Series Capacitors - Calculation Considerations (ART417E)

The determination of the capacitance equivalent to a series capacitor association offers the student of the professionalizing electronic courses the same issues found in determining the resistance equivalent to a parallel association. In fact, we can say that the formulas are equivalent, although they deal with different quantities. However, since the algebraic process which leads to the final result is the same for both cases, when solving problems, the student finds the same difficulties.


In this article we will give a brief explanation of the basic process of using the formula, and then some exercises in increasing order of difficulties will work to train the student in its use.


As we know, the formula for calculating the resistance equivalent to a series association of resistors is:


1/C = 1/C1 + 1/C2 + 1/C3 + …...... + 1/Cn


at where:

C = equivalent capacitance in microfarads, picofarads or nanofarads

C1, C2, C3, ... Cn = associated capacitances in the same unit where we want the equivalent capacitance.


Note that if we apply the formula to two capacitors:


1/C = 1/C1 + 1/C2


And then reduce the second equality member to the same denominator we will have:


1/C = (C2 + C1) / (C1 x C2) ? C = (C1 x C2) / (C1 + C2)


We get the formula much used directly by many technicians.

However, it should be noted that this is not a formula, but a particular expression in the case of two capacitors connected in series.

For the use of the basic formula, the best systematic, is demonstrated in the following example problem:


Example problem

Determine the capacitance equivalent to the association of three capacitors in series, whose values are: C1 = 4 uF; C2 = 6 uF; C3 = 12 uF



In this problem we then have:

C1 = 4 uF

C2 = 6 uF

C3 = 12 uF


We want to determine C and therefore apply the formula:


1/C = 1/C1 + 1/C2 + 1/C3


Substituting formula C1, C2 through C3 for their values in microfarads:


1/C = 1/4 + 1/6 + 1/12


We reduce the second equality member to the same denominator. For this we use the mmc (minimum common multiple) between 4, 6 and 12 which is 12.


1/C = 3/12 + 2/12 + 1/12


Then, we add the sum of the fractions that, as they have the same denominator, is given by the sum of the numerators and maintenance of the common denominator.


1/C = 6/12


We obtain in these conditions a result equivalent to the inverse of C, that is, 1/C.


In order to obtain the value of C, we must simply reverse both members of equality.

C = 12/6 ? 2 uF



Therefore, the equivalent capacitance will be C = 2 uF.

The inversion of both members of equality is in fact equivalent to the multiplication of numerators by C, and then by 12 which obviously does not change it, and then the division of both members by 6 is proceeded.


Problems for the student to train

1) Calculate the capacitance equivalent to the association of a 200 pF capacitor in series with a capacitor of 300 pF

2) Determine the capacitance equivalent to a capacitor of 60 nF in series with a capacitor of 120 nF

3) What is the capacitance obtained from the association of three capacitors in series with their values 20 uF, 30 uF and 60 uF ?

4) What capacitor we should associate in series, with a capacitor of 60 uF to get a total capacitance equivalent of 24 uF?

5) One capacitor has twice the capacitance of another. When coupled in series they are equivalent to a capacitor of 40 uF. Determine their capacitances.



1) 120 pF

2) 40 nF

3) 10 uF

4) 0 uF

5) 60 uF e 120 uF



Circuit Bench