Chapter 27, Problem 85PQ

The circuit in Figure P27.85 shows four capacitors connected to a battery. The switch S is initially open, and all capacitors have reached their final charge. The capacitances are C₁ = 6.00 μF, C₂ = 12.00 μF, C₃ = 8.00 μF, and C₄ = 4.00 μF. a. Find the potential difference across each capacitor and the charge stored in each. b. The switch is now closed. What is the new final potential difference across each capacitor and the new charge stored in each?

Figure P27.85

To determine

The potential difference across the capacitors and the charge stores in each capacitor.

Answer to Problem 85PQ

The charge stored in $C_1$, $C_2$ is $40.0\mu \text{C}$ and $C_3$, $C_4$ is $26.7\mu \text{C}$.

The potential difference across the capacitors $C_1$, $C_4$ is $6.67\text{V}$ and $C_2$, $C_3$ is $3.33\text{V}$.

Explanation of Solution

The circuit diagram,

Since, the capacitors $C_1$ and $C_2$ are connected in series; the charge on each will be same.

Write the expression to find the equivalent capacitance on $C_1$ and $C_2$ combination.

  $C_{eq}=\frac{C_1{C}_2}{C_1+C_2}$

Here, $C_{eq}$ is the equivalent capacitance.

Substitute $6.00\mu \text{F}$ for $C_1$ and $12.00\mu \text{F}$ for $C_2$ to find the equivalent capacitance on $C_1$ and $C_2$ combination.

  $\begin{array}{c}{C}_{eq}=\frac{\left(6.00\mu \text{F}\right)\left(12.00\mu \text{F}\right)}{6.00\mu \text{F}+12.00\mu \text{F}}\\ =4.00\mu \text{F}\end{array}$

Write the expression to find the charges on $C_1$ and $C_2$.

  $\begin{array}{c}{Q}_1=Q_2\\ =C_{eq}V\end{array}$

Here, $Q_1$ is the charge on $C_1$, $Q_2$ is the charge on $C_2$, $V$ is the voltage.

Substitute $4.00\mu \text{F}$ for $C_{eq}$ and $10.0\text{V}$ for $V$ to find the charges on $C_1$ and $C_2$.

  $\begin{array}{c}{Q}_1=Q_2\\ =\left(4.00\mu \text{F}\right)\left(10.0\text{V}\right)\\ =40.0\mu \text{C}\end{array}$

Write the expression to find the equivalent capacitance on $C_3$ and $C_4$ combination.

  $C_{eq}=\frac{C_3{C}_4}{C_3+C_4}$

Here, $C_{eq}$ is the equivalent capacitance.

Substitute $8.00\mu \text{F}$ for $C_3$ and $4.00\mu \text{F}$ for $C_4$ to find the equivalent capacitance on $C_3$ and $C_4$ combination.

  $\begin{array}{c}{C}_{eq}=\frac{\left(8.00\mu \text{F}\right)\left(4.00\mu \text{F}\right)}{8.00\mu \text{F}+4.00\mu \text{F}}\\ =2.67\mu \text{F}\end{array}$

Write the expression to find the charges on $C_3$ and $C_4$.

  $\begin{array}{c}{Q}_3=Q_4\\ =C_{eq}V\end{array}$

Here, $Q_3$ is the charge on $C_3$, $Q_4$ is the charge on $C_4$, $V$ is the voltage.

Substitute $2.67\mu \text{F}$ for $C_{eq}$ and $10.0\text{V}$ for $V$ to find the charges on $C_3$ and $C_4$.

  $\begin{array}{c}{Q}_3=Q_4\\ =\left(2.67\mu \text{F}\right)\left(10.0\text{V}\right)\\ =26.7\mu \text{C}\end{array}$

Write the equation to find the voltage across the capacitor.

  $V=\frac{Q}{C}$

Conclusion:

Substitute $40.0\mu \text{C}$ for $Q$ and $6.00\mu \text{F}$ for $C$ to find the voltage across the capacitor $C_1$.

  $\begin{array}{c}{V}_1=\frac{40.0\mu \text{C}}{6.00\mu \text{F}}\\ =6.67\text{V}\end{array}$

Substitute $40.0\mu \text{C}$ for $Q$ and $12.00\mu \text{F}$ for $C$ to find the voltage across the capacitor $C_2$.

  $\begin{array}{c}{V}_2=\frac{40.0\mu \text{C}}{12.00\mu \text{F}}\\ =3.33\text{V}\end{array}$

Substitute $26.7\mu \text{C}$ for $Q$ and $8.00\mu \text{F}$ for $C$ to find the voltage across the capacitor $C_3$.

