Chapter 27, Problem 27.63AP

A charge Q is placed on a capacitor of capacitance C. The capacitor is connected into the circuit shown in Figure P26.37, with an open switch, a resistor, and an initially uncharged capacitor of capacitance 3C. The switch is then closed, and the circuit comes to equilibrium. In terms of Q and C, find (a) the final potential difference between the plates of each capacitor, (b) the charge on each capacitor, and (c) the final energy stored in each capacitor. (d) Find the internal energy appearing in the resistor.

Figure P26.37

To determine

The final potential difference between the plates of each capacitor.

Answer to Problem 27.63AP

The final potential difference between the plates of each capacitor is $\frac{Q}{4C}$.

Explanation of Solution

Given information: The value of capacitor into the circuit is $C$ and initial uncharged capacitor of capacitance is $3C$, the charge placed on a capacitor is $Q$.

Explanation:

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

$C_{\text{eq}}=C_1+C_2$ (1)

Here,

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

$C_1$ is the value of capacitor into the circuit.

$C_2$ is the initial uncharged capacitor of capacitance.

Substitute $C$ for $C_1$, $3\text{C}$ for $C_2$ in equation (1) to find $C_{\text{eq}}$,

$\begin{array}{c}{C}_P=\text{C}+3\text{C}\\ =4C\end{array}$

Thus, the equivalent capacitance of the system when they are connected in parallel is $4C$.

Formula to calculate the final potential difference between the plates of each capacitor.

$V_f=\frac{Q}{C_{\text{eq}}}$ (2)

Here,

$V_f$ is the final potential difference between the plates of each capacitor.

$Q$ is the charge placed on a capacitor.

Substitute $4C$ for $C_{\text{eq}}$ in equation (2) to find $V_f$,

$V_f=\frac{Q}{4C}$

Thus, the final potential difference between the plates of each capacitor is $\frac{Q}{4C}$.

Conclusion:

Therefore, the final potential difference between the plates of each capacitor is $\frac{Q}{4C}$.

To determine

The charge on each capacitor.

Answer to Problem 27.63AP

The charge on capacitor $C$ is $\frac{Q}{4}$ and charge on capacitor $3C$ is $\frac{3Q}{4}$.

Explanation of Solution

Given information: The value of capacitor into the circuit is $C$ and initial uncharged capacitor of capacitance is $3C$, the charge placed on a capacitor is $Q$.

Explanation:

Formula to calculate the charge placed on a capacitor $C$.

$Q_C=C_1{V}_f$ (3)

Here,

$Q_C$ is the charge placed on a capacitor $C$.

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

$\begin{array}{c}{Q}_C=C\times \frac{Q}{4C}\\ =\frac{Q}{4}\end{array}$

Thus, the charge placed on a capacitor $C$ is $\frac{Q}{4}$.

Formula to calculate the charge placed on a capacitor $3C$.

$Q_{3C}=C_2{V}_f$ (4)

Here,

$Q_{3C}$ is the charge placed on a capacitor $3C$.

Substitute $3C$ for $C_2$ in equation (4) to find $Q_{3C}$,

$\begin{array}{c}{Q}_{3C}=3C\times \frac{Q}{4C}\\ =\frac{3Q}{4}\end{array}$

Thus, the charge placed on a capacitor $3C$ is $\frac{3Q}{4}$.

Conclusion:

Therefore, the charge on capacitor $C$ is $\frac{Q}{4}$ and charge on capacitor $3C$ is $\frac{3Q}{4}$.

To determine

The final energy stored in each capacitor.

Answer to Problem 27.63AP

The energy stored in the capacitor $C$ is $\frac{Q^2}{32C}$, energy stored in the capacitor $3C$ is $\frac{3Q^2}{32C}$.

Explanation of Solution

Given information: The value of capacitor into the circuit is $C$ and initial uncharged capacitor of capacitance is $3C$, the charge placed on a capacitor is $Q$.

Explanation:

Formula to calculate the energy stored in the capacitor $C$.

$E_C=\frac{1}{2}{C}_1{V}_f^2$ (5)

Here,

$E_C$ is the energy stored in the capacitor $C$.

Substitute $\frac{Q}{4C}$ for $V_f$, $C$ for $C_1$ in equation (5) to find $E_C$,

$\begin{array}{c}{E}_C=\frac{1}{2}\left(C\right)\times {\left(\frac{Q}{4C}\right)}^2\\ =\frac{Q^2}{32C}\end{array}$

Thus, the energy stored in the capacitor $C$ is $\frac{Q^2}{32C}$.

Formula to calculate the energy stored in the capacitor $3C$.

$E_{3C}=\frac{1}{2}{C}_2{V}_f^2$ (6)

Here,

$E_{3C}$ is the energy stored in the capacitor $3C$.

Substitute $\frac{Q}{4C}$ for $V_f$, $3C$ for $C_2$ in equation (6) to find $E_{3C}$,

$\begin{array}{c}{E}_{3C}=\frac{1}{2}\left(3C\right)\times {\left(\frac{Q}{4C}\right)}^2\\ =\frac{3Q^2}{32C}\end{array}$

Thus, the energy stored in the capacitor $3C$ is $\frac{3Q^2}{32C}$.

Conclusion:

Therefore, the energy stored in the capacitor $C$ is $\frac{Q^2}{32C}$, energy stored in the capacitor $3C$ is $\frac{3Q^2}{32C}$.

To determine

The internal energy appearing in the resistor.

Answer to Problem 27.63AP

The internal energy appearing in the resistor is $\frac{3Q^2}{8C}$.

Explanation of Solution

Given information: The value of capacitor into the circuit is $C$ and initial uncharged capacitor of capacitance is $3C$, the charge placed on a capacitor is $Q$.

Explanation:

Write the expression for the original energy in the resistor.

$E_{\text{original}}=\frac{Q^2}{2C}$ (7)

Here,

$E_{\text{original}}$ is the original energy in the resistor.

Formula to calculate the internal energy appearing in the resistor.

$E_r=E_{\text{original}}-\left(E_C+E_{3C}\right)$

Here,

$E_r$ is the internal energy appearing in the resistor.

Substitute $\frac{Q^2}{2C}$ for $E_{\text{original}}$, $\frac{Q^2}{32C}$ for $E_C$, $\frac{3Q^2}{32C}$ for $E_{3C}$ in equation (7) to find $E_r$,

$\begin{array}{c}{E}_r=\frac{Q^2}{2C}-\left(\frac{Q^2}{32C}+\frac{3Q^2}{32C}\right)\\ =\frac{Q^2}{8C}\end{array}$

Thus, the internal energy appearing in the resistor is $\frac{3Q^2}{8C}$.

Conclusion:

Therefore, the internal energy appearing in the resistor is $\frac{3Q^2}{8C}$.