Chapter 26, Problem 26.34P

Two capacitors, C₁ = 18.0 μF and C₂ = 36.0 μF, are connected in series, and a 12.0-V battery is connected across the two capacitors. Find (a) the equivalent capacitance and (b) the energy stored in this equivalent capacitance. (c) Find the energy stored in each individual capacitor. (d) Show that the sum of these two energies is the same as the energy found in part (b). (e) Will this equality always be true, or docs it depend on the number of capacitors and their capacitances? (f) If the same capacitors were connected in parallel, what potential difference would be required across them so that the combination stores the same energy as in part (a)? (g) Which capacitor stores more energy in this situation, C₁ or C₂?

To determine

The equivalent capacitance of the system.

Answer to Problem 26.34P

The equivalent capacitance of the system is $12.0\text{μF}$.

Explanation of Solution

Given information: The value of capacitor 1 is $18.0\text{μF}$, value of capacitor 2 is $36.0\text{μF}$, voltage of the battery is $12.0\text{V}$.

Explanation:

The capacitors $C_1\text{and}{C}_2$ are in series.

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

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

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.

$C_2$ is the value of capacitor 2.

Substitute $18.0\text{μF}$ for $C_1$, $36.0\text{μF}$ for $C_2$ in equation (1) to find $C_{\text{eq}}$,

$\begin{array}{c}\frac{1}{C_{\text{eq}}}=\frac{1}{18.0\text{μF}}+\frac{1}{36.0\text{μF}}\\ C_{\text{eq}}=12.0\text{μF}\end{array}$

Thus, the equivalent capacitance of the system is $12.0\text{μF}$.

Conclusion:

Therefore, the equivalent capacitance of the system is $12.0\text{μF}$.

To determine

The energy stored in this equivalent capacitance.

Answer to Problem 26.34P

The energy stored in this equivalent capacitance is $864\text{μJ}$.

Explanation of Solution

Given information: The value of capacitor 1 is $18.0\text{μF}$, value of capacitor 2 is $36.0\text{μF}$, voltage of the battery is $12.0\text{V}$.

Explanation:

Formula to calculate the energy stored in this equivalent capacitance.

$E=\frac{1}{2}C{V}^2$ (2)

Here,

$E$ is the energy stored in this equivalent capacitance.

$V$ is the voltage of the battery.

$C$ is the capacitance of the system.

Substitute $12.0\text{V}$ for $V$, $12.0\text{μF}$ for $C$ in equation (2) to find $E$,

$\begin{array}{c}E=\frac{1}{2}\left(12.0\text{μF}\right)\times {\left(12.0\text{V}\right)}^2\\ =864\text{μJ}\end{array}$

Thus, the energy stored in this equivalent capacitance is $864\text{μJ}$.

Conclusion:

Therefore, the energy stored in this equivalent capacitance is $864\text{μJ}$.

To determine

The energy stored in each individual capacitor.

Answer to Problem 26.34P

The energy stored in the capacitor 1 is $576\text{μJ}$, energy stored in the capacitor 2 is $288\text{μJ}$.

Explanation of Solution

Given information: The value of capacitor 1 is $18.0\text{μF}$, value of capacitor 2 is $36.0\text{μF}$, voltage of the battery is $12.0\text{V}$.

Explanation:

In series connection, the charge will be same in capactor 1 and capacitor 2,

$\begin{array}{c}{Q}_1=Q_2\\ C_1{V}_1=C_2{V}_2\end{array}$ (3)

It is given that the total voltage of the battery is $12.0\text{V}$.

Write the expression to calculate the voltage across capacitor 1.

$\begin{array}{c}{V}_1+V_2=12.0\text{V}\\ V_1=12.0\text{V}-V_2\end{array}$ (4)

Substitute $12.0\text{V}-V_2$ for $V_1$ in equation (3) to find $V_2$,

$V_2=\frac{C_1}{C_2}\left(12.0\text{V}-V_2\right)$ (5)

Substitute $18.0\text{μF}$ for $C_1$, $36.0\text{μF}$ for $C_2$ in equation (5) to find $V_2$,

$\begin{array}{c}{V}_2=\frac{18.0\text{μF}}{36.0\text{μF}}\left(12.0\text{V}-V_2\right)\\ =6-0.5V_2\\ 1.5V_2=6\\ V_2=4.0\text{V}\end{array}$

Thus, the voltage across capacitor 2 is $4.0\text{V}$.

Substitute $4.0\text{V}$ for $V$ in equation (4) to find $V_1$,

$\begin{array}{c}{V}_1=12.0\text{V}-4.0\text{V}\\ =\text{8}.0\text{V}\end{array}$

Thus, the voltage across capacitor 1 is $\text{8}.0\text{V}$.

Formula to calculate the energy stored in the capacitor 1.

$E_1=\frac{1}{2}{C}_1{V}^2$ (6)

Here,

$E_1$ is the energy stored in the capacitor 1.

Substitute $12.0\text{V}$ for $V$, $18.0\text{μF}$ for $C$ in equation (6) to find $E_1$,

$\begin{array}{c}{E}_1=\frac{1}{2}\left(18.0\text{μF}\right)\times {\left(8.0\text{V}\right)}^2\\ =576\text{μJ}\end{array}$

Thus, the energy stored in the capacitor 1 is $576\text{μJ}$.

Formula to calculate the energy stored in the capacitor 2.

$E_2=\frac{1}{2}{C}_2{V}^2$ (7)

Here,

$E_2$ is the energy stored in the capacitor 2.

