Trouble Calculating DC Resistance

One way to solve it would be to convert the R1,R2,R3 set from its current Y form to the equivalent delta form. See

Once that is done, the topology can be step-wise converted using the rules you should know.

(R1 R2 R4 + R1 R3 R4 + R2 R3 R4 + R1 R2 R5 + R1 R3 R5 + R2 R3 R5 + R1 R4 R5 + R2 R4 R5) / (R1 R2 + R1 R3 + R2 R3 + R2 R4 + R3 R4 + R1 R5 + R3 R5 + R4 R5)

Welcome.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

Assume:

If the ratios of R1/R2 and R4/R5 are equal, it becomes trivial because no current flows through R3 and you can eliminate it from your analysis.

But don't be too hard on yourself if they aren't equal, as this is the ever-popular unbalanced Wheatstone bridge problem and it takes an amount of algebra to solve it via simultaneous equations and a branch current analysis. (Or you can use mesh analysis, which is perhaps a little easier, algebraically speaking.)

Here is how a spice program might solve the problem. First, spice uses conductances for the resistors, instead of resistance values. (Which is a reason why spice hates resistances of 0 ohms.)

So it would do something like:

Treat the resistances as conductances,

G1 = 1/R1 G2 = 1/R2 G3 = 1/R3 G4 = 1/R4 G5 = 1/R5

Set known voltages,

V(1) = V V(4) = 0 (Spice needs a ground ref, so it might as well be here.)

Set up an equation for the rest (KCL/KVL):

V(2)*(G1+G2+G3)-V(3)*G3-V(1)*G1-V(4)*G2 = 0 V(3)*(G3+G4+G5)-V(2)*G3-V(1)*G4-V(4)*G5 = 0

V(2)*(G1+G2+G3)-V(3)*G3-V(1)*G1-V(4)*G2 = 0

I was thinking,

"Hmm. V(2) will spill current away based on the sum of the conductances leaving the node. This is G1, G2, and G3, so the current spilling away is simply V(2)*(G1+G2+G3). But, current is spilling into the node based on the three different voltages at the nearby nodes coming back through these same conductances, so this amount will be of opposite sign and will be -V(1)*G1, -(V3)*G3, and -V(4)*G2. Since the total current arriving into a node must be the same as the total current leaving it, the sum of these two must be zero -- or else we are in for big trouble!"

Similar thinking gets you the results for V(3), too. Later on, I'll apply this thinking to the node at V(1) [V(4) would work, as well] in order to compute the total current.

Okay, back to reality. Since V(4)=0 and V(1)=V, the above reduces slightly to:

V(2)*(G1+G2+G3)-V(3)*G3-V*G1 = 0 V(3)*(G3+G4+G5)-V(2)*G3-V*G4 = 0

In the above pair, you have two unknowns. These are V(2) and V(3). You also have two equations, luckily. So let's rewrite them into slightly different form which calls out the constants a little more clearly:

V(2)*[G1+G2+G3] + V(3)*[-G3] = [V*G1] V(2)*[-G3] + V(3)*[G3+G4+G5] = [V*G4]

which can be rewritten as:

a*x + b*y = c d*x + e*y = f

where x is V(2) and y is V(3) and with the obvious substitutions for a, b, c, d, e, and f.

The solution to such a pair of equations can be done through simple algebraic manipulation, but it is often shown through setting up a basic matrix form because it is easy to visualize without getting too caught up in the detailed manipulations:

[ a b ] [x] [c] [ ] * [ ] = [ ] [ d e ] [y] [f]

The solution is, of course:

[ a b ]-1 [c] [x] [ ] * [ ] = [ ] [ d e ] [f] [y]

So you just need to compute:

c*e - f*d (G1*(G3+G4+G5) + G3*G4) V(2) = x = --------- = V * ---------------------------- a*e - b*d (G1+G2+G3)*(G3+G4+G5) - G3^2

and,

f*a - c*d (G4*(G1+G2+G3) + G3*G1) V(3) = y = --------- = V * ---------------------------- a*e - b*d (G1+G2+G3)*(G3+G4+G5) - G3^2

To solve for the voltages at the two nodes. After you have the voltages, you should be able to find all the currents, of course.

You can look up matrix solutions in most any pre-calculus (preparation for linear systems) or decent algebra math book (solving intersections of two different lines is a very common task in algebra 2, I think.) Or you can work through the solution of a two-line intersection problem, keeping the terms abstract, and see it just as clearly.

(You might also want to visualize the actual "lines" involved and how they relate back to the diagram, but that's for another time.)

