Chapter 25, Problem 66SP

Referring to Fig. 25-13, what is the equivalent capacitance of the circuit across the 6.0-V battery? What are the voltages across $C_3,C_8,C_9,C_5,\text{ and }{C}_2$? How much energy is stored in the 2.0- $\mu \text{F}$ capacitor? [Hint: Redraw the circuit and watch for shorts. Remember that the charge on the equivalent capacitor representing a string of series elements is the same on each of those series capacitors.]

Fig. 25-13

To determine

The equivalent capacitanceacross the $6.0\text{ V}$ battery of the circuit shownin Fig. 25.13. Also calculate the voltages across $C_3,C_8,C_9,C_5$, and $C_2$, and the energy stored in the $2.0\mu \text{F}$ capacitor.

Answer to Problem 66SP

Solution:

${\text{C}}_{eq}=2.0\mu \text{F}$ and $\text{3 V, 0 V, 0 V, 2 V, 1 V}$;$\text{4 }\mu \text{J}$

Explanation of Solution

Given data:

Refer to the circuit mentioned in the Fig.25-13.

The potential difference of the battery is $6.0\text{ V}$.

Formula used:

Write the expression for equivalent capacitance, when the capacitors are connected in series.

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

Here, $C_{eq}$ is the equivalent capacitance, $C_1$ and $C_2$ are the individual capacitances.

Write the expression for equivalent capacitance, when the capacitors are connected in parallel.

$C_{eq}=C_1+C_2$

Write the expression for the charge on the capacitor.

$q=CV$

Here, $q$ is the charge on the capacitor, $C$ is the capacitance, and $V$ is the voltage.

Write the expression for the energy stored in the system of capacitors.

$E=\frac{1}{2}C{V}^2$

Here, $E$ is the energy stored in the system of capacitors, $C$ is the capacitance, and $V$ is the voltage of the system.

Explanation:

Consider the following figure 1.

From figure 1, the capacitors $C_1$ and $C_9$ are open circuited and the series combination of capacitors $C_7$ and $C_8$ is shorted, therefore, there is no voltage drop across this capacitor. Thus, the capacitors $C_1$, $C_9$, $C_7$, and $C_8$ are removable from the circuit.

Consider Figure 2, it is a reduced form of figure 1.

From figure 2, the capacitors $C_5,C_6$, and $C_4$ are connected in parallel.

The expression for equivalent capacitance, when the capacitors are connected in parallel is,

$C_{e{q}_1}=C_5+C_6+C_4$

Here, $C_{e{q}_1}$ is the equivalent capacitance of capacitors $C_5,C_6$, and $C_4$.

Substitute $2.0\mu \text{F}$ for $C_5$, $3.0\mu \text{F}$ for $C_6$, and $1.0\mu \text{F}$ for $C_4$

$\begin{array}{c}{C}_{e{q}_1}=2.0\mu \text{F}+3.0\mu \text{F}+\text{1}\text{.0 }\mu \text{F }\\ =6.\text{0 }\mu \text{F}\end{array}$

The resultant combination $C_{e{q}_1}$ is connected in series with $C_2$ and $C_3$.

The expression for equivalent capacitance, when the capacitors are connected in series is,

$\frac{1}{C_{e{q}_2}}=\frac{1}{C_{e{q}_1}}+\frac{1}{C_2}+\frac{1}{C_3}$

Here, $C_{e{q}_2}$ is the equivalent capacitance of capacitors $C_{e{q}_1}$, $C_2$, and $C_3$.

Substitute $6.0\mu \text{F }$ for $C_{e{q}_1}$, $12.0\mu \text{F}$ for $C_2$, and $4.0\mu \text{F}$ for $C_3$

$\begin{array}{c}\frac{1}{C_{e{q}_2}}=\frac{1}{6.0\mu \text{F}}+\frac{1}{12.0\mu \text{F}}+\frac{1}{4.0\mu \text{F}}\\ \frac{1}{C_{e{q}_2}}=\frac{6}{12.0\mu \text{F}}\\ C_{e{q}_2}=2.0\mu \text{F}\end{array}$

Thus, the equivalent capacitance across the $6.0\text{ V}$ battery is $2.0\mu \text{F}$.

Calculate the total charge flow in the circuit.

Write the expression for the charge on the capacitor.

$q=C_{eq}V$

Here, $q$ is the charge on the capacitor, $C_{eq}$ is the equivalent capacitance of the capacitor, and $V$ is the voltage.

Substitute $2.0\mu \text{F}$ for $C_{eq}$ and $6.0\text{ V}$ for $V$

$\begin{array}{c}q=\left(2.0\mu \text{F}\right)\left(6.0\text{ V}\right)\\ =\left(2.0\mu \text{F}\left(\frac{{10}^{-6}\text{ F}}{1\mu \text{F}}\right)\right)\left(6.0\text{ V}\right)\\ =1.2\times {10}^{-5}\text{ C}\end{array}$

If the capacitors are connected in series, then the charge flow across each capacitor is same.

