Chapter 26, Problem 22P

(a)

To determine

The equivalent capacitance between $a$ and $b$.

Answer to Problem 22P

The equivalent capacitance between $a$ and $b$ is $6.04\mu \text{F}$.

Explanation of Solution

Write the expression for equivalent capacitance for capacitor $C_1$ and $C_2$ connected in series.

    $\frac{1}{{\left(C_{12}\right)}_{\text{L}}}=\frac{1}{C_1}+\frac{1}{C_2}$                                  (I)

Here, ${\left(C_{12}\right)}_{\text{L}}$ is the equivalent capacitance on left side for series connection of capacitor $C_1$ and $C_2$.

Write the expression for equivalent capacitance for capacitor $C_1$ and $C_2$ connected in series.

    $\frac{1}{{\left(C_{12}\right)}_{\text{R}}}=\frac{1}{C_1}+\frac{1}{C_2}$                                   (II)

Here, ${\left(C_{12}\right)}_{\text{R}}$ is the equivalent capacitance on right side for series connection of capacitor $C_1$ and $C_2$.

Write the expression for equivalent capacitance for capacitor ${\left(C_{12}\right)}_{\text{R}}$, ${\left(C_{12}\right)}_{\text{L}}$ and $C_3$ connected in parallel.

    $C_{\text{P}}={\left(C_{12}\right)}_{\text{R}}+{\left(C_{12}\right)}_{\text{L}}+C_3$                               (III)

Here, $C_{\text{P}}$ is the equivalent capacitance for capacitor ${\left(C_{12}\right)}_{\text{R}}$, ${\left(C_{12}\right)}_{\text{L}}$ and $C_3$ connected in parallel.

In the $b$ portion, write the expression for the equivalent capacitance for capacitor $C_2$ connected in parallel.

    $C_{22}=C_2+C_2$                                        (IV)

Here, $C_{22}$ is the equivalent capacitance for capacitor $C_2$ connected in parallel.

Write the expression for the equivalent capacitance for $C_{22}$ and $C_{\text{P}}$ in series.

    $\frac{1}{C_{\text{eq}}}=\frac{1}{C_{\text{P}}}+\frac{1}{C_{22}}$                                 (V)

Here, $C_{\text{eq}}$ is the equivalent capacitance for $C_{22}$ and $C_{\text{P}}$ in series.

Conclusion:

Substitute $5.00\mu \text{F}$ for $C_1$ and $10.0\mu \text{F}$ for $C_2$ in equation (I) to solve for ${\left(C_{12}\right)}_{\text{L}}$.

    $\begin{array}{c}\frac{1}{{\left(C_{12}\right)}_{\text{L}}}=\frac{1}{5.00\mu \text{F}}+\frac{1}{10.0\mu \text{F}}\\ {\left(C_{12}\right)}_{\text{L}}=\frac{\left(5.00\mu \text{F}\right)\left(10.0\mu \text{F}\right)}{10.0\mu \text{F}+5.00\mu \text{F}}\\ {\left(C_{12}\right)}_{\text{L}}=\left(\frac{50}{15}\right)\mu \text{F}\\ {\left(C_{12}\right)}_{\text{L}}=3.33\mu \text{F}\end{array}$

Substitute $5.00\mu \text{F}$ for $C_1$ and $10.0\mu \text{F}$ for $C_2$ in equation (II) to solve for ${\left(C_{12}\right)}_{\text{R}}$.

    $\begin{array}{c}\frac{1}{{\left(C_{12}\right)}_{\text{R}}}=\frac{1}{5.00\mu \text{F}}+\frac{1}{10.0\mu \text{F}}\\ {\left(C_{12}\right)}_{\text{R}}=\frac{\left(5.00\mu \text{F}\right)\left(10.0\mu \text{F}\right)}{10.0\mu \text{F}+5.00\mu \text{F}}\\ {\left(C_{12}\right)}_{\text{R}}=\left(\frac{50}{15}\right)\mu \text{F}\\ {\left(C_{12}\right)}_{\text{R}}=3.33\mu \text{F}\end{array}$

Substitute $3.33\mu \text{F}$ for ${\left(C_{12}\right)}_{\text{R}}$, $2.00\mu \text{F}$ for $C_3$ and $3.33\mu \text{F}$ for ${\left(C_{12}\right)}_{\text{L}}$ in equation (III) to solve for $C_{\text{P}}$.