Anyway, what about the total current through the system? Well, you can apply the same reasoning I was applying before, to think like this:

"The total current flowing into node V(1) from the voltage source would give me what I need to know. So let's do up an equation for that node. The total current spilling into the node from the voltage source must be equal to the net of what happens when I only consider the other connections. So, current spilling out through R1 and R4 are simply V(1)*(G1+G4) and spilling back in through those resistances must be -V(2)*G1 and -V(3)*G4. What arrives through the only other connection, which is from the voltage source, must be this amount. So: I(total) = V(1)*(G1+G4) - V(2)*G1 - V(3)*G4"

Since you already now have V(2) and V(3), computing this becomes easy.

Let's select some exact values for the resistors.

R1 = 1200 R2 = 2700 R3 = 330 R4 = 1800 R5 = 680

and,

V = V(1) = 12V.

If you do your equations as I did (and if I didn't get all of this dead wrong), it should result in:

V(2) = 5.7881 V V(3) = 4.7872 V

And the total voltage source current is 9.184 mA. Net resistance of the bridge is about 1306.66 ohms.

Jon

Correction. This is:

c*e - f*b (G1*(G3+G4+G5) + G3*G4) V(2) = x = --------- = V * ---------------------------- a*e - b*d (G1+G2+G3)*(G3+G4+G5) - G3^2

Typo. In this case, it makes no difference in the calculation as 'b' and 'd' are the same variable and sign.

Jon

[The OP had asked, about a 5 resistor bridge:]

Also for the OP, in case he missed the first part of my post, I had suggested a perfectly workable way to solve the OP's problem: | One way to solve it would be to convert the R1,R2,R3 set | from its current Y form to the equivalent delta form. See |

| Once that is done, the topology can be step-wise converted | using the rules you should know. That suggestion, or its generalization into "apply transformations where they simplify analysis", ought to count as a fishing lesson.

Hmmm. They appear to have done the algebra I left the OP to work out, using the approach I suggested. But that was not the derivation of my expression, (which was exactly what the OP requested, no more and no less.)

Here is another fishing lesson for those ready to take it: (The blocks marked "in:" and "out:" represent input to and output from a Mathematica session.)

in: (* Setup node equations for the internal nodes. *) en1 = e1(1/R1 + 1/R3 + 1/R2) == et(1/R1) + e2(1/R3); en2 = e2(1/R4 + 1/R5 + 1/R3) == et(1/R4) + e1(1/R3); (* Solve them for the internal node voltages. *) cc = Simplify[Solve[ en1 && en2, { e1, e2 }]] out: {{e1 -> (et R2 (R3 R4 + R1 R5 + R3 R5 + R4 R5)) /

(R1 R2 R4 + R1 R3 R4 + R2 R3 R4 + R1 R2 R5 +

R1 R3 R5 + R2 R3 R5 + R1 R4 R5 + R2 R4 R5),

e2 -> -((et R3 (R1 R2 + R1 R3 + R2 R3 + R2 R4) R5) /

(R1 R2 R4 R5 - (R1 R2 + R1 R3 + R2 R3)

(R3 R4 + R3 R5 + R4 R5)))}} in: (* Using those node voltages, find resistance from top. *) Rt = Simplify[et / ((et-e1)/R1 + (et-e2)/R4) /. cc[[1]] ] out: (R1 R2 R4 + R1 R3 R4 + R2 R3 R4 + R1 R2 R5 + R1 R3 R5 +

R2 R3 R5 + R1 R4 R5 + R2 R4 R5) /

(R1 R2 + R1 R3 + R2 R3 + R2 R4 + R3 R4 + R1 R5 + R3 R5 +

R4 R5)

For Mr. Kirwan's amusement, after the above has been evaluated, the session can be extended as follows: in: (* Verify Kirwan's result with his inputs. *) Rt /. { R1->1200, R2->2700, R3->330, R4->1800, R5->680 } //N out:

1306.66

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

"Jonathan Kirwan" wrote in message news: snipped-for-privacy@4ax.com...

I suppose I agree in the "more is better" sense. When I've been paid to tutor individuals, that kind of attention made sense. But when I've got deadlines to meet and errands to run, I have to make a tradeoff in favor of letting an OP come back if more help is needed. From his post, I had no reason to think he would have any trouble once the delta-wye transformation was done. Doing his work or overkilling the suggestion did not make overall sense.

