# Trouble Calculating DC Resistance

Discussion in 'Electronic Basics' started by [email protected], Jun 19, 2005.

1. ### Guest

I was looking through some old electronics books of mine, and I can't
seem to figure out the equation to calculate the following circuit:

o
|
|
---------
| |
R R
1 4
| |
---R 3---
| |
R R
2 5
| |
---------
|
|
o

I just can't figure out how R3 fits in. I would guess that it is
somehow in parallel, but I'm not sure with what. Could someone write
out the equation for solving the total resistance of this circuit? I'm
very confused.

Thanks a bunch,

Danny H.

2. ### Larry BrasfieldGuest

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)

3. ### Jonathan KirwanGuest

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

One way of viewing the above setup for these two "= 0" equations (and
it's only one of the ways, but I think it may help here) is to look at
each node like this: Current _spills_ away from a node based on the
voltage at the node and each of the conductances away from it and
current _spills_ back into a node based on the voltages at the nearby
nodes and the conductances back in.

So when I wrote:

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

4. ### Jonathan KirwanGuest

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

5. ### Jonathan KirwanGuest

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:

http://mcraefamily.com/MathHelp/PhysicsElec.htm

Jon

6. ### Larry BrasfieldGuest

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

7. ### Jonathan KirwanGuest

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
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

8. ### Larry BrasfieldGuest

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.

9. ### Jonathan KirwanGuest

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

Jon

10. ### Jonathan KirwanGuest

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.
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

11. ### Larry BrasfieldGuest

Actually, coming after I explained that it made no sense to
take scarce time to give the OP more of an answer than he
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.

12. ### Jonathan KirwanGuest

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