Chapter 2, Problem 11P

The equivalent capacitance C of three capacitors connected in series-parallel as shown in Fig. P2.11 is given by

$C=25+\frac{C_1{C}_2}{C_1+C_2}$ (2.60)

If the total capacitance is $C=125\mu \text{F}$ and $C_2=C_1+100\mu \text{F}$, find $C_1$ and $C_2$ as follows:

1. Substitute the values of C and $C_2$
2. in equation (2.60) and obtain the quadratic equation for $C_1$.
3. Solve the quadratic equation for $C_1$
4. obtained in part (a) by completing the square and the quadratic formula. Also, compute the corresponding values of $C_2$.