You're getting things backwards here. I told the OP how to solve the problem. The link was merely a way to get him the transformation formula and, if necessary, get him to understand what the transformation was. If he had been willing to look at the link, and drawn the delta in place of the R1,R2,R3 wye, his problem would have been reduced to the kind EE and EET students tackle every day. I see no reason to draw out every step in that sequence.

Maybe. And if he needed clarification, he could easily have sought it.

It's all obvious once one decides to do the transformation.

Ok. Perhaps you would be interested in finding the dual of that bridge and seeing if it is more directly solvable.

In circuit analysis, it is sometimes useful to use a combination of nodal and loop equations rather than just one or the other.

Actually, the wye-delta transform is just an example of the more general "dual" concept.

Your comments make me wonder if you have to worry about time management.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

Just for the OP, the above completely lacks any analysis to help you understand it or apply your knowledge anywhere else. In biblical parable terms, you were "given a fish rather than being taught how to fish."

But for a straight-forward derivation of the above, you might look here, for example:

Jon

I suppose. But I think it's quite a bit better to make a general lesson clearer through at least one concrete example. Better still, is if the transition from abstract to concrete (or visa versa) is taken in several, leveled steps.

The link you provided was merely a hint, waving in the direction of somewhere. There was no exposition on how the OP could apply the abstraction, at all. It may have been a fishing lesson, but without the right directions to get to the right pond, the OP might have wound up fishing in the wrong place and quite frustrated by it, as well.

They also lead the horse to the water, so to speak, showing how to apply the Y-delta idea to the specific case at hand. I consider this a very important element to include, indeed.

...

By the way, what is more interesting to me is the sheer number of useful concepts that can be independently applied to this problem, with the same results. It's one of those problems that can be sliced from a number of perspectives. And gathering those perspectives, all of them, is very helpful in developing broad mental tools for attacking various problems later on. Since the problem can be "seen" from different perspectives, with the same answers arriving from each of them, this kind of problem provides a nice fulcrum for developing some skill in each and to gain a broader grasp.

I consider the general approach applied by spice to be useful over a broader range of problems (as is evidenced well by how accurately spice can deal with complex situations.) And while branch analysis is vital to master, the mesh analysis will often yield less complexity and should be mastered, as well. (Both fall under KCL, which of course should be understood, too.) Finally, if I had to pick one of these to _NOT_ memorize, it would be the y-delta transform since it falls out almost trivially as a consequence of more powerful approaches.

Of course, that's just my sense of things speaking as a hobbyist.

Jon

A non sequitur. Worse, it might tend to make someone feel badly about trying to help others, were I sensitive to such prying comments. (And I have no _other_ idea what your purpose was in saying so unless it were to add a negative element on a personal level.)

But this comment has no business being made here, Larry. None I can see. But you are welcome to worry about my situation. That's none of my business.

Jon

I wasn't complaining about your post, Larry. I just felt that if the OP wanted more exposition on the subject you brought up, there was another site where that exposition could be better found. This is no skin off of your neck. I see no reason why my addition is a problem to you, or needs any defense on your part. It was not an attack.

I was explaining why I added to your post, not criticizing it. So again, you are just being defensive where it isn't needed. I don't think it's necessary to defend what you did. However, I did explain to you why I felt the need to add some more. No harm, there.

Agreed.

Perhaps. I think that's a little presumptive, though.

Do you care if I do? Would this help anyone's current question?

What's really useful is having the encompassing mental tools that provide the flexibility to appropriately make those choices -- not just having the tools themselves, but the knowledge about those tools and about the unifying nature of them.

I just enjoy seeing them all.

Jon

Actually, coming after I explained that it made no sense to take scarce time to give the OP more of an answer than he has asked for, it is a sequitur. Your comments make it appear that you think it would be better to provide way more of an answer than appears needed, on the chance that it might do some good. To me, that attitude indicates somebody with plenty of time on their hands..

If you were sensitive, (which I had not guessed up to now), I apologize for triggering that reaction.

Not worried, just observing. As for appropriateness, I take no responsibility for speculative malign interpretations.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

The simple fact is that it *IS* better. You haven't disputed this, at all.

In other words, you are making this personal.

About your comment, I'm not. But that is because my circumstances are exactly as I have worked to make them and I'm quite satisfied, these days. Were it otherwise, were I still struggling about who I am for example, I might have taken your personal comment somewhat more poorly. The reason I bother calling you on it is simply a matter of my own perfunctoriness.

Frankly, I still cannot see a _positive_ reason for you to have made it. And you've not made it any better by your explanations. It was simply uncalled for.

Of course not. And I've not made any. It's simply that I cannot find any possible _positive_ interpretation for your comment.

Jon