Chapter 16, Problem 44P

Three capacitors are connected to a battery as shown in Figure P16.44. 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) Assume C₃ is increased. Explain what happens to the charge stored by each capacitor.

Figure P16.44

To determine

The equivalent capacitance.

Answer to Problem 44P

The equivalent capacitance is 2C.

Explanation of Solution

The capacitors $C_2$ and $C_3$ are connected in parallel combination. The equivalent capacitance is,

$C_p=C_2+C_3$

The capacitances $C_p$ and $C_1$ are in series combination. The total equivalent capacitance is,

$C_{eq}=\frac{C_p{C}_1}{C_p+C_1}$

Therefore,

$C_{eq}=\frac{C_1\left(C_2+C_3\right)}{C_1+C_2+C_3}$

Substitute $3C$ for $C_1$, $C$ for $C_2$ and $5C$ for $C_3$

$\begin{array}{c}{C}_{eq}=\frac{\left(3C\right)\left(C+5C\right)}{3C+C+5C}\\ =2C\end{array}$

On Re-arranging,

$\begin{array}{c}{C}_{eq}=\frac{8.00\text{μF}}{3}\\ =2.67\text{μF}\end{array}$

Conclusion:

The equivalent capacitance is 2C

To determine

The charge on each capacitor.

Answer to Problem 44P

 The charges on each capacitor are ranked as $Q_1>Q_3>Q_2$

Explanation of Solution

Formula to calculate the charge on $C_1$ is,

$Q_1=C_{eq}\left(\Delta V\right)$

• $\Delta V$ is the potential difference.

Substitute $2C$ for $C_{eq}$

$Q_1=2C\left(\Delta V\right)$

Formula to calculate the charge on $C_2$ is,

$Q_2=C_2\left(\frac{Q_1}{C_2+C_3}\right)$

Substitute $C$ for $C_2$ and $5C$ for $C_3$ and $2C\left(\Delta V\right)$ for $Q_1$

$\begin{array}{c}{Q}_2=\left(C\right)\left(\frac{2C\left(\Delta V\right)}{\left(C\right)+\left(5C\right)}\right)\\ =\frac{C\left(\Delta V\right)}{3}\end{array}$

Formula to calculate the charge on $C_3$ is,

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

Substitute $C$ for $C_2$ and $5C$ for $C_3$ and $2C\left(\Delta V\right)$ for $Q_1$

$\begin{array}{c}{Q}_2=\left(5C\right)\left(\frac{2C\left(\Delta V\right)}{\left(C\right)+\left(5C\right)}\right)\\ =\frac{5C\left(\Delta V\right)}{3}\end{array}$

Conclusion:

The charges on each capacitor are ranked as $Q_1>Q_3>Q_2$

To determine

The potential difference on each capacitor.

Answer to Problem 44P

The potential difference on each capacitor are ranked as $\left(\Delta V_1\right)>\left(\Delta V_2\right)=\left(\Delta V_3\right)$

Explanation of Solution

Formula to calculate the potential difference on $C_1$ is,

$\Delta V_1=\Delta V-\left(\frac{Q_1}{C_2+C_3}\right)$

Substitute $C$ for $C_2$ and $5C$ for $C_3$ and $2C\left(\Delta V\right)$ for $Q_1$

$\begin{array}{c}\Delta V_1=\Delta V-\left(\frac{2C\left(\Delta V\right)}{C+5C}\right)\\ =\frac{2\Delta V}{3}\end{array}$

Formula to calculate the potential difference on $C_2$ and $C_3$ is,

$\begin{array}{c}\Delta V_2=\Delta V_3\\ =\frac{Q_1}{C_2+C_3}\end{array}$

Substitute $C$ for $C_2$ and $5C$ for $C_3$ and $2C\left(\Delta V\right)$ for $Q_1$

$\begin{array}{c}\Delta V_2=\Delta V_3\\ =\frac{2C\left(\Delta V\right)}{C+5C}\\ =\frac{\Delta V}{3}\end{array}$

Conclusion:

The potential difference on each capacitor are ranked as $\left(\Delta V_1\right)>\left(\Delta V_2\right)=\left(\Delta V_3\right)$

To determine

The effect of increasing $C_3$

Answer to Problem 44P

$Q_1$ and $Q_3$ increases, $Q_2$ decreases.

Explanation of Solution

When $C_3$ is increased, the equivalent capacitance increases. As a result, charges on $C_1$ and $C_3$ increase. The charge on $C_2$ decreases.

Conclusion:

Thus, $Q_1$ and $Q_3$ increases, $Q_2$ decreases.