Chapter 26, Problem 26.20P

Three capacitors are connected to a battery as shown in Figure P25.10. Their capacitances are C₁ = 3C, C₂ = C, and C₃ = 5C. (a) What is the equivalent capacitance of this set of capacitors? (b) State the ranking of the capacitors according to the charge they store from largest to smallest. (c) Rank the capacitors according to the potential differences across them from largest to smallest. (d) What If? Assume C₃ is increased. Explain what happens to the charge stored by each capacitor.

Figure P25.10

To determine

The equivalent capacitance of this set of capacitors.

Answer to Problem 26.20P

The equivalent capacitance of this set of capacitors is .

Explanation of Solution

Given information: The value of capacitor 1 is $3C$, value of capacitor 2 is $C$, value of capacitor 3 is $5C$.

Explanation:

The capacitors $C_2\text{and}{C}_3$ are in parallel.

Formula to calculate the equivalent capacitance of the system when they are connected in parallel.

${C^{\prime }}_P=C_2+C_3$ (1)

Here,

${C^{\prime }}_P$ is the equivalent capacitance of the system when they are connected in parallel.

$C_2$ is the value of capacitor 2.

$C_3$ is the value of capacitor 3.

Substitute $C$ for $C_2$, $5C$ for $C_3$ in equation (1) to find ${C^{\prime }}_P$,

$\begin{array}{c}{C^{\prime }}_P=C+5C\\ =6C\end{array}$

Thus, the equivalent capacitance of the system when they are connected in parallel is $6C$.

The capacitors $C_1\text{and}{C^{\prime }}_P$ are in series.

Formula to calculate the equivalent capacitance of the system when they are connected in series.

$\frac{1}{C_{\text{eq}}}=\frac{1}{C_1}+\frac{1}{{C^{\prime }}_P}$ (2)

Here,

$C_{\text{eq}}$ is the equivalent capacitance of the system when they are connected in series.

$C_1$ is the value of capacitor 1.

Substitute $3C$ for $C_1$, $6C$ for ${C^{\prime }}_P$ in equation (2) to find $C_{\text{eq}}$,

$\begin{array}{c}\frac{1}{C_{\text{eq}}}=\frac{1}{3C}+\frac{1}{6C}\\ C_{\text{eq}}=2C\end{array}$

Thus, the equivalent capacitance of this set of capacitors is $2C$.

Conclusion:

Therefore, the equivalent capacitance of this set of capacitors is $2C$.

To determine

The ranking of the capacitors according to the charge they store from largest to smallest.

Answer to Problem 26.20P

The ranking of the capacitors according to the charge they store from largest to smallest is $Q_1>Q_3>Q_2$.

Explanation of Solution

Given information: The value of capacitor 1 is $3C$, value of capacitor 2 is $C$, value of capacitor 3 is $5C$.

Explanation:

Calculate the voltage across ${C^{\prime }}_P$.

$V_{{{\text{C}}^{\prime }}_{\text{P}}}=\frac{Q}{{C^{\prime }}_P}$ (3)

Here,

$V_{{{\text{C}}^{\prime }}_{\text{P}}}$ is the voltage across ${C^{\prime }}_P$.

Substitute $6C$ for ${C^{\prime }}_P$ in equation (3) to find $V_{{{\text{C}}^{\prime }}_{\text{P}}}$,

$V_{{{\text{C}}^{\prime }}_{\text{P}}}=\frac{Q}{6C}$

Calculate the charge for the capacitor $C_2$.

$Q_2=C_2{V}_{{{\text{C}}^{\prime }}_{\text{P}}}$ (4)

Here,

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

Substitute $C$ for $C_2$, $\frac{Q}{6C}$ for $V_{{{\text{C}}^{\prime }}_{\text{P}}}$ in equation (4) to find $Q_2$,

$\begin{array}{c}{Q}_2=C\times \frac{Q}{6C}\\ =\frac{Q}{6}\end{array}$

Thus, the charge for the capacitor $C_2$ is $\frac{Q}{6}$.

Calculate the charge for the capacitor $C_3$.

$Q_3=C_3{V}_{{{\text{C}}^{\prime }}_{\text{P}}}$ (5)

Here,

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

Substitute $5C$ for $C_3$, $\frac{Q}{6C}$ for $V_{{{\text{C}}^{\prime }}_{\text{P}}}$ in equation (5) to find $Q_3$,

$\begin{array}{c}{Q}_3=5C\times \frac{Q}{6C}\\ =\frac{5Q}{6}\end{array}$

Thus, the charge for the capacitor $C_3$ is $\frac{5Q}{6}$.