    $\begin{array}{c}{C}_{\text{P}}=3.33\mu \text{F}+3.33\mu \text{F}+2.00\mu \text{F}\\ =8.66\mu \text{F}\end{array}$

Substitute $10.0\mu \text{F}$ for $C_2$ in equation (IV) to solve for $C_{22}$.

    $\begin{array}{c}{C}_{22}=10.0\mu \text{F}+10.0\mu \text{F}\\ =20.0\mu \text{F}\end{array}$

Substitute $8.66\mu \text{F}$ for $C_{\text{P}}$ and $20.0\mu \text{F}$ for $C_{22}$ in equation (V) to solve for $C_{\text{eq}}$.

    $\begin{array}{c}\frac{1}{C_{\text{eq}}}=\frac{1}{8.66\mu \text{F}}+\frac{1}{20.0\mu \text{F}}\\ C_{\text{eq}}=\frac{\left(8.66\mu \text{F}\right)\left(20.0\mu \text{F}\right)}{20.0\mu \text{F}+8.66\mu \text{F}}\\ C_{\text{eq}}=6.04\mu \text{F}\end{array}$

Therefore, the equivalent capacitance between $a$ and $b$ is $6.04\mu \text{F}$.

(b)

To determine

The charge stored in $C_3$ for the given potential difference.

Answer to Problem 22P

The charge stored in $C_3$ for the given potential difference is $83.76\mu \text{C}$.

Explanation of Solution

Write the expression for the total charge stored in the circuit.

    $Q=C_{\text{eq}}V$                              (VI)

Here, $Q$ is the total charge and $V$ is the potential difference between points $a$ and $b$.

Write the expression to calculate the voltage in the lower part of the circuit.

    $V_{\text{L}}=\frac{Q}{C_{22}}$                              (VII)

Here, $V_{\text{L}}$ is the voltage in the lower part of the circuit.

Write the expression to calculate the total voltage.

    $V=V_{\text{U}}+V_{\text{L}}$                                (VIII)

Here, $V_{\text{U}}$ is the voltage in the upper part of the circuit.

Write the expression to calculate the charge stored in the capacitor $C_3$.

    $Q_3=C_3{V}_{\text{U}}$                         (IX)

Here, the charge stored in capacitor $C_3$ is $Q_3$.

Conclusion:

Substitute $6.04\mu \text{F}$ for $C_{\text{eq}}$ and $60.0\text{V}$ for $V$ in equation (VI) to solve for $Q$.

    $\begin{array}{c}Q=\left(6.04\mu \text{F}\times \frac{{10}^{-6}\text{F}}{1\mu \text{F}}\right)60.0\text{V}\\ =3.\text{624}\times {\text{10}}^{-4}\text{C}\end{array}$

Substitute $3.\text{624}\times {\text{10}}^{-4}\text{C}$ for $Q$ and $20.0\mu \text{F}$ for $C_{22}$ in equation (VII) to solve for $V_{\text{L}}$.

    $\begin{array}{c}{V}_{\text{L}}=\frac{3.\text{624}\times {\text{10}}^{-4}\text{C}}{20.0\mu \text{F}\times \frac{{10}^{-6}\text{F}}{1\mu \text{F}}}\\ =18.12\text{V}\end{array}$

Substitute $60.0\text{V}$ for $V$ and $18.12V$ for $V_{\text{L}}$ in equation (VIII) to solve for $V_{\text{U}}$.

    $\begin{array}{c}60.0\text{V}=V_{\text{U}}+18.12\text{V}\\ V_{\text{U}}=41.88\text{V}\end{array}$

Substitute $2.00\mu \text{F}$ for $C_3$ and $41.88\text{V}$ for $V_{\text{U}}$ in equation (IX) to solve for $Q_3$.

    $\begin{array}{c}{Q}_3=\left(2.00\mu \text{F}\times \frac{{10}^{-6}\text{F}}{1\mu \text{F}}\right)\left(41.88\text{V}\right)\\ =83.76\times {10}^{-6}\text{C}\times \frac{{10}^6\mu \text{C}}{1\text{C}}\\ =83.76\mu \text{C}\end{array}$

Therefore, the charge stored in $C_3$ for the given potential difference is $83.76\mu \text{C}$.