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Circuit with two cells

  1. Nov 4, 2013 #1

    BMW

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    1. The problem statement, all variables and given/known data
    What is the current in this circuit:
    http://img802.imageshack.us/img802/4646/8ekh.png [Broken]

    2. Relevant equations
    All potential differences in a closed portion of a circuit must add to 0.
    Terminal Voltage = EMF - I x internal resistance.

    3. The attempt at a solution
    I do not know how to do this, the answer says the current is 0A. I have no idea why.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 4, 2013 #2

    gneill

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    Staff: Mentor

    You forgot the Relevant Equations portion of the template. What circuit laws have you learned about?
     
  4. Nov 4, 2013 #3

    BMW

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    Sorry, I have added them in.
     
  5. Nov 4, 2013 #4

    gneill

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    Staff: Mentor

    Okay, so potential changes come in two flavors: Rises and drops. Typically sources are responsible for potential rises and resistors for potential drops.

    If you "walk" around an isolated loop in a given direction and sum up just the changes in potential due to the sources you'll arrive at the total amount of EMF that can drive current in that loop. What do you find when you do this "walk" around your loop?
     
  6. Nov 4, 2013 #5

    BMW

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    I'm getting confused with EMF, terminal voltage and internal resistance etc... isn't the 1.5V the EMF, which means that the actual potential rise (terminal voltage) will be less that 1.5V?

    If I were to add up the EMFs, I would be assuming no current is flowing.
     
  7. Nov 4, 2013 #6

    gneill

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    Staff: Mentor

    The EMF, or "Electromotive force", refers to the potential across the thing inside the battery that produces the potential difference, and ignored any internal losses due to internal resistance. It's the potential that you would measure across the cell if you were to employ a perfect voltmeter that draws no current (so there would be no drop across any internal resistances).

    In summing up the EMFs that are driving current in the circuit, ignore all resistances no matter where they are located and concentrate on the raw EMF values. You want to find out what the total EMF available to drive current is, not the effects of the current (not yet, anyways!).
     
  8. Nov 4, 2013 #7

    BMW

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    Aha! So the net EMF is what creates a current in a closed circuit or part of a circuit? And the net EMF in this circuit is zero so the current will also be zero?

    :D
     
  9. Nov 4, 2013 #8

    gneill

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    Staff: Mentor

    Exactly! :smile:
     
  10. Nov 4, 2013 #9

    BMW

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    Thank you for your time :biggrin:
     
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