Calculate the voltage across $C_{\text{eq}}$.

$V_{eq}=\frac{Q}{C_{\text{eq}}}$ (6)

Here,

$V_{eq}$ is the voltage across $C_{\text{eq}}$.

Substitute $2C$ for $C_{\text{eq}}$ in equation (6) to find $V_{eq}$,

$V_{eq}=\frac{Q}{2C}$

Calculate the charge for the capacitor $C_1$.

$Q_1=C_1{V}_{\text{eq}}$ (7)

Here,

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

Substitute $3C$ for $C_1$, $\frac{Q}{2C}$ for $V_{eq}$ in equation (7) to find $Q_1$,

$\begin{array}{c}{Q}_1=3C\times \frac{Q}{2C}\\ =\frac{3Q}{2}\end{array}$

Thus, the charge for the capacitor $C_1$ is $\frac{3Q}{2}$.

The ranking of the capacitors according to the charge they store from largest to smallest is $\frac{3Q}{2}>\frac{5Q}{6}>\frac{Q}{6}$.

Thus, the ranking of the capacitors according to the charge they store from largest to smallest is $Q_1>Q_3>Q_2$.

Conclusion:

Therefore, the ranking of the capacitors according to the charge they store from largest to smallest is $Q_1>Q_3>Q_2$.

To determine

The ranking of the capacitors according to the potential differences across them from largest to smallest.

Answer to Problem 26.20P

The ranking of the capacitors according to the potential differences across them from largest to smallest is $V_1>V_2=V_3$.

Explanation of Solution

Given information: The value of capacitor 1 is $3C$, value of capacitor 2 is $C$, value of capacitor 3 is $5C$.

Explanation:

Calculate the potential difference across $C_1$,

$V_1=\frac{Q}{C_1}$ (8)

Here,

$V_1$ is the potential difference across $C_1$.

Substitute $3C$ for $C_1$ in equation (8) to find $V_1$,

$V_1=\frac{Q}{3C}$

Thus, the potential difference across $C_1$ is $\frac{Q}{3C}$.

Calculate the potential difference across $C_2$,

$V_2=\frac{Q}{{C^{\prime }}_P}$ (9)

Here,

$V_2$ is the potential difference across $C_2$.

Substitute $6C$ for ${C^{\prime }}_P$ in equation (9) to find $V_2$,

$V_2=\frac{Q}{6C}$

Thus, the potential difference across $C_2$ is $\frac{Q}{6C}$.

Calculate the potential difference across $C_3$,

$V_3=\frac{Q}{{C^{\prime }}_P}$ (10)

Here,

$V_3$ is the potential difference across $C_3$.

Substitute $6C$ for ${C^{\prime }}_P$ in equation (10) to find $V_3$,

$V_3=\frac{Q}{6C}$

Thus, the potential difference across $C_3$ is $\frac{Q}{6C}$.

The ranking of the capacitors according to the potential differences across them from largest to smallest is $\frac{Q}{3C}>\frac{Q}{6C}=\frac{Q}{6C}$.

Thus, the ranking of the capacitors according to the potential differences across them from largest to smallest is $V_1>V_2=V_3$.

Conclusion:

Therefore, the ranking of the capacitors according to the potential differences across them from largest to smallest is $V_1>V_2=V_3$.

To determine

The effect acting on the charge stored by each capacitor if $C_3$ is increased.

Answer to Problem 26.20P

The effect acting on the charge stored by each capacitor if $C_3$ is increased, then the charge across capacitor 3 decreases, the charge across capacitor 2 decreases and the charge across capacitor 1 increases.

Explanation of Solution

Given information: The value of capacitor 1 is $3C$, value of capacitor 2 is $C$, value of capacitor 3 is $5C$.

Explanation:

If the value of capacitor 3 is increased, the total capacitance will increase which results in increasing the total charge.

Due to this, the charge across capacitor 1 increases.

Since, the charge across capacitor 1 is directly proportional to the voltage across the capacitor 1. So, the voltage across $C_1$ will increases. Hence, the voltage across the capacitor 2 and capacitor 3 will decreases.

Since the voltage across the capacitor 2 decreases, charge across the capacitor 2 also decreases and the voltage across the capacitor 3 decreases, charge across the capacitor 3 also decreases.

Thus, if $C_3$ is increased, then the charge across capacitor 3 decreases, the charge across capacitor 2 decreases and the charge across capacitor 1 increases.

Conclusion:

Therefore, if $C_3$ is increased, then the charge across capacitor 3 decreases, the charge across capacitor 2 decreases and the charge across capacitor 1 increases.