Chapter 2, Problem 9P

The equivalent capacitance C of two capacitors connected in series as n in Fig. P2.9 is given by

$C=\frac{C_1{C}_2}{C_1+C_2}$ (2.59)

If the total capacitance is $C=120$ HF and $C_2=C_1+100\mu \text{F}$, find C ₁ and C ₂ as follow:

(a) Substitute the values of C and C₂ in equation (2.59) and obtain the quadratic equation for C₁.

(b) Solve the quadratic equation for obtained in part (a) by each of the following method factoring, completing the square, and the quadratic formula. Also, compute the corresponding values of C ₂.