Chapter 19, Problem 41P

To determine

The equivalent capacitance of the circuit.

Answer to Problem 41P

The equivalent capacitance of the circuit is $\frac{C_1{C}_2+C_2{C}_3+C_3{C}_1}{C_2+C_3}$ .

Explanation of Solution

Given info:

The capacitance of the capacitors in series are, $C_2,C_3$ .

The capacitance of the capacitors in parallel is, $C_1$ .

Formula Used:

The expression to calculate the net capacitance of capacitors in series connection is,

  $C_s=\frac{C_2{C}_3}{C_2+C_3}$

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

  $C=C_1+C_s$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}C=C_1+\frac{C_2{C}_3}{C_2+C_3}\\ C=\frac{C_1{C}_2+C_2{C}_3+C_3{C}_1}{C_2+C_3}\end{array}$

Conclusion:

Therefore, the equivalent capacitance of the circuit is $\frac{C_1{C}_2+C_2{C}_3+C_3{C}_1}{C_2+C_3}$ .