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Compound circuit voltage drop along parallel component

  1. Nov 12, 2012 #1
    1. The problem statement, all variables and given/known data

    http://imageshack.us/a/img197/3519/circuith.png [Broken]

    Equivalent Resistance of circuit = 17.31 Ω


    2. Relevant equations

    V = I * R

    3. The attempt at a solution

    Total voltage of the circuit, if I'm doing this correctly should be:

    E = (3.00 A) * (17.31 Ω) = 51.93 V

    But something doesn't seem right about this number

    When we collapse the series in the top parallel circuit branch it becomes: 24.0 Ω + 6.00 Ω = 30.0 Ω

    Then we collapse the whole parallel circuit into a single resistor equal to: [itex] \frac{1}{\frac{1}{15.0 Ω} + \frac{1}{30.0 Ω}} = 10.0 Ω [/itex]


    I'm not quite sure where to go past this and I'm questioning the voltage I'm getting, so any pointers in the right direction would be awesome, thanks :)
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 12, 2012 #2

    lewando

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    What makes you question the voltage? It is in the ballpark.
     
  4. Nov 12, 2012 #3
    The thing that makes me question the voltage is the fact that all the voltages given to us so far in any problems were whole numbers and usually over 100V.

    And if that's the correct voltage, then I'm not quite sure where to go next.
     
  5. Nov 12, 2012 #4

    SammyS

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    What is it you are trying to find?

    The voltage value you found is correct.
     
    Last edited by a moderator: May 6, 2017
  6. Nov 12, 2012 #5
    I type that up and I managed to forget the actual question! Haha, sorry.

    The question:

    What is the voltage drop across the parallel arrangement in the upper branch?

    E = I * R

    E = (3.00 A) * 10.0 Ω = 30.0 V?

    That looks right to me. Anyone confirm for me please?
     
  7. Nov 12, 2012 #6

    SammyS

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    Yes, that's correct!
     
  8. Nov 12, 2012 #7
    Thanks SammyS, I appreciate your help.
     
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