Chapter 26, Problem 26P

(a)

To determine

The equivalent capacitance of the capacitors for the given system.

Answer to Problem 26P

The equivalent capacitance of the capacitors for the given system is $2.67\mu \text{F}$.

Explanation of Solution

Write the expression for equivalent capacitance for capacitor of $6.00\mu \text{F}$ and $2.00\mu \text{F}$ connected in parallel.

    $C_{\text{26}}=C_6+C_2$                                                                                                               (I)

Here, $C_{26}$ is the equivalent capacitance for capacitor of $6.00\mu \text{F}$ and $2.00\mu \text{F}$ connected in parallel, $C_6$ is the capacitor with $6.00\mu \text{F}$ capacitance and $C_2$ is the capacitor with $2.00\mu \text{F}$ capacitance

Write the expression for equivalent capacitance for capacitor of $8.00\mu \text{F}$ and $8.00\mu \text{F}$ and $C_{26}$ connected in series.

    $\frac{1}{C_{\text{eq}}}=\frac{1}{C_8}+\frac{1}{C_8}+\frac{1}{C_{26}}$                                                                                                   (II)

Here, $C_{\text{eq}}$ is equivalent capacitance for capacitor of $8.00\mu \text{F}$ and $8.00\mu \text{F}$ and $C_{26}$ connected in series and $C_8$ is the capacitor with $8.00\mu \text{F}$ capacitance.

Conclusion:

Substitute $2.00\mu \text{F}$ for $C_2$ and $6.00\mu \text{F}$ for $C_6$ in equation (I) to solve for $C_{26}$.

    $\begin{array}{c}{C}_{\text{26}}=6.00\mu \text{F}+2.00\mu \text{F}\\ =8.00\mu \text{F}\end{array}$

Substitute $8.00\mu \text{F}$ for $C_8$ and $8.00\mu \text{F}$ for $C_{26}$ in equation (II) to solve for $C_{\text{eq}}$.

    $\begin{array}{c}\frac{1}{C_{\text{eq}}}=\frac{1}{8.00\mu \text{F}}+\frac{1}{8.00\mu \text{F}}+\frac{1}{8.00\mu \text{F}}\\ C_{\text{eq}}=\frac{\left(8.00\mu \text{F}\right)}{3}\\ C_{\text{eq}}=2.67\mu \text{F}\end{array}$

Therefore, the equivalent capacitance of the capacitors for the given system is $2.67\mu \text{F}$.

(b)

To determine

The charge on each capacitor.

Answer to Problem 26P

The charge stored in $2\mu \text{F}$ is $6.00\mu \text{C}$, in $6\mu \text{F}$ is $18.0\mu \text{C}$ and in both $8\mu \text{F}$ is $24.0\mu \text{C}$.

Explanation of Solution

The charge remains the same in series connection.

Write the expression for charge stored in $C_{26}$ and $C_8$.

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

Here, $Q$ is the charge stored in $C_{26}$ and $C_8$.and $V$ is the potential difference.

The charge divides in parallel connection but the potential difference remains the same.

Write the expression to calculate the potential difference across $C_{26}$.

    $V_{26}=\frac{Q}{C_{26}}$                                                                                                                (IV)

Here, $V_{26}$ is the potential difference across $C_{26}$.

Write the expression to calculate the charge stored in the capacitor $C_6$.

    $Q_6=C_6{V}_{\text{26}}$                                                                                                                (V)

Here, the charge stored in capacitor $C_6$ is $Q_6$.

Write the expression to calculate the charge stored in the capacitor $C_2$.

    $Q_2=C_2{V}_{\text{26}}$                                                                                                               (VI)

Here, the charge stored in capacitor $C_2$ is $Q_2$.

Conclusion:

Substitute $2.67\mu \text{F}$ for $C_{\text{eq}}$ and $9.00\text{V}$ for $V$ in equation (III) to solve for $Q$.

    $\begin{array}{c}Q=\left(2.67\mu \text{F}\times \frac{{10}^{-6}\text{F}}{1\mu \text{F}}\right)9.00\text{V}\\ =\left(24.03\times {\text{10}}^{-6}\text{C}\right)\left(\frac{{10}^6\mu \text{C}}{1\text{C}}\right)\\ \approx 24.0\mu \text{C}\end{array}$

Substitute $24.0\times {\text{10}}^{-6}\text{C}$ for $Q$ and $8.00\mu \text{F}$ for $C_{26}$ in equation (IV) to solve for $V_{\text{26}}$.

    $\begin{array}{c}{V}_{\text{26}}=\frac{24.0\times {\text{10}}^{-6}\text{C}}{\left(8.00\mu \text{F}\right)\left(\frac{{10}^{-6}\text{F}}{1\mu \text{F}}\right)}\\ =\frac{24.0\times {\text{10}}^{-6}\text{C}}{8.00\times {10}^{-6}\text{F}}\\ =3.00\text{V}\end{array}$

Substitute $6.00\mu \text{F}$ for $C_6$ and $3.00\text{V}$ for $V_{\text{26}}$ in equation (V) to solve for $Q_6$.

    $\begin{array}{c}{Q}_6=\left(\left(6.00\mu \text{F}\right)\left(\frac{{10}^{-6}\text{F}}{1\mu \text{F}}\right)\right)\left(3.00\text{V}\right)\\ =18\times {10}^{-6}\text{C}\left(\frac{{10}^6\mu \text{C}}{1\text{C}}\right)\\ =18.0\mu \text{C}\end{array}$

Substitute $2.00\mu \text{F}$ for $C_2$ and $3.00\text{V}$ for $V_{\text{26}}$ in equation (VI) to solve for $Q_2$.

    $\begin{array}{c}{Q}_2=\left(\left(2.00\mu \text{F}\right)\left(\frac{{10}^{-6}\text{F}}{1\mu \text{F}}\right)\right)\left(3.00\text{V}\right)\\ =6\times {10}^{-6}\text{C}\left(\frac{{10}^6\mu \text{C}}{1\text{C}}\right)\\ =6.00\mu \text{C}\end{array}$

Therefore, the charge stored in $2\mu \text{F}$ is $6.00\mu \text{C}$, in $6\mu \text{F}$ is $18.0\mu \text{C}$ and in both $8\mu \text{F}$ is $24.0\mu \text{C}$.

(c)

To determine

The potential difference across each capacitor.

Answer to Problem 26P

The potential difference across each capacitor is $3.00\text{V}$.

Explanation of Solution

The potential difference remains same in parallel connection.

So, the potential difference across $C_2$ and $C_6$ will be equal as they are in parallel connection.

The capacitors $C_8$ are connected in series.

Write the expression to calculate the potential difference across $C_8$.

    $V_8=\frac{Q}{C_8}$                                                                                                                 (VII)

Here, $V_8$ is the potential difference across $C_8$.

Conclusion:

Substitute $24.0\times {\text{10}}^{-6}\text{C}$ for $Q$ and $8.00\mu \text{F}$ for $C_6$ in equation (VII) to solve for $V_{\text{8}}$.

    $\begin{array}{c}{V}_{\text{8}}=\frac{24.0\times {\text{10}}^{-6}\text{C}}{\left(8.00\mu \text{F}\right)\left(\frac{{10}^{-6}\text{F}}{1\mu \text{F}}\right)}\\ =\frac{24.0\times {\text{10}}^{-6}\text{C}}{8.00\times {10}^{-6}\text{F}}\\ =3.00\text{V}\end{array}$

Therefore, the potential difference across each capacitor is $3.00\text{V}$.