(90.0\text{V}\right)\\ =120\text{μC}\end{array}$

Thus, the charge on the first row of capacitor is $120\text{μC}$.0

The energy stored in capacitor $C_3$ is,

$U_3=\frac{Q_2{}^2}{2C_3}$

Here,

$U_2$ is the energy stored in capacitor $C_3$.

Substitute $180\text{μC}$ for $Q_2$ and $3.00\text{μF}$ for $C_3$ in above equation to find $U_3$.

$\begin{array}{c}{U}_3=\frac{{\left(120\text{μC}\left(\frac{{10}^{-6}\text{C}}{\text{1}\text{μC}}\right)\right)}^2}{2\left(2.00\text{μF}\left(\frac{{10}^{-6}\text{F}}{\text{1}\text{μF}}\right)\right)}\\ =3.6\times {10}^{-3}\text{J}\end{array}$

Thus, the energy stored in capacitor $C_3$ is $3.6\times {10}^{-3}\text{J}$.

The energy stored in capacitor $C_4$ is,

$U_4=\frac{Q_2{}^2}{2C_4}$

Here,

$U_4$ is the energy stored in capacitor $C_4$.

Substitute $180\text{μC}$ for $Q_2$ and $4.00\text{μF}$ for $C_4$ in above equation to find $U_4$.

$\begin{array}{c}{U}_4=\frac{{\left(120\text{μC}\left(\frac{{10}^{-6}\text{C}}{\text{1}\text{μC}}\right)\right)}^2}{2\left(4.00\text{μF}\left(\frac{{10}^{-6}\text{F}}{\text{1}\text{μF}}\right)\right)}\\ =1.8\times {10}^{-3}\text{J}\end{array}$

Thus, the energy stored in capacitor $C_4$ is $1.8\times {10}^{-3}\text{J}$.

Conclusion:

Therefore, the energy stored in capacitors $C_1$, $C_2$, $C_3$ and $C_4$ are $5.4\times {10}^{-3}\text{J}$, $2.7\times {10}^{-3}\text{J}$, $3.6\times {10}^{-3}\text{J}$ and $1.8\times {10}^{-3}\text{J}$ respectively.

To determine

The comparison of the sum of the energy in part (b) with energy of part (a) and explain it.

Answer to Problem 26.56AP

The sum of the energy of par (b) is same as the energy in part (a).

Explanation of Solution

Given info: The electric potential in the system is $90.0\text{V}$.

The sum of the energy of the part (b) is,

$U\text{'}=U_1+U_2+U_3+U_4$

Substitute $5.4\times {10}^{-3}\text{J}$ fir $U_1$, $2.7\times {10}^{-3}\text{J}$ for $U_2$, $3.6\times {10}^{-3}\text{J}$ for $U_3$ and $1.8\times {10}^{-3}\text{J}$ for $U_4$ in above equation to find $U\text{'}$.

$\begin{array}{c}U\text{'}=\left(5.4\times {10}^{-3}\text{J}\right)+\left(2.7\times {10}^{-3}\text{J}\right)+\left(3.6\times {10}^{-3}\text{J}\right)+\left(1.8\times {10}^{-3}\text{J}\right)\\ =1.35\times {10}^{-2}\text{J}\end{array}$

The sum of the energy of par (b) is same as the energy in part (a).

Conclusion:

Therefore, the sum of the energy of par (b) is same as the energy in part (a).