Chapter 19, Problem 40P

To determine

The capacitance to be added to produce the effect without removing existing circuit elements.

Answer to Problem 40P

The capacitance of capacitor that corrected the circuit without removing any element is $11\text{μF}$ .

Explanation of Solution

Given info:

The initial capacitance of the circuit is, $C_1=5\text{μF}$ .

The reduced capacitance of the circuit is, $C_2=16\text{pF}$ .

Formula Used:

The expression to calculate the capacitance of capacitor that corrected the circuit without removing any element is,

  $C=C_2-C_1$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}C=16\text{μF}-5\text{μF}\\ =11\text{μF}\end{array}$

Conclusion:

Therefore, the capacitance of capacitor that corrected the circuit without removing any element is $11\text{μF}$ .