Thus, charge flow across the capacitor $C_2$ and $C_3$ will be equal to $q$.

$q_2=1.2\times {10}^{-5}\text{ C}$

$q_3=1.2\times {10}^{-5}\text{ C}$

Here, $q$ is the total charge, $q_2$ and $q_3$ are the charges flow across the capacitors $C_2$ and $C_3$.

The expression for the charge across thecapacitor $C_3$, $q_3$ is,

$q_3=C_3{V}_3$

Here, $V_3$ is the voltage across the capacitor $C_3$.

Substitute $1.2\times {10}^{-5}\text{ C}$ for $q_3$ and $4.0\mu \text{F}$ for $C_3$

$\left(1.2\times {10}^{-5}\text{ C}\right)=\left(4.0\mu \text{F}\right)V_3$

Solve for $V_3$.

$\begin{array}{c}{V}_3=\frac{1.2\times {10}^{-5}\text{ C}}{4.0\mu \text{F}\left(\frac{{10}^{-6}\text{ F}}{1\mu \text{F}}\right)}\\ =3.0\text{ V}\end{array}$

Since the capacitors $C_7$ and $C_8$ are short, the voltage across the capacitors $C_7$ and $C_8$ is zero.

$\begin{array}{l}{V}_7=0\text{ V}\\ V_8=0\text{ V}\end{array}$

Here, $V_7$ is the voltage across the capacitor $C_7$ and $V_8$ is the voltage across the capacitor $C_8$.

The capacitor $C_9$ is open circuit, thus the voltage across the $C_9$ will be zero.

$V_9=0\text{V}$

Here, $V_9$ is the voltage across capacitor $C_9$

The charge across the parallel combination $C_{e{q}_1}$ is $1.2\times {10}^{-5}\text{ C}$ as it is connected in series with capacitors $C_2$ and $C_3$.

The expression for the charge on the parallel combination $C_{e{q}_1}$ is,

$q=C_{e{q}_1}V$

Substitute $1.2\times {10}^{-5}\text{ C}$ for $q$ and $6.0\mu \text{F}$ for $C_{e{q}_1}$

$1.2\times {10}^{-5}\text{ C}=\left(6.0\mu \text{F}\left(\frac{{10}^{-6}\text{ F}}{1\mu \text{F}}\right)\right)V$

Solve for $V$.

$\begin{array}{c}V=\frac{1.2\times {10}^{-5}\text{ C}}{6\times {10}^{-6}\text{ F}}\\ =2\text{ V}\end{array}$

Since, the capacitors $C_4$, $C_5$, and $C_6$ are connected in parallel, therefore, the voltage across each of them is equal.

The voltage across the capacitor $C_5$ is $2\text{ V}$.

$V_5=2\text{ V}$

Here, $V_5$ is the voltage across the capacitor $C_5$.

The expression for a charge on the capacitor $C_2$ is,

$q_2=C_2{V}_2$

Here, $V_2$ is the voltage across the capacitor $C_2$.

Substitute $1.2\times {10}^{-5}\text{ C}$ for $q_2$ and $12.0\mu \text{F}$ for $C_2$

$\left(1.2\times {10}^{-5}\text{ C}\right)=\left(12.0\mu \text{F}\right)V_2$

Solve for $V_2$.

$\begin{array}{c}{V}_2=\frac{1.2\times {10}^{-5}\text{ C}}{12.0\mu \text{F}\left(\frac{{10}^{-6}\text{ F}}{1\mu \text{F}}\right)}\\ =1.0\text{ V}\end{array}$

The $2.0\mu \text{F}$ capacitor connected in the circuit is $C_5$.

The expression for the energy stored in the capacitor $C_5$, $E_5$ is,

$E_5=\frac{1}{2}{C}_5{V}_5{}^2$

Substitute $2.0\mu \text{F}$ for $C_5$ and $2.0\text{ V}$ for $V_5$

$\begin{array}{c}{E}_5=\frac{1}{2}\left(2.0\mu \text{F}\left(\frac{{10}^{-6}\text{ F}}{1\mu \text{F}}\right)\right){\left(2.0\text{ V}\right)}^2\\ =\frac{\left(2\times {10}^{-6}\right)\left(4\right)}{2}\text{ J}\\ =\text{4}\times {\text{10}}^{-6}\text{ J}\left(\frac{1\mu \text{J}}{{10}^{-6}\text{ J}}\right)\\ =4\mu \text{J}\end{array}$

Conclusion:

The equivalent capacitance across the $6.0\text{ V}$ battery in the given circuit is $2.0\mu \text{F}$. The voltages across $C_3,C_8,C_9,C_5,and{C}_2$ are $3.0\text{ V}$, $0\text{ V}$, $0\text{ V}$, $2\text{ V}$, and $1\text{ V}$, respectively. The energy stored in the $2.0\mu \text{F}$ capacitor is $4.0\mu \text{J}$.