Chapter 26, Problem 56AP

(a)

To determine

The total energy stored in the system.

Answer to Problem 56AP

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

Explanation of Solution

Below figure shows the system of four capacitors.

Figure-(1)

Write the expression for equivalent capacitance for series capacitors.

    $\frac{1}{C_{\text{seq}}}=\frac{1}{C_1}+\frac{1}{C_2}$                                                                                                           (I)

Here, $C_{\text{seq}}$ is the equivalent capacitance for series capacitors.

Write the expression for equivalent capacitance for parallel capacitors.

    $C_{\text{peq}}=C_1+C_2$                                                                                                            (II)

Here, $C_{\text{peq}}$ is the for equivalent capacitance for parallel capacitors, $C_1$, and $C_2$ are the magnitude of the capacitors.

Write the expression for net equivalent capacitance.

    $C_{\text{eq}}=C_{\text{peq}}+C_{\text{seq}}$                                                                                                      (III)

Here, $C_{\text{eq}}$ is the net equivalent capacitance.

Write the expression for energy stored in the system.

    $U=\frac{1}{2}\left(C_{\text{eq}}\right){\left(\Delta V\right)}^2$                                                                                                  (IV)

Here, $U$ is the total energy stored in the capacitors and $\Delta V$ is the potential difference.

Conclusion:

Substitute, $3\text{μF}$ for $C_{\text{1}}$ and $6\text{μF}$ for $C_2$ in Equation (I) to calculate $C_{\text{r1}}$.

    $\begin{array}{l}\frac{1}{C_{\text{r1}}}=\frac{1}{3\text{μF}}+\frac{1}{6\text{μF}}\\ \frac{1}{C_{\text{r1}}}=0.333\text{μF}+0.166\text{μF}\\ \frac{1}{C_{\text{r1}}}=\text{0}\text{.499}\text{μF}\\ C_{\text{r1}}=\frac{1}{\text{0}\text{.499}\text{μF}}\end{array}$

Further solve the above equation.

    $C_{\text{r1}}=2\text{μF}$

Substitute, $2\text{μF}$ for $C_{\text{1}}$ and $4\text{μF}$ for $C_2$ in Equation (I) to calculate $C_{\text{r2}}$.

    $\begin{array}{l}\frac{1}{C_{\text{r2}}}=\frac{1}{2\text{μF}}+\frac{1}{4\text{μF}}\\ \frac{1}{C_{\text{r2}}}=0.5\text{μF}+0.25\text{μF}\\ \frac{1}{C_{\text{r2}}}=\text{0}\text{.75}\text{μF}\\ C_{\text{r2}}=\frac{1}{\text{0}\text{.75}\text{μF}}\end{array}$

Further solve the above equation.

    $C_{\text{r2}}=1.33\text{μF}$

Substitute $2\text{μF}$ for $C_{\text{r1}}$ and $1.33\text{μF}$ for $C_{\text{r2}}$ in equation (III) to calculate $C_{\text{eq}}$.

    $\begin{array}{c}{C}_{\text{eq}}=2\text{μF}+1.33\text{μF}\\ =3.33\text{μF}\end{array}$

Substitute $3.33\text{μF}$ for $C_{\text{eq}}$ and $90\text{V}$ for $\Delta V$ in equation (IV) to calculate $U$.

    $\begin{array}{c}U=\frac{1}{2}\left(3.33\text{μF}\right){\left(90\text{V}\right)}^2\left(\frac{1\times {10}^{-6}\text{F}}{1\text{μF}}\right)\\ =\frac{1}{2}\left(3.33\times {10}^{-6}\text{F}\right){\left(90\text{V}\right)}^2\\ =\frac{1}{2}\left(26973\times {10}^{-6}\right)\text{J}\\ =1.348\times {10}^{-2}\text{J}\end{array}$

Further solve the above equation.

    $U=1.35\times {10}^{-2}\text{J}$

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

(b)

To determine

The energy stored by each capacitor.

Answer to Problem 56AP

Energy stored at $3\text{μF}$ capacitor is $\text{5}\text{.4}\times {\text{10}}^3\text{J}$, $6\text{μF}$ capacitor is $\text{2}\text{.7}\times {\text{10}}^3\text{J}$, $2\text{μF}$ capacitor is $\text{3}\text{.6}\times {\text{10}}^3\text{J}$ and $4\text{μF}$ capacitor is $\text{1}\text{.8}\times {\text{10}}^3\text{J}$.

Explanation of Solution

Write the expression for charge in the capacitor.

