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Finding pd given r, R, and emf?

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data
    What potential difference is measured across
    a 23 ohm load resistor when it is connected
    across a battery of emf 7.99 V and internal
    resistance 0.555 ohms
    Answer in units of V


    2. Relevant equations
    V=IR
    V=emf - Ir


    3. The attempt at a solution
    V=IR
    emf-Ir=IR
    (emf/I)-r=R
    emf/I=R-r
    I=(emf/(R-r))
    I=(7.99/((23+.555)-.555)))
    I=0.34739

    V=IR
    V=(0.34739)(23.555)
    V=8.182802174
    INCORRECT
     
  2. jcsd
  3. Nov 1, 2011 #2

    Redbelly98

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    Welcome to Physics Forums.
    That should be R+r in that last equation, since you are adding +r to both sides of the equation before it.
    R is simply 23 ohms, not 23+.555
     
  4. Nov 1, 2011 #3
    so,
    I=7.99/22.445
    I=0.355598
    then,
    V=(0.355598)(23) ?
    or V=(0.355598)(23.555)

    I don't understand why I'm not supposed to use the total resistance as opposed to just the external resistance.
     
  5. Nov 1, 2011 #4
    I tried out both ways, and each answer was deemed incorrect.
    8.385139229
    and
    8.187569615
     
  6. Nov 1, 2011 #5
    I finally got! I didn't see the comment you made about the mistake in my simplifying! Thanks for the help
     
  7. Nov 1, 2011 #6

    Redbelly98

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    When you said that "V=emf - Ir", that V is the voltage across the 23 ohm resistor.

    And you also said "V=IR" -- the same V here. Since V is the voltage across the 23 ohm resistor, R must be referring to that resistor.

    You're welcome, glad it worked out. :smile:
     
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