Resistor Reduction

Basic Idea

This is a techniques for dealing with more complex circuit arrangements. Consider this problem: Based on what we have learned in the previous lectures this circuit is solvable. All we have to do is follow the reduction algorithm:

1 If any two or more resistors are in series combine them together and then replace it with an equivalent resistance.
2 If any two or more resistors are in parallel combine them together and then replace it with an equivalent resistance.
3 Continue doing this until you are left with only one resistor.
4 Solve for the source current/voltage (whichever is unknown)
5 systematically unwrap the circuit solving as you go along.

First example

Lets go through this example assume we want to know the current through each resistor given that R1=5ohms, R2= 20ohms, R3=20 ohms, and the source is producing 20 volts:

Step 1 Can not be done because there are no series resistors
Step 2 Can be done R1 can be combined with R2 to make an effective resistance. R1|R2 is a way of saying R1 in parallel with R2. In this case its is equal to 4 ohms:
Step 3 Well were not down to one resistor so we start again we combine the series resistors and get a 24 ohm equivalent resistance.

Step 4 So we know the voltage and the equivalent resistance lets calculate the current. By V=IR we get 5/6 amps.

step 5 Now we start to expand the circuit

Since we know the current for the combined resistance and the other resistor is in series we know the current through R3 is equal to 5/6 Amps.
Step 5 now we unwind the circuit completely and get the final result

V=IR, V=(5/6)(4)=20/6. So R1's current is equal to 20/6 divided by 5 which is 2/3 amps. And R2 is 20/6 divided by 20 which is equal to 1/6.

Review

So we see the goal of the technique. We can apply it to any number of circuit problems. Be forewarned this is a hard techniques to master. It is recommend that the student goes through many examples. Remember the basic steps. Combine all the resistors into one resistor. Solve for the current/voltage of that one resistor and then unreduce the circuit solving for each resistor along the way.

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