I’d like to take some time today to work through a common circuit analysis question to help develop an understanding of the concept of equivalent resistance. This is an easy technique you can use to simplify circuits you are working on, allowing you to turn something complex and intimidating into a straightforward problem which can be solved. We accomplish this by finding the equivalent resistance to a collection of resistors in our circuit and representing them as a single resistor. Note there is a similar process which can be done with capacitors which I will cover in an upcoming article.
There are two basic rules for equivalent resistors which pertain to how the resistors are connected in the circuit. First if two or more resistors are connected in series as shown below the resistance of the group of resistors is equivalent to the sum of the set of resistors.
The second rule comes into play when the resistors are placed in parallel as seen below. This time the equivalent resistance is equal to the inverse of the sum of the inverse of the resistances. Which sounds scary but isn’t as bad as you think. Basically what that means is that if you add up 1 over each resistance (1/2 for a 2 ohm resistor + 1/5 for a five ohm resistor) then divide 1 by that sum you will get the equivalent resistance.
If that all sounds weird and confusing at this point, don’t worry! I’m going to walk you through an equivalent resistance problem step by step and it should all become clear.
Here is the circuit we will be dealing with. I should note that this is not a practical circuit but was designed for the purpose of illustrating this concept. Still it is a valuable tool help to develop our understanding. Our task here is to find the current flowing out of the 12V power source. We know from Ohm’s Law that the current is equal to the voltage divided by the resistance but without knowing how much total resistance stands between the positive and negative terminals of the battery we find ourselves at an impasse. Luckily we can perform some fancy footwork and transform this seemingly complex system into something extremely simple.
We are going to start by looking at the far right side of the circuit, at resistors R4 and R6 specifically. You will notice these two resitors are connected directly to each other with no other branches splitting off between them. This means they are in series and we can use the first equivalence rule to combine them into one resistor. You can find this equivalent resistance by adding the two resistances together, 4 ohms plus 6 ohms equals 10 ohms. We will replace the two resistors with one 10 ohm resistor.
We aren’t there yet but already this circuit is looking (And hopefully feeling) a bit more manageable. We are going to continue working from the right this time focusing on R5 and R6 above. As seen on the circuit diagram these two resistors are connected at both the top and bottom meaning they are in parallel, This means it’s time to try out the second equivalence rule. So if we have two 10 ohm resistors we want to divide one by each of them and add them together. 1/10 + 1/10 = 2/10, which can be reduced to 1/5. If we then divide 1 by 1/5 we get 5 ohms which is the value of the combined resistance of R5 and R6. A handy rule you can keep in the back of your mind is that if you ever have two equal resistors in parallel (such as these two 10 ohm resistors) the equivalent resistance will be equal to half the resistance of either resistor.
Now we’ve used both our equivalence rules and the circuit continues to shrink. You may already see on the right side that the new circuit looks similar to where we started this time with resistors R2 and R5 in series. By adding these together we can create a combined resistor with a value of 6 ohms.
Next we take resistors R3 and R5 in parallel. Dividing one by each resistor we get 1/3 + 1/6. We can convert this to 2/6 + 1/6 which equals 3/6 or 1/2. By dividing one by 1/2 we get 2 ohms as the value for the new resistor.
And finally we will combine R1 and R3 as series resistors by adding together 4 and 2 ohms.
And there we have it! a 6 resistor circuit with several loops and branches is reduced to a single equivalent resistor. Now our original question of how much current is flowing out of the battery can easily be solved using ohms law. Current is equal to voltage divided by resistance so the current would be 12V / 6 ohms which gives us 2A.