Chapter 19, Problem 38P

To determine

The voltage across each capacitor in the circuit.

Answer to Problem 38P

The net capacitance of the circuit is $3.71\text{μF}$ .

Explanation of Solution

Given info:

The capacitance of the capacitors in series are, $C_1,C_2=3\text{μF},4\text{μF}$ .

The capacitance of the capacitors in parallel is, $C_3=2\text{μF}$ .

The voltage in the circuit is, $V=26\text{V}$ .

Formula Used:

The expression to calculate the net capacitance of the circuit is,

  $C=\frac{C_1{C}_2}{C_1+C_2}+C_3$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}C=\frac{\left(3\text{μF}\right)\left(4\text{μF}\right)}{3\text{μF}+4\text{μF}}+\left(2\text{μF}\right)\\ =3.71\text{μF}\end{array}$

Conclusion:

Therefore, the net capacitance of the circuit is $3.71\text{μF}$ .