Adding Components in Series and Parallel

How do components add in series and parallel?

When two or more passive components are connected in series or parallel, their values add differently, depending upon what type of components are being connected.

Adding resistors in series and parallel:

Resistor values add normally when connected in series, but add in reciprocal when connected in parallel.  For example, if three resistors of values 1K, 2.2K, and 3.9K are connected in series, the total resistance would be:

R = R1 + R2 + R3 = 1K + 2.2K + 3.9K = 7.1K Note that the total series resistance is always greater than the largest resistance.

If the three resistors are connected in parallel, however, they add in reciprocal, as follows:

1/R = 1/R1 + 1/R2 + 1/R3 = 1/1K + 1/2.2K + 1/3.9K = 1/584.5, therefore R = 584.5 ohms

or, solving for R to make life easier by explicitely showing the recipocal instead of
defining the equation in terms of 1/R,  the equation would become:

R = 1/(1/R1+1/R2+1/R3) = 1/(1/1K+1/2.2K+1/3.9K) = 1/0.0017 = 584.5 ohms Note that the total parallel resistance is always less than the smallest resistance.

There is a special parallel case where all the resistors are the same value.  In this case, the total resistance is the common resistance value divided by the number of resistors being connected in parallel.  For example, if five 10K resistors are connected in parallel, the total resistance is equal to 10K/5, or 2K.  The reciprocal formula will also work, but this method is quicker.

Another method of determining the value of two resistors in parallel is by using the following equation:

R = (R1*R2)/(R1+R2)
For example, if a 470K resistor is connected in parallel with a 220K resistor, the total resistance is equal to (470K*220K)/(470K+220K) = 149.9K.  This result could have been obtained using the reciprocal equation: 1/R = 1/470K + 1/220K = 1/149.9K, therefore R = 149.9K.  Either equation may be used, but one or the other may be desirable if symbolic equations are being solved, or if the result is part of a larger equation.

All of these equations hold true for impedances as well as resistances.

Adding capacitors in series and parallel:

Capacitor values add normally when connected in parallel, but add in reciprocal when connected in series, exactly the opposite of resistors.  For example, if three capacitors of values 0.1uF, 0.022uF, and 0.01uF are connected in parallel, the total capacitance would be:

C = C1 + C2 + C3 = 0.1uF + 0.022uF + 0.01uF = 0.132uF Note that the total parallel capacitance is always greater than the largest capacitance.  Note also that the voltage rating of the parallel capacitors is only as high as the smallest of the two, if they aren't equal.  If they are equal, the voltage rating is the same as that of the single capacitor.

If the three capacitors are connected in series, however, they add in reciprocal, as follows:

1/C = 1/C1 + 1/C2 + 1/C3 = 1/0.1uF + 1/0.022uF + 1/0.01uF = 1/0.0064, therefore C = 0.0064uF Note that the total series capacitance is always less than the smallest capacitance. Also note that the voltage rating of the series capacitors is equal to the sum of the voltage ratings of the individual capacitors if they are the same value.  If the capacitors are of different values, and are used in an AC circuit, the voltage division will not be equal.  See the paper entitled "The voltage divider rule" for more information on this.

There is a special series case where all the capacitors are the same value.  In this case, the total capacitance is the common capacitance value divided by the number of capacitors being connected in series.  For example, if three 1uF capacitors are connected in series, the total capacitance is equal to 1uF/3, or 0.333uF.  The reciprocal formula will also work, but this method is quicker.

Another method of determining the value of two capacitors in series is by using the following equation:

C = (C1*C2)/(C1+C2)
For example, if a 470pF capacitor is connected in series with a 47pF capacitor, the total capacitance is equal to (470pF*47pF)/(470pF+47pF) = 42.7pF.  This result could have been obtained using the reciprocal equation: 1/C = 1/470pF + 1/47pF = 1/42.7pF, therefore C = 42.7pF.  Again, either equation may be used, depending upon which is more convenient.

Adding inductors in series and parallel:

Inductor values add normally when connected in series, but add in reciprocal when connected in parallel, just like resistors.  For example, if three inductors of values 1mH, 10mH, and 33mH are connected in series, the total inductance would be:

L = L1 + L2 + L3 = 1mH + 10mH + 33mH = 44mH Note that the total series inductance is always greater than the largest inductance.

If the three inductors are connected in parallel, however, they add in reciprocal, as follows:

1/L = 1/L1 + 1/L2 + 1/L3 = 1/1mH + 1/10mH + 1/33mH = 1/0.885mH, therefore L = 0.885mH or 885uH Note that the total parallel inductance is always less than the smallest inductance.

There is a special parallel case where all the inductors are the same value.  In this case, the total inductance is the common inductance value divided by the number of inductors being connected in parallel.  For example, if four 1mH inductors are connected in parallel, the total inductance is equal to 1mH/4, or 250uH.  The reciprocal formula will also work, but this method is quicker.

Another method of determining the value of two inductors in parallel is by using the following equation:

L = (L1*L2)/(L1+L2)
For example, if a 22mH inductor is connected in parallel with a 33mH inductor, the total inductance is equal to (22mH*33mH)/(22mH+33mH) = 13.2mH.  This result could have been obtained using the reciprocal equation: 1/L = 1/22mH + 1/33mH = 1/13.2mH, therefore L = 13.2mH.  Again, either equation may be used, depending upon which is more convenient.

Copyright © 1999,  Randall Aiken.  May not be reproduced in any form without written approval from Aiken Amplification.

Revised 02/18/14