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Complex Circuit Question

  1. Aug 27, 2014 #1
    1. The problem statement, all variables and given/known data
    circuit.jpg
    Solve for all voltage drops and currents at the resistors and battery


    2. Relevant equations
    V=IR
    RT=R1+R2+...+Rn
    1/RT=1/R1+1/R2+...+1/Rn
    IT=I1+I2+...+In

    3. The attempt at a solution

    I just wanted to check if my solution is correct

    ------Voltage----Current----Resistance
    R1-----150--------3.75---------40
    R2-----150--------3.75---------40
    R3-----45---------0.75---------60
    R4-----15-----------3------------5
    R5-----30-----------2-----------15
    R6-----20-----------1-----------20
    R7-----10-----------1-----------10
    Total---345--------3.75---------92
     
    Last edited: Aug 27, 2014
  2. jcsd
  3. Aug 27, 2014 #2

    mfb

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    This would be easier if the resistor numbers and current directions would be marked in the diagram...

    You can simply check Kirchhoff's laws for every node and loop as cross-check.
     
  4. Aug 27, 2014 #3

    BvU

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    Is there something about V that you seem to know and we don't ?
    So far I can only stamp an approval stamp on R6 and R7 (they happen to be the only ones with 10 and 20 Ohm).
     
  5. Aug 27, 2014 #4
    Because the voltage drop though resistors R6(20ohms) and R7(10ohms) causes a total voltage drop of 30V any parallel circuits to it will also have a voltage drop of 30 V. therefore R5(15ohms) must have a voltage drop of 30V.
     
  6. Aug 27, 2014 #5
    Because the voltage drop though resistors R6(20ohms) and R7(10ohms) causes a total voltage drop of 30V any parallel circuits to it will also have a voltage drop of 30 V. therefore R5(15ohms) must have a voltage drop of 30V.
     
  7. Aug 27, 2014 #6

    mfb

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    The voltage drops of 30 V on the right side are fine, but the voltage of the voltage source is unclear.
     
  8. Aug 27, 2014 #7
    The total is what I calculated the voltage source to be but I'm not sure its correct.
     
  9. Aug 27, 2014 #8

    mfb

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    Which total?
    You cannot solve this problem if you don't know the voltage drop at the voltage source.
     
  10. Aug 27, 2014 #9

    NascentOxygen

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    It appears the ammeter is displaying a reading of 1.0A, meaning there is sufficient information to determine V.
     
  11. Aug 28, 2014 #10

    BvU

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    My mistake. I saw a current source instead of an ammeter. :redface:
     
  12. Aug 28, 2014 #11

    mfb

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    Same here.
    Okay, then it is fine.
     
  13. Aug 28, 2014 #12

    BvU

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    So herewith a stamp of approval goes to post #1. And now I run the risk of being chastized by the spirits that float over and guard PF: the forum isn't really intended for that kind of thing at all. With good reasons.
    Still, it was nice to look into this exercise...
     
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