Chapter 19, Problem 39P

To determine

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

Answer to Problem 39P

The added capacitance is $73.26\text{pF}$ .

Explanation of Solution

Given info:

The initial capacitance of the circuit is, $C_1=4800\text{pF}$ .

The reduced capacitance of the circuit is, $C_r=2900\text{pF}$ .

Formula Used:

The expression to calculate the added capacitance is,

  $C_2=\frac{C_1{C}_r}{C_1-C_r}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{C}_2=\frac{\left(4800\text{pF}\right)\left(4800\text{pF}\right)}{4800\text{pF}-2900\text{pF}}\\ =73.26\text{pF}\end{array}$

Conclusion:

Therefore, the added capacitance is $73.26\text{pF}$ .