Chapter 19, Problem 47P

To determine

The value of the capacitance and method of connection for capacitor added for changing the energy stored in single capacitor.

Answer to Problem 47P

The capacitor of the capacitance $500\text{pF}$ should be added in parallel with single capacitor to do the energy stored in single capacitor three times.

Explanation of Solution

Given info:

The capacitance of the single capacitor is, $C_1=250\text{pF}$ .

The capacitance added to triple the energy is, $C_2$ .

Formula Used:

The expression for the energy stored in the capacitor is,

  $E=\frac{1}{2}C{V}^2$

Here,

• $C$ is the total capacitance of the combination.
• $V$ is the potential difference across capacitor.

Calculation:

The energy stored in the capacitor depends on the capacitance of the capacitor. The energy of the capacitor becomes three times if the capacitance of the becomes three times the capacitance of the capacitor. One capacitor is given and to increase the energy three times either the capacitor should replace by the another capacitor of capacitance three times the initial capacitor or add another capacitor of capacitance two times of single capacitor in parallel combination.

The capacitor added with single capacitor increase the energy stored in the capacitor is given as,

  $C_2=2C_1$

Substitute all the values in the above expression.

  $\begin{array}{c}{C}_2=2\left(250\text{pF}\right)\\ =500\text{pF}\end{array}$

Conclusion:

Thus, the capacitor of the capacitance $500\text{pF}$ should be added in parallel with single capacitor to do the energy stored in single capacitor three times.