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Thread: 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 $\displaystyle R_1$ (is it away from node a?). Dtermine the c) magnitude and d) direction of the current through $\displaystyle R_2$ (is it away from node b?). e) Find $\displaystyle V_{ab}$.

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

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

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

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

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

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

    Therefore, the magnitude of the current going through $\displaystyle I_{R_1}=1.5A-1A=0.5A$. 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.

    $\displaystyle R_{eq}=5.875\Omega$

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

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

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

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

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

    e) I find this one very confusing too. $\displaystyle V_{ab} = V_a-V_b=(4V+1.915V)-(1.277V+6V)=-1.36V$
    Attached Thumbnails Attached Thumbnails Circuit Analysis-figure-2.jpg  
    Last edited by quantoembryo; Mar 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; Mar 2nd 2011 at 04:47 PM.
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  3. #3
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    Let's call the resistor in the middle $\displaystyle R_{3}.$ 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 $\displaystyle V_{1}=4\,\text{V}$ source. Call the left mesh current $\displaystyle i_{1},$ going in the clockwise direction. Call the right mesh $\displaystyle i_{2},$ again going in the clockwise direction.

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

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

    The system becomes

    $\displaystyle 7i_{1}-3i_{2}=-6$

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

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

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