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Trouble Calculating DC Resistance

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

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  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:

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

    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. 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. 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

    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

    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


    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

    "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


    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.

  4. 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.

  5. 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

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

  6. [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.)

    (* 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 }]]
    {{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)))}}
    (* Using those node voltages, find resistance from top. *)
    Rt = Simplify[et / ((et-e1)/R1 + (et-e2)/R4) /. cc[[1]] ]
    (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:
    (* Verify Kirwan's result with his inputs. *)
    Rt /. {
    R1->1200, R2->2700, R3->330, R4->1800, R5->680
    } //N
  7. 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

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

  8. 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.
  9. 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.

  10. 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.

  11. 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.
  12. The simple fact is that it *IS* better. You haven't disputed this, at
    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.

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