Chapter 16, Problem 50P

Two capacitors, C₁ = 18.0 μF and C₂ = 36.0 μF, are connected in series, and a 12.0-V battery is connected across them. (a) Find the equivalent capacitance, and the energy contained in this equivalent capacitor. (b) Find the energy stored in each individual capacitor. Show that the sum of these two energies is the same as the energy found in part (a). Will this equality always be true, or does it depend on the number of capacitors and their capacitances? (c) 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)? Which capacitor stores more energy in this situation, C₁ or C₂?

To determine

The equivalent capacitance and the energy stored in the capacitor.

Answer to Problem 50P

The equivalent capacitance is $12.0\text{μF}$

The energy stored in the capacitor is $8.64\times {10}^{-4}\text{J}$.

Explanation of Solution

Given info: The capacitors $C_1=18.0\text{μF}$ and $C_2=36.0\text{μF}$ are connected in series. The potential difference applied by the battery is 12.0 V, The capacitors $C_1=18.0\text{μF}$ and $C_2=36.0\text{μF}$ are connected in series. The potential difference applied by the battery is 12.0 V.

Explanation:

Formula to calculate the equivalent capacitance is,

$C_{eq}=\frac{C_1{C}_2}{C_1+C_2}$

Substitute $18.0\text{μF}$ for $C_1$ and $36.0\text{μF}$ for $C_2$.

$\begin{array}{c}{C}_{eq}=\frac{\left(18.0\text{μF}\right)\left(36.0\text{μF}\right)}{\left(18.0\text{μF}\right)+\left(36.0\text{μF}\right)}\\ =12.0\text{μF}\end{array}$

Formula to calculate the energy stored is,

$E=\frac{1}{2}{C}_{eq}{\left(\Delta V\right)}^2$

• $\Delta V$ is the potential difference.

Substitute $12.0\text{μF}$ for $C_{eq}$ and 12.0 V for $\Delta V$

$\begin{array}{c}E=\frac{1}{2}\left(12.0\text{μF}\right){\left(12.0\text{V}\right)}^2\\ =\frac{1}{2}\left(12.0\times {10}^{-6}\text{F}\right){\left(12.0\text{V}\right)}^2\\ =8.64\times {10}^{-4}\text{J}\end{array}$

Conclusion:

The equivalent capacitance is $12.0\text{μF}$

The energy stored in the capacitor is $8.64\times {10}^{-4}\text{J}$.

To determine

The energy stored in each capacitor.

Answer to Problem 50P

The energy stored in $C_1$ is $5.76\times {10}^{-4}\text{J}$

The energy stored in $C_2$ is $2.88\times {10}^{-4}\text{J}$.

Explanation of Solution

Given info: The capacitors $C_1=18.0\text{μF}$ and $C_2=36.0\text{μF}$ are connected in series. The potential difference applied by the battery is 12.0 V, The capacitors $C_1=18.0\text{μF}$ and $C_2=36.0\text{μF}$ are connected in series. The potential difference applied by the battery is 12.0 V.

Explanation:

The charge on each capacitor is,

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

Substitute $12.0\text{μF}$ for $C_{eq}$ and 12.0 V for $\Delta V$

$\begin{array}{c}Q=\left(12.0\text{μF}\right)\left(12.0\text{V}\right)\\ =144\text{μC}\end{array}$

Formula to calculate the energy stored is,

$E_1=\frac{Q^2}{2C_1}$

Substitute $18.0\text{μF}$ for $C_1$ and $144\text{μC}$ for Q.

$\begin{array}{c}{E}_1=\frac{{\left(144\text{μC}\right)}^2}{2\left(18.0\text{μF}\right)}\\ =5.76\times {10}^{-4}\text{J}\end{array}$

Formula to calculate the energy stored is,

$E_2=\frac{Q^2}{2C_2}$

Substitute $36.0\text{μF}$ for $C_2$ and $144\text{μC}$ for Q.

$\begin{array}{c}{E}_2=\frac{{\left(144\text{μC}\right)}^2}{2\left(36.0\text{μF}\right)}\\ =2.88\times {10}^{-4}\text{J}\end{array}$

Conclusion:

The energy stored in $C_1$ is $5.76\times {10}^{-4}\text{J}$

The energy stored in $C_2$ is $2.88\times {10}^{-4}\text{J}$

To determine

The potential difference.

Answer to Problem 50P

The potential difference is 5.66 V.

Explanation of Solution

Given info: The capacitors $C_1=18.0\text{μF}$ and $C_2=36.0\text{μF}$ are connected in parallel. The potential difference applied by the battery is 12.0 V.

Explanation:

Formula to calculate the potential difference is,

$\Delta V=\sqrt{\frac{2E}{C_{eq}}}$

Formula to calculate the equivalent capacitance is,

$C_{eq}=C_1+C_2$

Substitute $18.0\text{μF}$ for $C_1$ and $36.0\text{μF}$ for $C_2$.

$\begin{array}{c}{C}_{eq}=\left(18.0\text{μF}\right)+\left(36.0\text{μF}\right)\\ =54.0\text{μF}\end{array}$

Substitute $54.0\text{μF}$ for $C_{eq}$ and $8.64\times {10}^{-4}\text{J}$ for E.

$\begin{array}{c}\Delta V=\sqrt{\frac{2\left(8.64\times {10}^{-4}\text{J}\right)}{54.0\text{μF}}}\\ =5.66\text{V}\end{array}$

Conclusion:

The potential difference is 5.66 V. The total energy stored in a capacitor is directly related to the individual capacitances. Thus, capacitor $C_2$ will have more energy.