Chapter 19, Problem 44P

To determine

The energy stored in network of the capacitors.

Answer to Problem 44P

The energy stored in the complete network of the capacitors is $32.854\text{kJ}$ .

Explanation of Solution

Given info:

The capacitance of the first capacitor is, $C_1=7.2\text{μF}$ .

The capacitance of the second capacitor is, $C_2=7.2\text{μF}$ .

The capacitance of the third capacitor is, $C_3=7.2\text{μF}$ .

The voltage in the circuit is, $V=78\text{V}$ .

Formula Used:

The expression to calculate the energy stored in the complete network of the capacitors is,

  $E=\frac{1}{2}\left[\frac{C_1{C}_2+C_2{C}_3+C_3{C}_1}{C_2+C_3}\right]V^2$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}E=\frac{1}{2}\left[\frac{\left(7.2\text{μF}\right)\left(7.2\text{μF}\right)+\left(7.2\text{μF}\right)\left(7.2\text{μF}\right)+\left(7.2\text{μF}\right)\left(7.2\text{μF}\right)}{7.2\text{μF}+7.2\text{μF}}\right]{\left(78\text{V}\right)}^2\\ =32853.6\text{J}\\ =32.854\text{kJ}\end{array}$

Conclusion:

Therefore, the energy stored in the complete network of the capacitors is $32.854\text{kJ}$ .