calculating currents in resistive network

Discussion in 'Homework Help' started by Hurt_it_Circuit, Oct 2, 2012.

  1. Hurt_it_Circuit

    Thread Starter Member

    Oct 2, 2012
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    I do not understand the approach to this problem. Initially I combined R6 and R7 using the current division equation then I added R8 and then I used current division equation to add R3 to R678. Then I added R4 to the total and finally I then used that total in current division equation to combine with R2. Then I tried to calculate I2 but my calculated current exceeded the current source of 9.25mA. My wrong calculated value was 33.59mA. What am I doing wrong please help!!!!!!!!!!!!
     
  2. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    You would need to show your work for us to determine where you went wrong.

    One aid I used to keep me on track is ..
     
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    Last edited: Oct 3, 2012
  3. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Ok ... here are the questions ...

    (1) What resistance did you get for the parallel combination of R6 and R7?
    (2) What resistance did you get for the parallel combination of R3 and everything to the right of R3?
    (3) What resistance did you get for the parallel combination of R2 and everything to the right of R2?
     
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  4. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    Proceedure:

    Combine the "right hand" resistors together in parallel series combinations until you get to these resistors: R1, R2, R5, and the combined equivalent value across R2.

    The current from the source (Is) is known (and is flowing through R1 and R5) , so it splits between R2 and it's parallel equivalent based on the inverse of the resistance values. That gives you I2. The difference between Is and I2 feeds all the circuits to the right.

    Open up the equivalent resistor network one side at a time and solve for the parallel current the same way each time.
     
    Last edited: Oct 3, 2012
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  5. Hurt_it_Circuit

    Thread Starter Member

    Oct 2, 2012
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    when i combined R6 and R7 i got a resistence of 45.9.
    Am I able to add resistor 8 to R67=45.9 if I do that I get 99.9.. OR am I only able to add the parallel circuits together?
     
  6. Hurt_it_Circuit

    Thread Starter Member

    Oct 2, 2012
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    Ok disregard previous comment. I found the resistance equivalent of R234678 =R(eq) to be 4.79 kOhms. So after doing this i am left with a series current only involving R1, R(eq) and R5. Is this the correct process? If so do I then use the current division equation to solve for the currents I2, I3, and I6 with my equivalent resistance being 4.79 kOhms? How do I go about calculating I8?
     
  7. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    The Req of R2 and everything to the right of it is not 4.79k.

    Answer my questions (1), (2), and (3) so I can see where you went astray. Please compute to whole units plus three decimal places.

    You are correct with the end result being a simple series circuit of R1, Req, and R5.

    Break the circuit down to the simplest form, then expand it, using KVL and KCL solving all your questions.
     
    Last edited: Oct 3, 2012
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