Chapter 26, Problem 26.77CP

Calculate the equivalent capacitance between points a and b in Figure P26.77. Notice that this system is not a simple series or parallel combination. Suggestion: Assume a potential difference Δv between [joints a and b. Write expressions for Δv_ab in terms of the charges and capacitances for the various possible pathways from a to b and require conservation of charge for those capacitor plates that are connected to each other.

To determine

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

Answer to Problem 26.77CP

The equivalent capacitance between points $a$ and $b$ is $\text{3}\text{.00}\text{μF}$.

Explanation of Solution

Given info: The potential difference between the points $a$ and $b$ is $\Delta V$.

From given Figure, the capacitor $C=8.00\text{μF}$ acts as the common capacitor due to there is no potential across it.

Since, capacitors $C_1$, $C_2$ and $C$ are connected with potential $V_1$ and capacitors $C_3$, $C_4$ and $C$ are connected with potential $V_2$.

Formula to calculate the charge on the capacitor is,

$Q=C\Delta V_{\text{ab}}$

Here,

$Q$ is the charge on the capacitor.

$\Delta V_{\text{ab}}$ is the electric potential difference between points $a$ and $b$.

$C$ is the equivalent capacitance of the capacitor.

The charge on the capacitor $C_1$ is,

$Q_1=C_1\left(V_{\text{a}}-V_1\right)$

Here,

$Q_1$ is the charge on the capacitor $C_1$.

The charge on the capacitor $C_2$ is,

$Q_2=C_2\left(V_1-V_{\text{b}}\right)$

Here,

$Q_2$ is the charge on the capacitor $C_2$.

$V_{\text{B}}$ is the  electric potential at point $b$.

$V_1$ is the electric potential across the capacitor $C_1$ and $C_2$.

The charge on the capacitor $C_3$ is,

$Q_3=C_3\left(V_{\text{A}}-V_2\right)$

Here,

$Q_3$ is the charge on the capacitor $C_3$.

$V_{\text{A}}$ is the  electric potential at point $a$.

$V_2$ is the electric potential across the capacitor $C_3$ and $C_4$.

The charge on the capacitor $C_4$ is,

$Q_4=C_4\left(V_2-V_{\text{b}}\right)$

Here,

$Q_4$ is the charge on the capacitor $C_4$.

Since, the charge on $C_1$ and $C_2$ are same.

$Q_1=Q_2$

Substitute $C_1\left(V_{\text{a}}-V_1\right)$ for $Q_1$ and $C_2\left(V_1-V_{\text{B}}\right)$ for $Q_2$.

$\begin{array}{c}{C}_1\left(V_{\text{a}}-V_1\right)=C_2\left(V_1-V_{\text{b}}\right)\\ V_1=\frac{C_1{V}_{\text{a}}+C_2{V}_{\text{b}}}{C_1+C_2}\end{array}$

Since, the charge on $C_3$ and $C_4$ are same.

$Q_3=Q_4$

Substitute $C_3\left(V_{\text{A}}-V_2\right)$ for $Q_3$ and $C_4\left(V_2-V_{\text{b}}\right)$ for $Q_4$ in above equation.

$\begin{array}{c}{C}_3\left(V_{\text{b}}-V_2\right)=C_4\left(V_2-V_{\text{b}}\right)\\ V_2=\frac{C_3{V}_{\text{a}}+C_4{V}_{\text{b}}}{C_3+C_4}\end{array}$

The potential across the capacitor $C$ is zero.

$\begin{array}{c}{V}_1-V_2=0\\ V_1=V_2\end{array}$

Substitute $\frac{C_1{V}_{\text{a}}+C_2{V}_{\text{b}}}{C_1+C_2}$ for $V_1$ and $\frac{C_3{V}_{\text{a}}+C_4{V}_{\text{b}}}{C_3+C_4}$ for $V_2$ in above equation.

$\begin{array}{c}\frac{C_1{V}_{\text{a}}+C_2{V}_{\text{b}}}{C_1+C_2}=\frac{C_3{V}_{\text{a}}+C_4{V}_{\text{b}}}{C_3+C_4}\\ C_1{C}_2\left(V_{\text{a}}-V_{\text{b}}\right)=C_2{C}_3\left(V_{\text{a}}-V_{\text{b}}\right)\\ \frac{C_1}{C_2}=\frac{C_3}{C_4}\end{array}$ (1)

The ratio of the capacitors $C_1$ and $C_2$ is,

$p=\frac{C_1}{C_2}$

Here,

$p$ is the ratio of capacitors $C_1$ and $C_2$.

Substitute $4\text{μF}$ for $C_1$ and $4\text{μF}$ for $C_2$ in above equation.

$\begin{array}{c}p=\frac{C_1}{C_2}\\ =\frac{4.00\text{μF}}{4.00\text{μF}}\\ =1\end{array}$ (2)

The ratio of the capacitors $C_3$ and $C_4$ is,

$r=\frac{C_3}{C_4}$

Here,

$r$ is the ratio of the capacitors $C_3$ and $C_4$.

Substitute $2.00\text{μF}$ for $C_3$ and $2.00\text{μF}$ for $C_4$ in above equation.

$\begin{array}{c}r=\frac{C_3}{C_4}\\ =\frac{2.00\text{μF}}{2.00\text{μF}}\\ =1\end{array}$ (3)

Equate the left hand side of equation (2) and equation (3).

$p=r$

Thus, the potential difference across the capacitor $8\text{μF}$ is zero.

Thus, the modified electrical circuit diagram is shown below.

Figure (1)

The upper part of the capacitors is in series. So, the equivalent capacitance for series connection is,

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

Here,

$C_{\text{eq}}$ is the equivalent capacitance of upper part for series connection.

Substitute $4.00\text{μF}$ for $C_1$ and $4.00\text{μF}$ for $C_2$ in above equation.

$\begin{array}{c}\frac{1}{C_{\text{eq}}}=\frac{1}{4.00\text{μF}}+\frac{1}{4.00\text{μF}}\\ C_{\text{eq}}=2.00\text{μF}\end{array}$

Thus, the equivalent capacitance for series connection is $2.00\text{μF}$.

The lower part of the capacitors is in series. So, the equivalent capacitance for series connection is,

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

Here,

$C_{\text{eq}}\text{'}$ is the equivalent capacitance of lower part for series connection.

Substitute $2.00\text{μF}$ for $C_3$ and $2.00\text{μF}$ for $C_4$ in above equation.

$\begin{array}{c}\frac{1}{C_{\text{eq}}\text{'}}=\frac{1}{2.00\text{μF}}+\frac{1}{2.00\text{μF}}\\ C_{\text{eq}}\text{'}=1\text{μF}\end{array}$

Thus, the equivalent capacitance of lower part for series connection is $1.00\text{μF}$.

Now the equivalent capacitances of the upper and lower parts are in parallel. Hence, the equivalent capacitance of the system is,

$C=C_{\text{eq}}+C_{\text{eq}}\text{'}$

Substitute $1.00\text{μF}$ for $C_{\text{eq}}\text{'}$ and $2.00\text{μF}$ for $C_{\text{eq}}$ in above equation to find $C$.

$\begin{array}{c}C=2.00\text{μF}+1.00\text{μF}\\ \text{=3}\text{.00}\text{μF}\end{array}$

Conclusion:

Therefore, the equivalent capacitance between points $a$ and $b$ is $\text{3}\text{.00}\text{μF}$.