# Complicated Equiv Resistance Problem

Discussion in 'Homework Help' started by sjgallagher2, Mar 31, 2015.

1. ### sjgallagher2 Thread Starter Active Member

Feb 6, 2013
128
12
I need some help, after taking a stab at this circuit to analyze and come up with an equivalent resistance, I realized I couldn't do it. I built the circuit and know the resistance is ~480 ohms. Here's the circuit, all resistors are 1k, and I need the equivalent resistance between the two ports.

If anyone can solve, and briefly describe what strategy you used to get there, that'd be so great! Thanks. ~Sam Gallagher

2. ### MikeML AAC Fanatic!

Oct 2, 2009
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Should have been posted in "Homework Help".

First principles: Mesh analysis, Kirchoff, and Ohm.

3. ### blocco a spirale AAC Fanatic!

Jun 18, 2008
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(R1+R4)||(R2+R6)||R3

Brackets...

Last edited: Mar 31, 2015
4. ### WBahn Moderator

Mar 31, 2012
24,559
7,696
That's probably going a bit too far in just doling out a solution to a homework problem.

But what's done is done.

@sjgallagher2 Do you see how sometimes just redrawing the circuit can make the solution pretty obvious when otherwise it appears you are looking at some really complicated mess?

Also, while nothing official says so, the parallel operator generally binds tighter than the series operator, so this would likely be interpreted as

R1+(R4||R2)+(R6||R3)

But even if that weren't the case it's better to be explicit

(R1+R4)||(R2+R6)||R3

Also note that this solution is NOT general. It is valid for this circuit ONLY because of another piece of information you provided. Can you see what that information is and why this solution would not be valid without it?

5. ### sjgallagher2 Thread Starter Active Member

Feb 6, 2013
128
12
This isn't for homework to be honest, its just a wheatstone bridge I was analyzing with KVL and KCL for practice, then wanted to make a thev-equiv because I don't practice that much, so I simplified into a resistor network, redrew/simplified the circuit a few times, changed the values to all 1k for simplicity, and realized I couldn't get any further! Thanks so much for the advice, by the way. But how come R5 doesn't affect the equivalent resistance? The original circuit had different values, is the fact that all the resistors are 1k the piece that makes this solution not general? And then my last question, assuming all names actually indicate resistances (R1 = 1 ohm, R2 = 2 ohm, etc) what is the solution? I appreciate the questions from you too, helpful! ~Sam Gallagher

6. ### blocco a spirale AAC Fanatic!

Jun 18, 2008
1,535
412
Notice that there is no potential difference across R5 and, therefore, no current through it. In other words, it has no effect on the circuit and may as well not be there.

If all the resistor values are different then R5 does come into play and analysis becomes more difficult...... post#2 holds the answer.

Last edited: Mar 31, 2015
7. ### WBahn Moderator

Mar 31, 2012
24,559
7,696
If the resistors are chosen such that the voltage across R5 is 0V, then no current will flow through it no matter what value it is. In fact, it could be replaced with either an open circuit or a short circuit.

Here are a couple of interesting and useful exercises for you. Knowing that the key to being able to ignore R5 is that the voltage dividers on both sides of it need to produce the same voltage on each side, what are the constraints that the R1, R2, R4, and R6 have to satisfy in order to render R5 inconsequential? You might be able to write the answer down by inspection, but don't get too quick to draw. Do NOT just say that R1=R2 and R4=R6 or something similar. Those might be "sufficient" conditions, but they aren't "necessary" conditions, meaning that I could find a set of values that render R5 moot while not satisfying those equations.

Once you have that, then approach figuring that same set of constraints out by a different, more subtle, and trickier path. Recall that I said that you could either replace R5 with a short or with an open and you wouldn't be able to tell the difference? Well, figure out the effective resistance of those four resistors when R5 is replaced by an open and then figure out the resistance if it is replaced by a short and, since I said your could tell the difference, set them equal to each other.

In the general case, where R5 can't be ignored, you can do two things (off the top of my head). You can use delta-wye transforms to create a circuit that consists of just series/parallel combinations, or you can apply a test voltage to the terminals and analyze the circuit using whatever technique you want to find the test current and then leverage the fact that, by definition, the equivalent resistance is the ratio of the test voltage to the test current.

8. ### sjgallagher2 Thread Starter Active Member

Feb 6, 2013
128
12
Thanks so much for that reply, you're really great. I looked into delta-wye and learned something new. I'll think about those exercises too!