Chapter 19, Problem 36P

To determine

The maximum and minimum capacitance of the combinations of three capacitors.

Answer to Problem 36P

The maximum capacitance is $20700\text{pF}$ and the minimum capacitance is $18.33\text{pF}$ .

Explanation of Solution

Given info:

The capacitance of the first capacitor is, $C_1=3200\text{pF}$ .

The capacitance of the second capacitor is, $C_2=7500\text{pF}$ .

The capacitance of the third capacitor is, $C_3=0.01\text{μF}=10000\text{pF}$ .

Formula Used:

The expression to calculate the maximum capacitance is,

  $C_{\max }=C_1+C_2+C_3$

The expression to calculate the minimum capacitance is,

  $C_{\min }=\frac{C_1{C}_2{C}_3}{C_1{C}_2+C_2{C}_3+C_1{C}_3}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{C}_{\max }=3200\text{pF}+7500\text{pF}+10000\text{pF}\\ =20700\text{pF}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{C}_{\min }=\frac{\left(3200\text{pF}\right)\left(7500\text{pF}\right)\left(10000\text{pF}\right)}{\left(3200\text{pF}\right)\left(7500\text{pF}\right)+\left(7500\text{pF}\right)\left(10000\text{pF}\right)+\left(3200\text{pF}\right)\left(10000\text{pF}\right)}\\ =18.33\text{pF}\end{array}$

Conclusion:

Therefore, the maximum capacitance is $20700\text{pF}$ and the minimum capacitance is $18.33\text{pF}$ .