Chapter 27, Problem 28PQ

(a)

To determine

The equivalent capacitance for the network given.

Answer to Problem 28PQ

The equivalent capacitance of the combination is $1.71\mu \text{F}$.

Explanation of Solution

Write the formula for equivalent capacitance of capacitors in series.

    $\frac{1}{C_S}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\dots$

Here, $C_S$ is the equivalent capacitance of the series combination of capacitors.

Write the formula for equivalent capacitance of capacitors in parallel.

    $C_P=C_1+C_2+C_3+\dots$

Here, $C_P$ is the equivalent capacitance of the parallel combination of capacitors.

The capacitor $C_A$ and $C_B$ are in parallel to each other.

Write the formula for the resultant of $C_A$ and $C_B$.

  $C_{AB}=C_A+C_B$

Here, $C_{AB}$ is the resultant capacitance of $C_A$ and $C_B$.

The capacitance $C_{AB}$ and $C_C$ are in series.

Write the formula for the equivalent capacitance of the given combination.

    $\begin{array}{c}\frac{1}{C_e}=\frac{1}{C_{AB}}+\frac{1}{C_C}\\ =\frac{1}{C_A+C_B}+\frac{1}{C_C}\end{array}$

Here, $C_e$ is the equivalent capacitance of the combination.

Solve the above equation to get expression for $C_e$.

    $C_e={\left[\frac{1}{C_A+C_B}+\frac{1}{C_C}\right]}^{-1}$

Conclusion:

Substitute $1.00\mu \text{F}$ for $C_A$, $2.00\mu \text{F}$ for $C_B$, $4.00\mu \text{F}$ for $C_C$ to determine $C_e$.

    $\begin{array}{c}{C}_e={\left[\frac{1}{1.00\mu \text{F}+2.00\mu \text{F}}+\frac{1}{4.00\mu \text{F}}\right]}^{-1}\\ =1.71\mu \text{F}\end{array}$

The equivalent capacitance of the combination is $1.71\mu \text{F}$.

(b)

To determine

The charge stored in each of the capacitors.

Answer to Problem 28PQ

The charge on capacitor $C_A$ is $5.14\mu \text{C}$. The charge on capacitor $C_B$ is $10.3\mu \text{C}$. The charge on capacitor $C_C$ is $15.4\mu \text{C}$.

Explanation of Solution

The capacitors in series $C_{AB}$ and $C_C$ will be same and equal to the charge on equivalent capacitor of the combination.

Write the formula for the charge on equivalent capacitor.

    $Q_e=C_e{V}_s$

Here, $C_e$ is the equivalent capacitance of the combination, $Q_e$ is the charge on the equivalent capacitor, and $V_s$ is the voltage provided by the supply.

Write the formula for the voltage across capacitor $C_A$ and $C_B$.

    $\begin{array}{c}{V}_{AB}=\frac{Q_{AB}}{C_{AB}}\\ =\frac{Q_{AB}}{C_A+C_B}\end{array}$

Here, $V_{AB}$ is the voltage across capacitor $C_A$ and $C_B$, $Q_{AB}$ is the charge on equivalent capacitance of capacitor $C_A$ and $C_B$, and $C_{AB}$ is the equivalent capacitance of capacitor $C_A$ and $C_B$.

Write the formula for the charge across $C_A$.

    $Q_A=C_A{V}_{AB}$

Here, $Q_A$ is the charge across $C_A$.

Write the formula for the charge across $C_B$.

    $Q_B=C_B{V}_{AB}$

Here, $Q_B$ is the charge across $C_B$.

Conclusion:

Substitute $1.71\mu \text{F}$ for $C_e$, $9.00\text{V}$ for $V_s$ to determine $Q_e$.

    $\begin{array}{c}{Q}_e=\left(1.71\mu \text{F}\right)\left(9.00\text{V}\right)\\ =15.4\mu \text{C}\end{array}$

Substitute $15.4\mu \text{C}$ for $Q_{AB}$, $1.00\mu \text{F}$ for $C_A$, $2.00\mu \text{F}$ for $C_B$ to determine $V_{AB}$.

    $\begin{array}{c}{V}_{AB}=\frac{15.4\mu \text{C}}{1.00\mu \text{F}+2.00\mu \text{F}}\\ =5.14\text{V}\end{array}$

Substitute $5.14\text{V}$ for $V_{AB}$, $1.00\mu \text{F}$ for $C_A$ to determine $Q_A$.

    $\begin{array}{c}{Q}_A=\left(1.00\mu \text{F}\right)\left(5.14\text{V}\right)\\ =5.14\mu \text{C}\end{array}$

Substitute $5.14\text{V}$ for $V_{AB}$, $2.00\mu \text{F}$ for $C_B$ to determine $Q_B$.

    $\begin{array}{c}{Q}_B=\left(2.00\mu \text{F}\right)\left(5.14\text{V}\right)\\ =10.3\mu \text{C}\end{array}$

The charge on capacitor $C_A$ is $5.14\mu \text{C}$. The charge on capacitor $C_B$ is $10.3\mu \text{C}$. The charge on capacitor $C_C$ is $15.4\mu \text{C}$.

(c)

To determine

The voltage across each of the capacitors.

Answer to Problem 28PQ

The voltage across $C_A$ and $C_B$ is $5.14\text{V}$. The voltage across $C_C$ is $3.86\text{V}$.

Explanation of Solution

The capacitors in series $C_{AB}$ and $C_C$ will be same and equal to the charge on equivalent capacitor of the combination.

Write the formula for the charge on equivalent capacitor.

    $Q_e=C_e{V}_s$

Here, $C_e$ is the equivalent capacitance of the combination, $Q_e$ is the charge on the equivalent capacitor, and $V_s$ is the voltage provided by the supply.

Write the formula for the voltage across capacitor $C_A$ and $C_B$.

    $\begin{array}{c}{V}_{AB}=\frac{Q_{AB}}{C_{AB}}\\ =\frac{Q_{AB}}{C_A+C_B}\end{array}$

Here, $V_{AB}$ is the voltage across capacitor $C_A$ and $C_B$, $Q_{AB}$ is the charge on equivalent capacitance of capacitor $C_A$ and $C_B$, and $C_{AB}$ is the equivalent capacitance of capacitor $C_A$ and $C_B$.

Write the formula for the voltage across $C_C$.

    $V_C=\frac{Q_C}{C_C}$

Here, $Q_C$ is the charge across $C_C$, $V_C$ is the voltage across $C_C$.

Conclusion:

Substitute $1.71\mu \text{F}$ for $C_e$, $9.00\text{V}$ for $V_s$ to determine $Q_e$.

    $\begin{array}{c}{Q}_e=\left(1.71\mu \text{F}\right)\left(9.00\text{V}\right)\\ =15.4\mu \text{C}\end{array}$

Substitute $15.4\mu \text{C}$ for $Q_{AB}$, $1.00\mu \text{F}$ for $C_A$, $2.00\mu \text{F}$ for $C_B$ to determine $V_{AB}$.

    $\begin{array}{c}{V}_{AB}=\frac{15.4\mu \text{C}}{1.00\mu \text{F}+2.00\mu \text{F}}\\ =5.14\text{V}\end{array}$

Substitute $15.4\mu \text{C}$ for $Q_C$, $4.00\mu \text{F}$ for $C_C$ to determine $V_C$.

    $\begin{array}{c}{V}_C=\frac{15.4\mu \text{C}}{4.00\mu \text{F}}\\ =3.86\text{V}\end{array}$

The voltage across $C_A$ and $C_B$ is $5.14\text{V}$. The voltage across $C_C$ is $3.86\text{V}$.