    $Q=C_{\text{i}}V$                                                                                                                    (V)

Here, $Q$ is the charge on the capacitor, $C_{\text{i}}$ is the equivalent capacitance.

Write the expression for energy stored in the capacitor.

    $U=\frac{Q^2}{2C_{\text{i}}}$                                                                                                                 (VI)

Conclusion:

Substitute $2\text{μF}$ for $C_{\text{i}}$ and $90\text{V}$ for $V$ in Equation (IV) to calculate $Q$.

    $\begin{array}{c}Q=\left(2\text{μF}\right)\left(90\text{V}\right)\\ =180\text{μC}\end{array}$

Substitute $180\text{μC}$ for $Q$ and $3\text{μF}$ for $C_{\text{i}}$ in Equation (V) to calculate $U_1$.

    $\begin{array}{c}{U}_1=\frac{{\left(180\text{μC}\right)}^2}{2\left(3\text{μF}\right)}\\ =\frac{{\left(180\text{μC}\right)}^2}{\left(6\text{μF}\right)}\\ =5400\text{J}\\ =\text{5}\text{.4}\times {\text{10}}^3\text{J}\end{array}$

Substitute $180\text{μC}$ for $Q$ and $6\text{μF}$ for $C_{\text{i}}$ in Equation (V) to calculate $U_2$.

    $\begin{array}{c}{U}_2=\frac{{\left(180\text{μC}\right)}^2}{2\left(6\text{μF}\right)}\\ =\frac{{\left(180\text{μC}\right)}^2}{\left(12\text{μF}\right)}\\ =2700\text{J}\\ =\text{2}\text{.7}\times {\text{10}}^3\text{J}\end{array}$

Substitute $1.33\text{μF}$ for $C_{\text{i}}$ and $90\text{V}$ for $V$ in Equation (IV) to calculate $Q$.

    $\begin{array}{c}Q=\left(1.33\text{μF}\right)\left(90\text{V}\right)\\ =120\text{μC}\end{array}$

Substitute $180\text{μC}$ for $Q$ and $2\text{μF}$ for $C_{\text{i}}$ in Equation (V) to calculate $U_3$.

    $\begin{array}{c}{U}_3=\frac{{\left(120\text{μC}\right)}^2}{2\left(2\text{μF}\right)}\\ =\frac{{\left(120\text{μC}\right)}^2}{\left(4\text{μF}\right)}\\ =3600\text{J}\\ =\text{3}\text{.6}\times {\text{10}}^3\text{J}\end{array}$

Substitute $180\text{μC}$ for $Q$ and $4\text{μF}$ for $C_{\text{i}}$ in Equation (V) to calculate $U_4$.

    $\begin{array}{c}{U}_4=\frac{{\left(120\text{μC}\right)}^2}{2\left(4\text{μF}\right)}\\ =\frac{{\left(120\text{μC}\right)}^2}{\left(8\text{μF}\right)}\\ =1800\text{J}\\ =\text{1}\text{.8}\times {\text{10}}^3\text{J}\end{array}$

Therefore, energy stored at $3\text{μF}$ capacitor is $\text{5}\text{.4}\times {\text{10}}^3\text{J}$, $6\text{μF}$ capacitor is $\text{2}\text{.7}\times {\text{10}}^3\text{J}$, $2\text{μF}$ capacitor is $\text{3}\text{.6}\times {\text{10}}^3\text{J}$ and $4\text{μF}$ capacitor is $\text{1}\text{.8}\times {\text{10}}^3\text{J}$.

(c)

To determine

The observation from part (a) and part (b).

Answer to Problem 56AP

Sum of the answers of part (a) is same as part (b).

Explanation of Solution

Write the expression for total energy stored.

    $U=U_1+U_2+U_3+U_4$                                                                                            (VI)

Conclusion:

Substitute $\text{5}\text{.4}\times {\text{10}}^3\text{J}$ for $U_1$, $\text{2}\text{.7}\times {\text{10}}^3\text{J}$ for $U_2$, $\text{3}\text{.6}\times {\text{10}}^3\text{J}$ for $U_3$ and $\text{1}\text{.8}\times {\text{10}}^3\text{J}$ for $U_4$ in equation (VI) to calculate $U$.

    $\begin{array}{c}U=\text{5}\text{.4}\times {\text{10}}^3\text{J}+\text{2}\text{.7}\times {\text{10}}^3\text{J}+\text{3}\text{.6}\times {\text{10}}^3\text{J}+\text{1}\text{.8}\times {\text{10}}^3\text{J}\\ \text{=13}\text{.5}\times {\text{10}}^3\text{J}\end{array}$

Therefore, sum of the answers of part (a) is same as part (b).