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Current after passing all resistors

  1. Sep 10, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the current at point I:

    [PLAIN]http://img690.imageshack.us/img690/8508/circuit.png [Broken]


    2. Relevant equations
    V=IR


    3. The attempt at a solution
    It seems there should be no current at point I, since there is no voltage. The total resistance in the circuit is 77 ohms. 10.0 V - (10.0 V/77 ohms)*(77)=0. However, the answer is supposedly 0.13 A. What is wrong with my reasoning?

    Thanks in advance.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 10, 2011 #2

    kuruman

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    OK, so the the total resistance is 77 Ohms. If you put a 10V battery across 77 Ohms, how much current is provided by the battery? That's what the problem is asking.
     
  4. Sep 10, 2011 #3
    Oh, I see. But at that point, there would be no current, right? If there is no voltage, how can there be a current?
     
  5. Sep 10, 2011 #4

    vk6kro

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    No.

    The current leaves the positive terminal of the battery and travels around the circuit through all the resistors (splitting equally when equal resistors are in parallel) and then exactly the same current arrives back at the negative terminal of the battery.
     
  6. Sep 10, 2011 #5

    kuruman

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    Voltage is a potential difference between two points in a circuit. Between which two points are you saying there is no voltage? Also, if you agree that the battery provides a current, where do the electrons that leave the battery go?
     
  7. Sep 10, 2011 #6
    Shouldn't there be no potential difference between point I and the negative terminal of the battery since there is nothing to provide resistance or potential? In a real circuit there would be some resistance in the wire.

    If there is no potential difference between these two points, how can there be a current (0=IR)?
     
  8. Sep 10, 2011 #7

    vk6kro

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    You have drawn these as two different points for convenience, but they are electrically the same point.

    The current leaves the positive terminal of the battery and travels around the circuit through all the resistors (splitting equally when equal resistors are in parallel) and then exactly the same current arrives back at the negative terminal of the battery.
     
  9. Sep 10, 2011 #8
    Thank you, I think I understand now.
     
  10. Sep 11, 2011 #9
    V = IR so I = V/R, on substituting the values, we get 10/77, which 0.129A, which is the answer.
     
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