Substitute $12.0\text{V}$ for $V$, $36.0\text{μF}$ for $C$ in equation (7) to find $E_2$,

$\begin{array}{c}{E}_2=\frac{1}{2}\left(36.0\text{μF}\right)\times {\left(4.0\text{V}\right)}^2\\ =288\text{μJ}\end{array}$

Thus, the energy stored in the capacitor 2 is $288\text{μJ}$.

Conclusion:

Therefore, the energy stored in the capacitor 1 is $576\text{μJ}$, energy stored in the capacitor 2 is $288\text{μJ}$.

To determine

To show: The sum of these two energies is the same as the energy found in part (b).

Answer to Problem 26.34P

The sum of these two energies is the same as the energy found in part (b) is $864\text{μJ}$.

Explanation of Solution

Given information: The value of capacitor 1 is $18.0\text{μF}$, value of capacitor 2 is $36.0\text{μF}$, voltage of the battery is $12.0\text{V}$.

Explanation:

The energy stored in this equivalent capacitance is $864\text{μJ}$.

The energy stored in the capacitor 1 is $576\text{μJ}$.

The energy stored in the capacitor 2 is $288\text{μJ}$.

Formula to calculate the sum of these two energies.

$E=E_1+E_2$ (8)

Here,

$E$ is the sum of these two energies.

Substitute $576\text{μJ}$ for $E_1$, $288\text{μJ}$ for $E_2$ in equation (8) to find $E$,

$\begin{array}{c}E=576\text{μJ}+288\text{μJ}\\ =864\text{μJ}\end{array}$

Thus, the sum of these two energies is the same as the energy found in part (b).

Conclusion:

Therefore, the sum of these two energies is the same as the energy found in part (b) is $864\text{μJ}$.

To determine

The reason that this equality will always be true, or the reason that it depends on the number of capacitors and their capacitances.

Answer to Problem 26.34P

This equality will always be true because the energy stored in series and parallel for the capacitors is same.

Explanation of Solution

Given information: The value of capacitor 1 is $18.0\text{μF}$, value of capacitor 2 is $36.0\text{μF}$, voltage of the battery is $12.0\text{V}$.

Explanation:

Formula to calculate the energy stored by the capacitor in series.

$E_S=E_1+E_2$

Here,

$E_S$ is the energy stored by the capacitor in series.

Formula to calculate the energy stored by the capacitor in parallel.

$E_P=E_1+E_2$

Here,

$E_P$ is the energy stored by the capacitor in parallel.

The value of the energy stored by the capacitor in series and the energy stored by the capacitor in parallel are equal so, this equality will always be true.

Thus, this equality will always be true because the energy stored in series and parallel for the capacitors is same.

Conclusion:

Therefore, this equality will always be true because the energy stored in series and parallel for the capacitors is same.

To determine

The required potential difference across them so that the combination stores the same energy as in part (b).

Answer to Problem 26.34P

The required potential difference across them so that the combination stores the same energy as in part (b) is $5.656\text{V}$.

Explanation of Solution

Given information: The value of capacitor 1 is $18.0\text{μF}$, value of capacitor 2 is $36.0\text{μF}$, voltage of the battery is $12.0\text{V}$.

Explanation:

If the same capacitors are connected in parallel.

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

$C_{\text{eq}}=C_1+C_2$ (9)

Here,

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

The energy stored in this equivalent capacitance is $864\text{μJ}$.

Formula to calculate the required potential difference across them so that the combination stores the same energy as in part (b).

$\begin{array}{c}E=\frac{1}{2}{C}_{\text{eq}}{V}^2\\ V=\sqrt{\frac{2E}{C_{\text{eq}}}}\end{array}$ (10)

Substitute $864\text{μJ}$ for $E$, $C_1+C_2$ for $C_{\text{eq}}$ in equation (10) to find $V$,

$V=\sqrt{\frac{2\times 864\text{μJ}}{C_1+C_2}}$ (11)

Substitute $18.0\text{μF}$ for $C_1$, $36.0\text{μF}$ for $C_2$ in equation (11) to find $V$,

$\begin{array}{c}V=\sqrt{\frac{2\times 864\text{μJ}}{18.0\text{μF}+36.0\text{μF}}}\\ =5.656\text{V}\end{array}$

Thus, the required potential difference across them so that the combination stores the same energy as in part (b) is $5.656\text{V}$.

Conclusion:

Therefore, the required potential difference across them so that the combination stores the same energy as in part (b) is $5.656\text{V}$.

To determine

The capacitor stores more energy $C_1\text{or}{C}_2$.

Answer to Problem 26.34P

The capacitor $C_1$ stores more energy because the energy store by the capacitor 1 i.e, $C_1$ is more than the energy stored by the capacitor 2 i.e, $C_2$.

Explanation of Solution

Given information: The value of capacitor 1 is $18.0\text{μF}$, value of capacitor 2 is $36.0\text{μF}$, voltage of the battery is $12.0\text{V}$.

Explanation:

The capacitor $C_1$ stores more energy because the energy store by the capacitor 1 i.e, $C_1$ is more than the energy stored by the capacitor 2 i.e, $C_2$.

Thus, the capacitor $C_1$ stores more energy.

Conclusion:

Therefore, the capacitor $C_1$ stores more energy because the energy store by the capacitor 1 i.e, $C_1$ is more than the energy stored by the capacitor 2 i.e, $C_2$.