Chapter 24, Problem 54P

(a)

To determine

To Find: The potential difference across each capacitor.

Answer to Problem 54P

  $V=4.50\text{V}$

Explanation of Solution

Given:

Capacitor 1, $C_1=4\mu \text{F = }4\times {10}^{-6}\text{F }$

Capacitor 2, $C_2=12\mu \text{F = 12}\times {10}^{-6}\text{F }$

Potential difference, $V=12\text{V}$

Formula used:

Potential difference

  $V=\frac{Q}{C}$

Where,

  $Q$ is the charge

  $C$ is the capacitance

Calculation:

Charge on the capacitor when they are initially connected in series can be calculated as:

  $\begin{array}{l}Q=C_{eq}V\\ Q=\left(\frac{C_1{C}_2}{C_1+C_2}\right)V\\ Q=\left(\frac{4\times {10}^{-6}\times \text{12}\times {10}^{-6}}{\left(4\times {10}^{-6}\right)+\left(\text{12}\times {10}^{-6}\right)}\right)\times 12\\ Q=\left(\frac{48\times {10}^{-12}}{\left(16\times {10}^{-6}\right)}\right)\times 12\\ Q=3\times {10}^{-6}\times 12\\ Q=36\times {10}^{-6}\text{C}\\ Q=36\mu \text{C}\end{array}$

When they are connected in parallel, net charge is:

  $\begin{array}{l}{Q}_{net}=2\times 36\\ Q_{net}=72\mu \text{C}\end{array}$

Equivalent capacitor when connected in parallel is

  $\begin{array}{l}{C}_{eq}^{\text{'}}=C_1+C_2\\ C_{eq}^{\text{'}}=\left(4\times {10}^{-6}\right)+\left(\text{12}\times {10}^{-6}\right)\\ C_{eq}^{\text{'}}=\text{16}\times {10}^{-6}\text{C}\\ C_{eq}^{\text{'}}=\text{16}\mu \text{C}\end{array}$

In parallel combination, potential difference is same across all the capacitors.

Potential difference can be calculated as:

  $\begin{array}{l}V\text{'}=\frac{Q_{net}}{C_{eq}^{\text{'}}}\\ V\text{'}=\frac{72\times {10}^{-6}}{16\times {10}^{-6}}\\ V\text{'}=4.5\text{V}\end{array}$

Conclusion:

Potential difference across each capacitor is $4.5\text{V}$

(b)

To determine

To Find: The energy stored in capacitor before disconnection and after reconnection.

Answer to Problem 54P

  $E_i=216\times {10}^{-6}\text{J}$

  $E_f=162\times {10}^{-6}\text{J}$

Explanation of Solution

Given:

Capacitor 1, $C_1=4\mu \text{F = }4\times {10}^{-6}\text{F }$

Capacitor 2, $C_2=12\mu \text{F = 12}\times {10}^{-6}\text{F }$

Potential difference, $V=12\text{V}$

Formula used:

Energy stored

  $E=\frac{1}{2}C{V}^2$

Where,

  $C$ is the capacitance

  $V$ is the potential difference

Calculation:

Equivalent capacitance when capacitors are connected in series:

  $\begin{array}{c}{C}_{eq}=\left(\frac{C_1{C}_2}{C_1+C_2}\right)\\ =\left(\frac{4\times {10}^{-6}\times \text{12}\times {10}^{-6}}{\left(4\times {10}^{-6}\right)+\left(\text{12}\times {10}^{-6}\right)}\right)\\ =\frac{48\times {10}^{-12}}{\left(16\times {10}^{-6}\right)}\\ =3\times {10}^{-6}\text{C}\end{array}$

Initially energy stored in capacitor is:

  $\begin{array}{l}E=\frac{1}{2}{C}_{eq}{V}^2\\ E=\frac{1}{2}\times 3\times {10}^{-6}\times {\left(12\right)}^2\\ E_i=216\times {10}^{-6}\text{J}\end{array}$

Equivalent capacitance when capacitors connected in parallel:

  $\begin{array}{l}{C}_{eq}^{\text{'}}=C_1+C_2\\ C_{eq}^{\text{'}}=\left(4\times {10}^{-6}\right)+\left(\text{12}\times {10}^{-6}\right)\\ C_{eq}^{\text{'}}=\text{16}\times {10}^{-6}\text{C}\\ C_{eq}^{\text{'}}=\text{16}\mu \text{C}\end{array}$

Potential difference can be calculated as:

  $\begin{array}{l}V\text{'}=\frac{Q_{net}}{C_{eq}^{\text{'}}}\\ V\text{'}=\frac{72\times {10}^{-6}}{16\times {10}^{-6}}\\ V\text{'}=4.5\text{V}\end{array}$

Final energy stored in capacitor is:

  $\begin{array}{l}{E}_f=\frac{1}{2}{C}_{eq}^{\text{'}}V{\text{'}}^2\\ E_f=\frac{1}{2}\times 16\times {10}^{-6}\times {\left(4.5\right)}^2\\ E_f=162\times {10}^{-6}\text{J}\end{array}$

Conclusion:

Energy stored in capacitor before disconnection is $E_i=216\times {10}^{-6}\text{J}$ and after reconnection is $E_f=162\times {10}^{-6}\text{J}$ .