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Possible confusion regarding application of Kirchoff's Laws?

  1. Oct 15, 2016 #1
    1. The problem statement, all variables and given/known data

    Determine Currents ##I_{1}, I_{2}, I_{3}##

    2. Relevant equations

    Kirchoff's Rules:

    $$\sum_{i=1}^n I_{i} = 0$$

    $$\sum_{i=1}^n V_{i} = 0$$


    3. The attempt at a solution

    zJe4NbGl.jpg

    Basically, I'm not sure if I'm applying the Loop Law correctly. For example, in Loop A, going clockwise from the upper leftmost corner of the circuit, it's easy enough to see that I'd have ##-\epsilon_{1}## and ## +I_{1}R_{1}## based on the direction I'm "tracing" the circuit in, but when I come to the first junction, ##J_{1}##, I'm not sure what to do.

    Applying Kirchoff's Junction rule to it based on the entirety of the circuit shows that

    $$\sum_{i=1}^n I_{i} = I_{2} + I_{3} - I_{1} = 0 \rightarrow I_{1} = I_{2} + I_{3}$$

    But ##I_{3}## isn't in Loop A... so then

    $$\sum_{i=1}^n I_{i} = I_{2} - I_{1} = 0 \rightarrow I_{1} = I_{2}$$

    I used a similar argument to set up my equation for Loop B.

    Is the way I have my system of equations set up (in the image above) correct? If not, would I have to calculate the voltage drop using ##I_{2} = I_{1} - I_{3}##?

    Using my current system of equations, my values for the currents are

    ##I_{1} = \frac{935}{626} A , I_{2} = \frac{205}{313} A, I_{3} = \frac{525}{626} A##

    Is that correct?

    Thanks in advance.
     
  2. jcsd
  3. Oct 16, 2016 #2

    andrewkirk

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    It looks correct to me.

    The metaphor I use is about swimming up or downstream. Imagine that the stream is flowing in the direction of the arrow that indicates the direction of each current ##I_k##. Start at one point in the loop and go around the loop, adding a positive or negative term for every item we encounter. For a resistor, we add (subtract) the term ##I_kR_j## if we are swimming upstream (downstream) because that is how much the potential increases as we pass through the resistor from the downstream end to the upstream end. For a power cell, we add (subtract) the voltage of the cell if we enter at the - (+) terminal and leave at the + (-) terminal.

    EDIT: When I say it looks correct I'm referring only to the image of the hand-written calcs. As per @LemmeThink's post below, I cannot follow your reasoning in the bit he quoted, which is not in the hand-written calcs. Nor do you need that bit. You already have one equation from the Kirchoff Junction rule, and on the handwritten page you get two more equations from the loop rules. So you have three equations and three unknowns, which can be solved.
     
    Last edited: Oct 16, 2016
  4. Oct 16, 2016 #3
    Hi!
    Your equations seem to have been setup correctly. However, the above quoted text doesn't seem to make sense. How did you get this?
     
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