Question

Two capacitors give an equivalent capacitance of 9.00 x10^(-6)F when connected in parallel and an equivalent capacitance of 2.00 x10^(-9) F when connected in series. What is the capacitance of each capacitor?

Answer 1

The equivalent capacitances of two capacitors connected in series and in parallel can be written as follows:

$$C_{eq,s}=\dfrac{C_1C_2}{C_1+C_2},$$ $$C_{eq,p}=C_1+C_2.$$

Let's express $C_1$ from the last equation in terms of $C_{eq,p}$ and $C_2$:

$$C_1=C_{eq,p}-C_2.$$

Substituting $C_1$into the first equation and simplifying, we get:

$$C_2^2-C_{eq,p}C_2+C_{eq,s}C_{eq,p}=0,$$

Let's substitute the numbers and solve the quadratic equation:

$$C_2^2-9.0\cdot10^{-6}C_2+18\cdot10^{-15}=0.$$

This quadratic equation has two roots: $C_1=9\ \mu F$ and $C_2=2\ nF$. Therefore, the capacitance of each capacitor: $C_1=9\ \mu F, C_2=2\ nF.$

Answer:

$C_1=9\ \mu F, C_2=2\ nF.$