Question

Two capacitors connected in parallel produce an equivalent capacitance of 40.0 μF but when connected in series the equivalent capacitance is only 7.1 μF .

What is the individual capacitance of each capacitor?

Answer 1

Let be capacitances of the two capacitors. When connected in parallel, the equivalent capacitance is given by

Ci and ca

$C_1+C_2=40\ \mu F\ \ \ \ \ \ \ \ (1)$

When connected in series the equivalent capacitance is given by

$\frac{C_1C_2}{C_1+C_2}=7.1\ \mu F$

Using equation (1) in the denominator of LHS

C1C2 40 uF = 7.1 MF

$C_1C_2=40\times7.1$

Again using equation (1) we get $C_1=40-C_2$ ,

$(40-C_2)C_2=284$

$40C_2-C_2^2=284$

$C_2^2-40C_2+284=0$

This is a quadratic equation in C2, the roots of a quadratic equation $ax^2+bx+c=0$ are

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

The roots of the above quadratic equation are

$C_2=\frac{-(-40)\pm \sqrt{(-40)^2-4\times 1\times 284}}{2\times 1}$

$C_2=30.77\ \mu F,\ 9.23\ \mu F$

Using (1) the other capacitance for each case is

$C_1= 9.23\ \mu F,\ 30.77\ \mu F$

Therefore, the two capacitors are $9.23\ \mu F,\ 30.77\ \mu F$ .