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Simplifying resistor networks

A systematic approach to simplify a complicated resistor network by looking for series and parallel resistor patterns. Created by Willy McAllister.

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• If you can sometimes collapse a resistor network down to a single resistor why would you start with so many resistors in the first place?
• When we do a simplification problem like this the first thing we do is pick a location in the circuit we care about. In this case, I picked the far left end. Then the question is, "What does the rest of the circuit 'look like' from the far left end?" If we wanted to know about a different location, the answer would turn out different. The resistor network I worked with is contrived. It doesn't do anything. The point of the video is to show how you can apply the resistor series and parallel formulas over and over. (It's sort of like I was teaching you how add and subtract multiple numbers, but I didn't tell you what numbers were for.) When we get to more complicated circuits, the method in this video will be useful.
• @ you said the voltage source is putting out enough current to drive 3ohms. What does it mean to "drive" 3ohms? How does this differ from a resistor "absorbing" (V/3) amps? And even more confusingly, what does it mean for a resistor to "draw" (V/3) amps?
• "It (the voltage source) is putting out enough current to drive 3 ohms". I use the word "drive" here to refer to the current the voltage source is required to provide. For the sake of illustration, let's say the voltage source is set to V = 3V. Ohm's Law tells us the amount of current is 3V/3ohm or 1 amp. That 1 amp current is provided by (or "driven by") the voltage source. A common way to say this is "the voltage source has to drive 1 amp into the resistor if we expect the voltage across the resistor to get up to 3V."

We can say something similar from the "point of view" of the resistor, the resistor "demands" or "pulls" or "draws" 1A from the 3V source. We might say "When connected to a 3V source, our 3 ohm resistor draws 1A of current from the source." We use the word "draw" in the sense of "withdraw".

These terms are examples of EE jargon creeping into my descriptions. Please let me know if you come across other places where I do this.
• What do you mean, at , by "pulling out of voltage source"?
(1 vote)
• If you connect something (a resistor, a bunch of resistors, anything that conducts current), it causes a current to flow out of the current source. So "pulling current out of a voltage source" is just a casual turn of phrase I used to give the sense that the resistors connected to the voltage source are "demanding" or "pulling" current. If the resistors were not there, there would be an open circuit, and therefore no current. When the resistors are connected, in a sense they are causing the current to flow.
• how can we decide the equivalence resistor position?!
• Hi Trần,

Here are my recommendations:

1) redraw your circuit so it has a left to right flow as Willy has done in this video

2) start at the far right and work to combine the series / parallel branches

3) take your time to redraw the circuit

Personally I find this operation easy but oh so time consuming with many potential errors.

Regards,

APD
• At 06.31 you get from 1/2 to 2 but saying (1)%(1/2) it get the right result but somehow I dont really get it. We have the formula 1/Rp = 1/R1 + 1/R2... so in the end we get (1/Rp)%(1/Rn) Anyway dont know if that is correct. Im studying to become an electrician so math is not my strongest subject
(1 vote)
• There are a lot of reciprocals in the general equation for parallel resistors. The tricky one is the one on the left side. I'll do the math slowly...

1/Rp = 1/R1 + 1/R2 + 1/R3

1/Rp = 1/12 + 1/4 + 1/6 = 1/12 + 3/12 + 2/12

1/Rp = (1 + 3 + 2)/12 = 6/12

1/Rp = 1/2

so now you have one more reciprocal to get Rp...

Rp = 2

Maybe it helps to write it like this...

1/Rp = 1/2 = 0.5

Rp = 1/0.5 = 2
• How about simplifying 3-D models of a circuit? Our professor gave that as a quiz one time. I got so low there :(
(1 vote)
• can you please show me how can I find the current and voltage at each resistors?
(1 vote)
• Good question. To find the current in each resistor and voltage at each node, start with the simplified circuit and work backwards.

It might help to assign a real value to the voltage source, like V = 3v.

The whole resistor network simplifies down to 3 ohms. Therefore, the current in the voltage source is 3V / 3ohm = 1 amp.

Then back up a step to in the video where you have 1 ohm in series with 2 ohms. This circuit lets you find the voltage at the node between the two resistors. That voltage is V (1ohm)/(1ohm + 2ohm) = 3 volts * 1/3 = 1 volt.

Next, back up to where the 2 ohm resistor is three parallel resistors. You know the voltage across the parallel resistors, and you know the resistor values. Use Ohm's Law to find the currents. Hint: they should add up to 1 amp.

Keep going backwards, alternating between finding currents and voltages. Notice how you are careful to only compute values for nodes or resistors that are part of the starting circuit. You never need to compute current in a resistor that is a simplification of several others.
• For Three resistors in parallel I prefer to work with Rp= (R1*R2*R3)/(R1*R2+R1*R3+R2*R3). Beyond three resistors this equivalent idea may get a bit cumbersome. Is this a common way of teaching this or is it considered more of a pain than it is a help?
(1 vote)
• From the author:You derive this formula by starting with the reciprocal version of three parallel resistors,
Rp = 1( 1/R1 + 1/R2 + 1/R3)
Then multiply the expression by R1R2R3/R1R2R3 to get the one you like to use.
I've not seen this taught, but if you can keep it in your head then by all means use it.
The reason it's not taught is probably because it is a special case for 3 in parallel (2 in parallel is far more common, so its special case is always taught). The general reciprocal-filled formula is fully general purpose so it's the one thing you have to memorize to master all parallel circuits.

Your formula reminds me of the conversions for Delta-Wye resistor networks: