Chapter 26, Problem 26.26P

Find (a) the equivalent capacitance of the capacitors in Figure P26.26, (b) the charge on each capacitor, and (c) the potential difference across each capacitor.

To determine

The equivalent capacitance of the capacitors.

Answer to Problem 26.26P

The equivalent capacitance of the capacitors in the system is $2.67\text{μF}$.

Explanation of Solution

Given: The figure given below shows the arrangement of capacitors.

Figure 1

Write the equation for equivalent capacitance in the series combination.

$\frac{1}{C_{\text{eq}}}=\frac{1}{C_1}+\frac{1}{C_2+C_3}+\frac{1}{C_4}$

Here,

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

$C_{\text{1}}$ is the capacitance of the first capacitor.

$C_{\text{2}}$ is the capacitance of the second capacitor in the upper branch.

$C_{\text{3}}$ is the capacitance of the third capacitor in the lower branch.

$C_{\text{4}}$ is the capacitance of the fourth capacitor.

Substitute $8.00\text{μF}$ for $C_{\text{1}}$ and $6.00\text{μF}$ for $C_{\text{2}}$, $2.00\text{μF}$ for $C_{\text{3}}$ and $8.00\text{μF}$ for $C_{\text{4}}$ in above equation to find $C_{\text{eq}}$.

$\begin{array}{c}\frac{1}{C_{\text{eq}}}=\frac{1}{\left(8.00\text{μF}\right)}+\frac{1}{\left(6.00\text{μF}\right)+\left(2.00\text{μF}\right)}+\frac{1}{\left(8.00\text{μF}\right)}\\ C_{\text{eq}}=2.67\text{μF}\end{array}$

Conclusion:

Therefore, the equivalent capacitance of the capacitors in the system is $2.67\text{μF}$.

To determine

The charge on each capacitor.

Answer to Problem 26.26P

The charge on $8.00\text{μF}$ capacitor to the left, $6.00\text{μF}$ capacitor, $2.00\text{μF}$ capacitor and $8.00\text{μF}$ capacitor to the right are $24.0\text{μC}$, $18.0\text{μC}$, $6.00\text{μC}$ and $24.0\text{μC}$ respectively.

Explanation of Solution

Write the equation for charge on capacitor $C_1$.

$Q_{\text{1}}=C_{\text{eq}}\left(\Delta V\right)$

Here,

$Q_{\text{1}}$ is the charge on capacitor $C_1$.

$\Delta V$ is the potential difference.

Substitute $2.67\text{μF}$ for $C_{\text{eq}}$ and $9.00\text{V}$ for $\Delta V$ in above equation to find $Q_{\text{1}}$.

$\begin{array}{c}{Q}_{\text{1}}=\left(2.67\text{μF}\right)\left(9.00\text{V}\right)\\ =24.0\text{μC}\end{array}$

Thus, the charge on $8.00\text{μF}$ capacitor to the left is $24.0\text{μC}$.

The charge on $8.00\text{μF}$ capacitor to the left and $8.00\text{μF}$ capacitor to the right is same due to the same potential difference.

The charge on capacitor $C_4$ is,

$Q_{\text{4}}=24.0\text{μC}$

Thus, the charge on $8.00\text{μF}$ capacitor to the right is $24.0\text{μC}$.

Write the equation for charge on capacitor $C_2$.

$Q_{\text{2}}=C_{\text{2}}\left(\frac{Q_1}{C_2+C_3}\right)$

Here,

$Q_{\text{2}}$ is the charge on capacitor $C_2$.

Substitute $6.00\text{μF}$ for $C_2$, $24.0\text{μC}$ for $Q_1$ and $2.00\text{μF}$ for $C_3$ in above equation to find $Q_{\text{2}}$.

$\begin{array}{c}{Q}_{\text{2}}=\left(6.00\text{μF}\right)\left(\frac{\left(24.0\text{μC}\right)}{\left(6.00\text{μF}\right)+\left(2.00\text{μF}\right)}\right)\\ =18.0\text{μC}\end{array}$

Thus, the charge on $6.00\text{μF}$ capacitor is $18.0\text{μC}$.

Write the equation for charge on capacitor $C_3$.

$Q_3=C_{\text{3}}\left(\frac{Q_1}{C_2+C_3}\right)$

Here,

$Q_{\text{3}}$ is the charge on capacitor $C_3$.

Substitute $2.00\text{μF}$ for $C_3$, $24.0\text{μC}$ for $Q_1$ and $6.00\text{μF}$ for $C_2$ in above equation to find $Q_{\text{3}}$.

$\begin{array}{c}{Q}_{\text{3}}=\left(2.00\text{μF}\right)\left(\frac{\left(24.0\text{μC}\right)}{\left(6.00\text{μF}\right)+\left(2.00\text{μF}\right)}\right)\\ =6.00\text{μC}\end{array}$

The charge on $2.00\text{μF}$ capacitor is $6.00\text{μC}$.

Conclusion:

Therefore, the charge on $8.00\text{μF}$ capacitor to the left, $6.00\text{μF}$ capacitor, $2.00\text{μF}$ capacitor and $8.00\text{μF}$ capacitor to the right are $24.0\text{μC}$, $18.0\text{μC}$, $6.00\text{μC}$ and $24.0\text{μC}$ respectively.

To determine

The potential difference across each capacitor.

Answer to Problem 26.26P

The potential difference across all the capacitors are same and equal to $3.00\text{V}$.

Explanation of Solution

Write the equation for potential difference across capacitor $C_1$.

$\Delta V_1=\frac{Q_1}{C_1}$

Here,

$\Delta V_1$ is the potential difference across $8.00\text{μF}$ to the left.

Substitute $24.0\text{μC}$ for $Q_1$ and $8.00\text{μF}$ for $C_1$ in above equation to find $\Delta V_1$.

$\begin{array}{c}\Delta V_1=\frac{\left(24.0\text{μC}\right)}{\left(8.00\text{μF}\right)}\\ =3.00\text{V}\end{array}$

Thus, the potential difference across $8.00\text{μF}$ to the left is $3.00\text{V}$.

The potential difference across $8.00\text{μF}$ capacitor to the left and $8.00\text{μF}$ capacitor to the right is same due to the same charge and capacitance.

The potential difference across capacitor $C_4$ is,

$\Delta V_4=3.00\text{V}$

Thus, the potential difference across $8.00\text{μF}$ to the right is $3.00\text{V}$.

Write the equation for potential difference across capacitor $C_2$.

$\Delta V_2=\frac{Q_1}{C_2+C_3}$

Here,

$\Delta V_2$ is the charge on capacitor $C_2$.

Substitute $24.0\text{μC}$ for $Q_1$, $6.00\text{μF}$ for $C_2$ and $2.00\text{μF}$ for $C_3$ in above equation to find $\Delta V_2$.

$\begin{array}{c}\Delta V_2=\frac{\left(24.0\text{μC}\right)}{\left(6.00\text{μF}\right)+\left(2.00\text{μF}\right)}\\ =3.00\text{V}\end{array}$

Thus, the potential difference across $6.00\text{μF}$ is $3.00\text{V}$.

The potential difference across the $2.00\text{μF}$ capacitor is same as the potential difference across $6.00\text{μF}$ due to the parallel combination.

The potential difference across capacitor $C_3$ is,

$\Delta V_3=3.00\text{V}$

The potential difference across $2.00\text{μF}$ is $3.00\text{V}$.

Conclusion:

Therefore, the potential difference across all the capacitors are same and equal to $3.00\text{V}$.