Chapter 26, Problem 19P

(a)

To determine

The equivalent capacitance of the system.

Answer to Problem 19P

The equivalent capacitance of the system is $3.33\mu \text{F}$.

Explanation of Solution

Write the expression for equivalent capacitance for upper capacitor connected in series.

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

Here, $C_{\text{U}}$ is the equivalent capacitance on left side for series connection of capacitor $C_1$ and $C_2$.

Write the expression for equivalent capacitance for lower capacitor $C_3$ and $C_4$ connected in series.

    $\frac{1}{C_{\text{L}}}=\frac{1}{C_3}+\frac{1}{C_4}$                                  (II)

Here, $C_{\text{L}}$ 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 $C_{\text{L}}$ and $C_{\text{U}}$ connected in parallel.

    $C_{\text{eq}}=C_{\text{U}}+C_{\text{L}}$                               (III)

Here, $C_{\text{P}}$ is the equivalent capacitance for capacitor $C_{\text{L}}$ and $C_{\text{U}}$ connected in parallel.

Conclusion:

Substitute $3.00\mu \text{F}$ for $C_1$ and $6.00\mu \text{F}$ for $C_2$ in equation (I) to solve for $C_{\text{U}}$.

    $\begin{array}{c}\frac{1}{C_{\text{U}}}=\frac{1}{3.00\mu \text{F}}+\frac{1}{6.00\mu \text{F}}\\ C_{\text{U}}=\frac{\left(3.00\mu \text{F}\right)\left(6.00\mu \text{F}\right)}{6.00\mu \text{F}+3.00\mu \text{F}}\\ C_{\text{U}}=\left(\frac{18}{9}\right)\mu \text{F}\\ C_{\text{U}}=2\mu \text{F}\end{array}$

Substitute $2.00\mu \text{F}$ for $C_1$ and $4.00\mu \text{F}$ for $C_2$ in equation (II) to solve for $C_{\text{L}}$.

    $\begin{array}{c}\frac{1}{C_{\text{L}}}=\frac{1}{4.00\mu \text{F}}+\frac{1}{2.00\mu \text{F}}\\ C_{\text{L}}=\frac{\left(4.00\mu \text{F}\right)\left(2.00\mu \text{F}\right)}{4.00\mu \text{F}+2.00\mu \text{F}}\\ C_{\text{L}}=\left(\frac{8}{6}\right)\mu \text{F}\\ C_{\text{L}}=1.33\mu \text{F}\end{array}$

Substitute $1.33\mu \text{F}$ for $C_{\text{L}}$, and $2\mu \text{F}$ for $C_{\text{U}}$ in equation (III) to solve for $C_{\text{eq}}$.

    $\begin{array}{c}{C}_{\text{e}}=1.33\mu \text{F}+2.00\mu \text{F}\\ =3.33\mu \text{F}\end{array}$

Therefore, the equivalent capacitance for the system is $3.33\mu \text{F}$.

(b)

To determine

The charge on each capacitor.

Answer to Problem 19P

The charge in capacitor $2.00\mu \text{F}$ and $4.00\mu \text{F}$ is $119.7\mu \text{C}$ and on $3.00\mu \text{F}$ and $6.00\mu \text{F}$ is $180\mu \text{C}$.

Explanation of Solution

For series connection the charge remains constant.

Write the expression for the charge in the upper row.

    $Q_{\text{U}}=C_{\text{U}}V$                              (IV)

Here, $Q_{\text{U}}$ is the charge on the upper capacitors and $V$ is the potential difference

Write the expression for the charge in the lower row.

    $Q_{\text{L}}=C_{\text{L}}V$                              (V)

Here, $Q_{\text{L}}$ is the charge on the lower capacitors and $V$ is the potential difference

Conclusion:

Substitute $2.00\mu \text{F}$ for $C_{\text{U}}$ and $90.0\text{V}$ for $V$ in equation (IV) to solve for $Q_{\text{U}}$.

