Finding current through R4

Discussion in 'Homework Help' started by metelskiy, Nov 22, 2010.

  1. metelskiy

    Thread Starter Member

    Oct 22, 2010
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    I need to find I4 in this problem:
    Practice 1.png
    I'm not sure how to start this problem because i'm getting confused with 2 power sources. Would i first find V between R1 and R2? I came out with 14V using voltage divider. What happens to 14V and 24V, do I add them to make it 38V? And I need help how to find current through R4. WOuld it be 38V splitting into R3 and R4? Thanks.
     
  2. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    metelskiy likes this.
  3. metelskiy

    Thread Starter Member

    Oct 22, 2010
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    Thanks, we learned superposition but forgot to apply it to this problem.
    So first i shorted 24V and found I4'=0.7A
    Then shorted 42V and found I4''=2.4A
    Total I4=3.1A If someone got time to check it, would be thankful.
     
  4. hgmjr

    Moderator

    Jan 28, 2005
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    That is not the answer I obtained using Millman's Theorem. Perhaps it would be best if you would scan your calculations and attach them here so that we can review you technique.

    I am assuming that I4 is the current flowing through R4.

    hgmjr
     
  5. metelskiy

    Thread Starter Member

    Oct 22, 2010
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    Here are my all calculations, yes I4 is current through R4. I used Superposition.
    P1070183 [1280x768].JPG
     
  6. hgmjr

    Moderator

    Jan 28, 2005
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    You have correctly calculated the current in R4 due to the 42 volt source. However, you need to revisit your calculation for the current in R4 due to the -24V source.

    hgmjr
     
  7. metelskiy

    Thread Starter Member

    Oct 22, 2010
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    I'm confused with +24V and -24V? Were i suppose to use -24V source instead of 24V? When I shorted 42V I redrew circuit with 24V left. As seen in original circuit, R3 and R4 connected to positive side of 24V and R1&R2 to negative side of 24V. Hint me on my mistake.
     
  8. hgmjr

    Moderator

    Jan 28, 2005
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    The first thing you need to do is calculate the parallel combination of the two resistor pairs. This will permit you to calculate the total current being drawn from the -24V source. Then you will be in a position to calculate the portion of voltage from -24V that is dropped across R4.

    hgmjr
     
  9. metelskiy

    Thread Starter Member

    Oct 22, 2010
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    hgmjr,
    I'm sorry i might look or sound stupid but i totally didn't understand what you meant by your last explanation.. I don't see where you got -24V source. :( If you could get into it more deep that's would be more helpful to me. Thanks.
     
  10. hgmjr

    Moderator

    Jan 28, 2005
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    I am referring to the diagram in your attached worksheet.

    The figure at the bottom of the page contains the 24V source and the two pairs of resistors in parallel. R3||R4 and R1||R2 are the two pairs of parallel resistors to which I was referring.

    hgmjr
     
  11. metelskiy

    Thread Starter Member

    Oct 22, 2010
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    3
    Ok, I found R3||R4=6Ω and R1||R2=6Ω, From this:
    Rt=12Ω Current total=24V/12Ω=2A
    Now since i know total current, thats when i need to determine how much current goes to R4 and how much to R3?
     
  12. hgmjr

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    Jan 28, 2005
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    Remember that the 2A current is split up between the 15 ohm and the 10 resistor. You will be better off if you determine the voltage across the 15 ohm and the 10 resistor. Then the current in R4 will be that voltage divided by 10 ohms. Right?

    hgmjr
     
  13. metelskiy

    Thread Starter Member

    Oct 22, 2010
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    3
    ok using current divider i figured out I4:
    (R3,4/R4)*It=(6Ω/10Ω)*2A=1.2A

    In other way as you mentioned to determine V across 15Ω and 10Ω. Wouldn't it be 24V since they are in parallel and connected to positive 24V? This is the part I'm confused about. Is my circuit with 24V where I shorted 42V correct or wrong?
     
  14. hgmjr

    Moderator

    Jan 28, 2005
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    The above calculation is correct for the current flowing in R4 due to the -24V power source.
    The calculation is correct as I indicated above.

    hgmjr
     
  15. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Do you see the error you made in the original calculation?
     
  16. metelskiy

    Thread Starter Member

    Oct 22, 2010
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    Not really, I believe i had mistake in second circuit where I shorted 42V and left 24V as a power source but i don't see exactly the error. Could you explain please. Thanks.
     
  17. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Your answer in Post 13 for the resistance doesn't match what you show in your work.
     
  18. renotenz

    New Member

    Oct 25, 2010
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    Nope... it's more like there's a +24 V difference between the voltage across R3-R4 node and R1-R2 node.

    Just parallel R1-R2 and R3-R4 as you've done before, then draw a new diagram consists of both parallel-equivalent resistance and the 24 V voltage source. Since both parallel-equivalent resistor are connected to ground, you can short their bottom terminals. Then use voltage divider to see the voltage of each resistor.
     
  19. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Here was your mistake ...
     
  20. metelskiy

    Thread Starter Member

    Oct 22, 2010
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    JoeJester, Thank you so much for taking time to explain. I see the mistake, but from other hand looking at the circuit (if it is drawn correctly) isn't R3&R4 connected to 24V, from that: Voltage in parallel is the same and to find current through I4=V4/R4
     
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