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Math Help - Circuit Analysis

  1. #1
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    Circuit Analysis

    I'd like to apologize for not finishing yesterday, as my computer was not cooperating. Anyways, he we go:

    Consider the following circuit. Using superposition determine a) the magnitude and b) the direction of the current going through R_1 (is it away from node a?). Dtermine the c) magnitude and d) direction of the current through R_2 (is it away from node b?). e) Find V_{ab}.

    a) First, I will find the equivalent resitance, and then short each of the voltage supplies and determine the current separately.

    R_{eq}=4\Omega+[(1/3\Omega)+(1/5\Omega)]^{-1}=5.875\Omega

    I_{R_{1,1}}=(R_{eq})(R_{1})^{-1}(I_1)

    I_{R_{1,1}}=(5.875\Omega)(4\Omega)^{-1}(4V/5.875\Omega)=1 A away from node a.

    Due to the other voltage source, we have I_{R_{1,2}}=(R_{eq})(R_{1})^{-1}(I_2)

    I_{R_{1,2}}=(5.875\Omega)(4\Omega)^{-1}(6V/5.875\Omega)=1.5 A towards node a.

    Therefore, the magnitude of the current going through I_{R_1}=1.5A-1A=0.5A. For b) it is directed towards a.

    Next, this is where I get rather confused, and because of the ground location.. Let's say for a minute that the current goes through R1 and heads towards R2, what is stopping it from going through the ground rather than the resistor? Anyways, here is my attempt.

    R_{eq}=5.875\Omega

    I_{R_{2,1}}=(R_{eq})(R_{2})^{-1}(I_1)

    I_{R_{2,1}}=(5.875\Omega)(5\Omega)^{-1}(4V/5.875\Omega)=0.8 A away from node b.

    Due to the other voltage source, we have I_{R_{2,2}}=(R_{eq})(R_{2})^{-1}(I_2)

    I_{R_{2,2}}=(5.875\Omega)(5\Omega)^{-1}(6V/5.875\Omega)=1.2 A away from b.

    c) 1.2 A+ 0.8A= 2 A d) away from b.

    e) I find this one very confusing too. V_{ab} = V_a-V_b=(4V+1.915V)-(1.277V+6V)=-1.36V
    Attached Thumbnails Attached Thumbnails Circuit Analysis-figure-2.jpg  
    Last edited by quantoembryo; March 2nd 2011 at 06:35 AM.
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  2. #2
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    There we go.

    Moderator edit: Due to computer issues the OP was unable to post the complete question until now. The date stamp can't be changed to accomadate this so it is reasonable on this occassion that the OP 'bump' this thread.
    Last edited by mr fantastic; March 2nd 2011 at 04:47 PM.
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  3. #3
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    Let's call the resistor in the middle R_{3}. I would definitely go with KVL analysis (mesh current analysis) on this one, since you have three nodes but only two meshes.

    First Case: Short out the V_{1}=4\,\text{V} source. Call the left mesh current i_{1}, going in the clockwise direction. Call the right mesh i_{2}, again going in the clockwise direction.

    Mesh 1: i_{1}R_{1}+V_{2}+(i_{1}-i_{2})R_{3}=0, and

    Mesh 2: i_{2}R_{2}+(i_{2}-i_{1})R_{3}-V_{2}=0.

    The system becomes

    7i_{1}-3i_{2}=-6

    -3i_{1}+8i_{2}=6, with solution i_{1}=-0.638,\;i_{2}=0.51.

    Similar analysis with V_{2} shorted out yields, with the same mesh current definitions, i_{1}=0.681,\; i_{2}=0.255.

    So I get i_{1}=0.043,\; i_{2}=0.765. The i_{1} current is directed away from a, and the i_{2} current is directed away from b.

    I'll need to refresh my memory on the passive sign convention in order to finish the problem.

    Does this help?
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  4. #4
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    THanks
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  5. #5
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    You're welcome!
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