    $\begin{array}{c}{Q}_{\text{U}}=\left(2.00\mu \text{F}\times \frac{{10}^{-6}\text{F}}{1\mu \text{F}}\right)90.0\text{V}\\ =180\times {\text{10}}^{-6}\text{C}\times \frac{1\mu \text{C}}{1\text{C}}\\ =180\mu \text{C}\end{array}$

The charge on $3.00\mu \text{F}$ and $6.00\mu \text{F}$ capacitor will be same.

Substitute $1.33\mu \text{F}$ for $C_{\text{U}}$ and $90.0\text{V}$ for $V$ in equation (V) to solve for $Q_{\text{L}}$.

    $\begin{array}{c}{Q}_{\text{L}}=\left(1.33\mu \text{F}\times \frac{{10}^{-6}\text{F}}{1\mu \text{F}}\right)90.0\text{V}\\ =119.7\times {\text{10}}^{-6}\text{C}\times \frac{1\mu \text{C}}{1\text{C}}\\ =119.7\mu \text{C}\\ \approx 120\mu \text{C}\end{array}$

The charge on $2.00\mu \text{F}$ and $4.00\mu \text{F}$ capacitor will be same.

Therefore, the charge in capacitor $2.00\mu \text{F}$ and $4.00\mu \text{F}$ is $120\mu \text{C}$ and on $3.00\mu \text{F}$ and $6.00\mu \text{F}$ is $180\mu \text{C}$.

(c)

To determine

The potential difference across each capacitor.

Answer to Problem 19P

The potential difference in capacitor $2.00\mu \text{F}$ and $3.00\mu \text{F}$ is $60.0\text{V}$ and on $6.00\mu \text{F}$ and $4.00\mu \text{F}$ is $30.0\text{V}$.

Explanation of Solution

Write the expression to calculate the potential difference across $3.00\mu \text{F}$ capacitor.

    $V_1=\frac{Q_{\text{U}}}{3.00\mu \text{F}}$                             (VI)

Here, $V_1$ is potential difference across $3.00\mu \text{F}$ capacitor.

Write the expression to calculate the potential difference across $6.00\mu \text{F}$ capacitor.

    $V_2=\frac{Q_{\text{U}}}{6.00\mu \text{F}}$                             (VII)

Here, $V_2$ is potential difference across $6.00\mu \text{F}$ capacitor.

Write the expression to calculate the potential difference across $2.00\mu \text{F}$ capacitor.

    $V_3=\frac{Q_{\text{L}}}{2.00\mu \text{F}}$                             (VIII)

Here, $V_3$ is potential difference across $2.00\mu \text{F}$ capacitor.

Write the expression to calculate the potential difference across $4.00\mu \text{F}$ capacitor.

    $V_4=\frac{Q_{\text{L}}}{4.00\mu \text{F}}$                             (VI)

Here, $V_4$ is potential difference across $4.00\mu \text{F}$ capacitor.

Conclusion:

Substitute $180\mu \text{C}$ for $Q_{\text{U}}$ in equation (VI) to solve for $V_1$.

    $\begin{array}{c}{V}_1=\frac{180\mu \text{C}}{3.00\mu \text{F}}\\ =60.0\text{V}\end{array}$

Substitute $180\mu \text{C}$ for $Q_{\text{U}}$ in equation (VII) to solve for $V_2$.

    $\begin{array}{c}{V}_2=\frac{180\mu \text{C}}{6.00\mu \text{F}}\\ =30.0\text{V}\end{array}$

Substitute $120\mu \text{C}$ for $Q_{\text{L}}$ in equation (VIII) to solve for $V_3$.

    $\begin{array}{c}{V}_3=\frac{120\mu \text{C}}{2.00\mu \text{F}}\\ =60.0\text{V}\end{array}$

Substitute $120\mu \text{C}$ for $Q_{\text{L}}$ in equation (IX) to solve for $V_4$.

    $\begin{array}{c}{V}_3=\frac{120\mu \text{C}}{4.00\mu \text{F}}\\ =30.0\text{V}\end{array}$

Therefore, the potential difference in capacitor $2.00\mu \text{F}$ and $3.00\mu \text{F}$ is $60.0\text{V}$ and on $6.00\mu \text{F}$ and $4.00\mu \text{F}$ is $30.0\text{V}$.