  $\begin{array}{c}{V}_3=\frac{26.7\mu \text{C}}{8.00\mu \text{F}}\\ =3.33\text{V}\end{array}$

Substitute $26.7\mu \text{C}$ for $Q$ and $4.00\mu \text{F}$ for $C$ to find the voltage across the capacitor $C_4$.

  $\begin{array}{c}{V}_3=\frac{26.7\mu \text{C}}{4.00\mu \text{F}}\\ =6.67\text{V}\end{array}$

Therefore, the charge stored in $C_1$, $C_2$ is $40.0\mu \text{C}$ and $C_3$, $C_4$ is $26.7\mu \text{C}$.

The potential difference across the capacitors $C_1$, $C_4$ is $6.67\text{V}$ and $C_2$, $C_3$ is $3.33\text{V}$.

To determine

The potential difference across the capacitors and the charge stores in each capacitor when the switch is closed.

Answer to Problem 85PQ

The charge stored in $C_1$, $C_2$, $C_3$, $C_4$ is $74.7\mu \text{C}$.

The potential difference across the capacitors $C_1$, $C_3$ is $5.33\text{V}$ and $C_2$, $C_4$ is $4.67\text{V}$.

Explanation of Solution

The circuit diagram,

When the switch is closed, the capacitors $C_1$ and $C_3$ is in parallel connection and it is in series connection with $C_3$ and $C_4$.

Write the expression to find the equivalent capacitance on $C_1$ and $C_3$ combination.

  $C_{13}=C_1+C_3$

Here, $C_{13}$ is the equivalent capacitance on $C_1$ and $C_3$ combination.

Substitute $6.00\mu \text{F}$ for $C_1$ and $8.00\mu \text{F}$ for $C_3$ to find the equivalent capacitance on $C_1$ and $C_3$ combination.

  $\begin{array}{c}{C}_{13}=6.00\mu \text{F}+8.00\mu \text{F}\\ =14.00\mu \text{F}\end{array}$

Write the expression to find the equivalent capacitance on $C_2$ and $C_4$ combination.

  $C_{24}=C_2+C_4$

Here, $C_{24}$ is the equivalent capacitance on $C_2$ and $C_4$ combination.

Substitute $12.00\mu \text{F}$ for $C_2$ and $4.00\mu \text{F}$ for $C_4$ to find the equivalent capacitance on $C_2$ and $C_4$ combination.

  $\begin{array}{c}{C}_{24}=12.00\mu \text{F}+4.00\mu \text{F}\\ =16.00\mu \text{F}\end{array}$

Write the expression to find the equivalent capacitance of the circuit.

  $C_{eq}=\frac{C_{13}{C}_{24}}{C_{13}+C_{24}}$

Here, $C_{eq}$ is the equivalent capacitance.

Substitute $14.00\mu \text{F}$ for $C_{13}$ and $16.00\mu \text{F}$ for $C_{24}$ to find the equivalent capacitance of the circuit.

  $\begin{array}{c}{C}_{eq}=\frac{\left(14.00\mu \text{F}\right)\left(16.00\mu \text{F}\right)}{14.00\mu \text{F+}16.00\mu \text{F}}\\ =7.47\mu \text{F}\end{array}$

Write the expression to find the charges on the capacitors.

  $\begin{array}{c}{Q}_{13}=Q_{24}\\ =C_{eq}V\end{array}$

Here, $V$ is the voltage.

Substitute $7.47\mu \text{F}$ for $C_{eq}$ and $10.0\text{V}$ for $V$ to find the charges on capacitors.

  $\begin{array}{c}{Q}_{13}=Q_{24}\\ =\left(7.47\mu \text{F}\right)\left(10.0\text{V}\right)\\ =74.7\mu \text{C}\end{array}$

Write the equation to find the voltage across the capacitor.

  $V=\frac{Q}{C}$

Conclusion:

Substitute $74.7\mu \text{C}$ for $Q$ and $14.00\mu \text{F}$ for $C$ to find the voltage across the capacitor $C_1$ and $C_3$.

  $\begin{array}{c}{V}_{13}=V_1=V_3\\ =\frac{74.7\mu \text{C}}{14.00\mu \text{F}}\\ =5.33\text{V}\end{array}$

Substitute $74.7\mu \text{C}$ for $Q$ and $16.00\mu \text{F}$ for $C$ to find the voltage across the capacitor $C_2$ and $C_4$

  $\begin{array}{c}{V}_{24}=V_2=V_4\\ =\frac{74.7\mu \text{C}}{16.00\mu \text{F}}\\ =4.67\text{V}\end{array}$

Therefore, the charge stored in $C_1$, $C_2$, $C_3$, $C_4$ is $74.7\mu \text{C}$.

The potential difference across the capacitors $C_1$, $C_3$ is $5.33\text{V}$ and $C_2$, $C_4$ is $4.67\text{V}$.