Mesh Current Method

Discussion in 'Homework Help' started by schutnik, Dec 19, 2005.

  1. schutnik

    Thread Starter New Member

    Dec 19, 2005
    4
    0
    Hi guys,

    I'm new to the forum, and new to electronics. I'm trying to learn on my own using the books, which are fabulous btw! I was thinking about the example for the Mesh Currents method given in chap 10 of vol 1. The example I'm confused about is the Wheatstone bridge example. I don't have a good way to submit a picture of this electronically (don't have the skills to draw it yet), but I'll try to describe my question.

    The bridge circuit is given as:

    ..........................
    + R1 R2
    24v ... R3 ...
    - R4 R5
    ..........................

    where R4 and R5 are connected to R1 and R2 with R3 bridging the two parallel connections (R4--R1 // R5--R2), and the junction of R4 and R5 is connected to the - on the battery and the junction of R1 and R2 is connected to the + of the battery.

    R1 = 150R, R2 = 50R, R3 = 100R, R4 = 300R, and R5 = 250R (R = Ohms)

    Now, I understand how the solution is found in the book, but my question is why can't the solution be found with only two currents defined as:

    I1: from battery - to R4, R3, R2, and back to + of battery
    I2: from battery - to R5, R3, R1, and back to - of battery.

    I understand this is a wierd way to try to solve the problem, but I didn't see any reason in the description of this method why it shouldn't work. I understand why it can't, since it would force the current calculated for R5 and R1 to be equal and the current for R4 and R2 to be equal, which according to the solution found, is not the case. However, I can't figure out why this "method" of solution would not be legal as the technique is described in the book. What rule am I missing for the application of the Mesh Current method? Is it that only one of the currents can go through a given source? I didn't see that written anywhere, but that rule, if imposed, would certainly disallow my "method".

    Thanks guys and gals. Looking forward to your help!
     
  2. n9352527

    AAC Fanatic!

    Oct 14, 2005
    1,198
    4
    I think you pin out the reason why it didn't work accurately. It is not because you have two loops going through the source, but because you have only one loop for two different legs in the circuit. Each distinct leg should have its own loop(s), either one distinct loop or combination of two or more loops. Otherwise, we are effectively adding one or more constraints in the circuit, which in your example are current through R1 is equal to current through R5 and current through R2 is equal to current through R4. These constraints are not true.

    It has been so long since I did this and I don't remember all the rules, but I think the mesh method is derived from KVL and superposition theorem. In order not to violate these, the loop has to be independent or in other words doesn't have another unassigned loop inside it or pass through more than one legs on its own.
     
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