Reduce and Return

Discussion in 'Homework Help' started by RyanKim, Oct 5, 2011.

  1. RyanKim

    Thread Starter Member

    Sep 18, 2011
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    Just wondering if anyone can help me here....either im doing it completely wrong or my simulation in circuit maker isnt working properly to which I dont know why.
    So i was able to deduct the Source current as 4.017 AMPS. my reasoning is that little square with the 3 resistors in it is in series with the 3.3 and 2.2 resistors. From there I know that the equivalent is in parallel with the 8.2k. (Also I know the 1k and 2.7k are in series then that equivalent is parallel with the 5.6K). So im trying to find the Voltage at B so I said to myself well that I can just get by finding out whatever voltage the 2.2 resistor is drawing. So i need to find that current in that branch...so i took that branches total R and the voltage of 20 to get an current value that is not consistent with CCT maker........so perhaps you guys can help me out here. For my current in that branch with the b in it I get 2.5888A where CCT maker is getting 2.068 A.


    *just realized I mean to say mA not A.
     
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    You got the correct source current. Next, calculate the voltage at the top of the 8.2k resistor (20V - 4.017V drop across 1k). To get the current through the 2.2k resistor, you have to divide this voltage by the total series resistance in the loop that includes node b. Ignore the ground symbol in this calculation. It is only a reference for the voltage at node b.
     
  3. RyanKim

    Thread Starter Member

    Sep 18, 2011
    37
    1
    Ahh i see. I think in my head I was doing 20V equals the potential across each branch cause I wasnt thinking straight for some reason I wasn't considering the 1k resistor. Thank you very much.
     
  4. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    De nada.:D
     
  5. RyanKim

    Thread Starter Member

    Sep 18, 2011
    37
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    Oh one last thing...for I1 is that the same as saying the current coming out of the 5.6k resistor? Just denoted in the opposite direction or something?
     
  6. RyanKim

    Thread Starter Member

    Sep 18, 2011
    37
    1
    Sorry to double post but I do have another main concern that I can't really grasp. When watching the currents through a circuit, I have a tough time seeing what happens. For instance with I1...im confused does it split off to the 5.6 k resistor and then towards the negative side of the source? Doesnt a current also split off after the 8.2K resistor that goes through the 3 resistors in series in the square and conflict with I1?
     
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Does this help?

    EDIT: Regarding I1... In the original drawing, it is the sum of the currents through the 8.2k and the 2.7k. In my drawing, it is just the current through the battery.
     
    Last edited: Oct 6, 2011
  8. RyanKim

    Thread Starter Member

    Sep 18, 2011
    37
    1
    Yes it does. However I do have some general questions. I see those 2 loops currents and i think to myself....doesnt current only flow in one direction? So depending on the magnitudes of the loops am i right in saying in the middle branch the current will flow in the direction of which ever one is bigger? That is to say if Loop 1 = 5A and Loop 2 = 2A the current in the R2 branch is 3A downwards? So does that also imply that the current through R12 is simply the I2 current?
     
  9. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    I1 will always be greater than I2 (in this ckt), because it is the battery current. Therefore, the current through the middle branch may be less than that of the right branch, but it will never be negative.
    I(R2)=I1-I2, where I1>I2.
     
  10. RyanKim

    Thread Starter Member

    Sep 18, 2011
    37
    1
    Ok ok I think i get it now....thanks for all that. I can now goto sleep soundly tonight :)
     
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