Chapter 19, Problem 68GP

(a)

To determine

The energy needs to charge the capacitors in parallel connection.

Answer to Problem 68GP

The energy of the battery to fully charge the capacitors in parallel connection is $4.05\times {10}^{-3}\text{J}$ .

Explanation of Solution

Given info:

The capacitance of the first capacitor is, $C_1=0.40\text{μF}=0.40\times {10}^{-6}\text{F}$ .

The capacitance of the second capacitor is, $C_2=0.60\text{μF}=0.60\times {10}^{-6}\text{F}$ .

The voltage of the battery is, $V=45\text{V}$ .

Formula Used:

The expression to calculate the equivalent capacitance of parallel connection of capacitors is,

  $C_p=C_1+C_2$

The expression to calculate the energy of the battery to fully charge the capacitors in parallel connection is,

  $E_p=\frac{1}{2}{C}_p{V}^2$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{C}_p=0.40\times {10}^{-6}\text{F}+0.60\times {10}^{-6}\text{F}\\ =\text{1}\times {10}^{-6}\text{F}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{E}_p=\frac{1}{2}\left(\text{1}\times {10}^{-6}\text{F}\right){\left(45\text{V}\right)}^2\\ =4.05\times {10}^{-3}\text{J}\end{array}$

Conclusion:

Thus, the energy of the battery to fully charge the capacitors in parallel connection is $4.05\times {10}^{-3}\text{J}$ .

(b)

To determine

The energy needs to charge the capacitors in series connection.

Answer to Problem 68GP

The energy of the battery to fully charge the capacitors in series connection is $2.43\times {10}^{-4}\text{J}$ .

Explanation of Solution

Formula Used:

The expression to calculate the equivalent capacitance of series connection of capacitors is,

  $C_s=\frac{C_1\cdot C_2}{C_1+C_2}$

The expression to calculate the energy of the battery to fully charge the capacitors in series connection is,

  $E_s=\frac{1}{2}{C}_s{V}^2$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{C}_s=\frac{\left(0.40\times {10}^{-6}\text{F}\right)\left(0.60\times {10}^{-6}\text{F}\right)}{0.40\times {10}^{-6}\text{F}+0.60\times {10}^{-6}\text{F}}\\ =0.24\times {10}^{-6}\text{F}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{E}_s=\frac{1}{2}\left(0.24\times {10}^{-6}\text{F}\right){\left(45\text{V}\right)}^2\\ =2.43\times {10}^{-4}\text{J}\end{array}$

Conclusion:

Thus, the energy of the battery to fully charge the capacitors in series connection is $2.43\times {10}^{-4}\text{J}$ .

(c)

To determine

The charge flowed from battery in parallel and series connections of the capacitors.

Answer to Problem 68GP

The charge flowed from battery in parallel connection is $45\times {10}^{-6}\text{C}$ and the charge flowed from battery in series connection is $1.08\times {10}^{-5}\text{C}$ .

Explanation of Solution

Formula Used:

The expression to calculate the charge flowed from battery in parallel connection is,

  $q_p=C_p\cdot V$

The expression to calculate the charge flowed from battery in series connection is,

  $q_s=C_s\cdot V$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{q}_p=\left(\text{1}\times {10}^{-6}\text{F}\right)\left(45\text{V}\right)\\ =45\times {10}^{-6}\text{C}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{q}_s=\left(0.24\times {10}^{-6}\text{F}\right)\left(45\text{V}\right)\\ =1.08\times {10}^{-5}\text{C}\end{array}$

Conclusion:

Thus, the charge flowed from battery in parallel connection is $45\times {10}^{-6}\text{C}$ and the charge flowed from battery in series connection is $1.08\times {10}^{-5}\text{C}$ .