Simplify Parallel Calculations

The total resistance in a parallel circuit is difficult to calculate because it cannot simply be added as in a series circuit. However, the total current flow in a parallel circuit is easy to calculate. It is simply the sum of the currents for all the individual loads. The famous parallel formula for resistances in parallel uses this concept.  Dividing the resistance of each load into 1 calculates the current draw for each load at 1 volt. Adding these fractions gives the total current draw of the entire circuit at 1 volt. Since resistance can be calculated by dividing current into voltage, dividing the total circuit current at 1 volt into 1 volt gives the total resistance. I confess that I have used and taught this formula for many years without knowing why it works. Once I understood the reason behind the formula, it seemed far more logical and less imposing.  Did you know that this works for any voltage you choose? We can use this to create a simpler parallel calculation. 


I find that many of today’s digital students struggle with the fractions, common denominators, and all the trappings that go with fractions. One solution is to convert the fractions to decimals. But the decimal numbers you get when dividing whole numbers into one are always several decimal places to the right of 0, causing another own sort of confusion. My suggestion is to choose a voltage at least equal to the highest resistance; preferably, twice the highest resistance. That way, all the answers will have at least one whole number. This not only makes the answers easier for the students to deal with, it makes the decimals to the right of 0 less significant, so lopping off a few won’t have as big an impact on the final answer.


 For example, with resistances of 15, 20, and 30 ohms choose 60 volts. Instead of trying to add 1/15  + 1/20 + 1/30  or their decimal equivalents, you are adding 60/30 + 60/20 + 60/15. Most people can add 2 + 3 + 4. Then, divide this total current, 9 amps, into the voltage we used, 60 volts, to get 6.6 ohms total resistance. OK, so I stacked the deck a bit. Another teacher’s trick – if you get to choose the questions, why not choose questions that make your life easier. After your students are adept at these “easy” problems, have them measure the resistance of some actual devices and do the calculations. This is where choosing to make the calculations easier really pays off. Anybody can make simple things difficult – the trick to teaching is to make difficult things simple.