# 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 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.