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Homework Help: Calculating the epislon ε and current intensity? help please

  1. May 14, 2010 #1
    1. The problem statement, all variables and given/known data

    a battery connected to a light bulb that has electrical resistance of (6 Ω) and the current intensity that gets through the bulb counts (12 A), if the internal resistance of the battery was (2 Ω):-
    1- what is the Driving force of the battery (ε)?
    2- if the light bulb was exchanged with another bulb that has resistance of (4 Ω) what would the current intensity getting through equal?


    2. Relevant equations
    ε= VR + Vr
    ε=I(R+r)
    ε=W/q
    r=ε-VR / I
    VR = I×R
    Vr = I×r
    (there might be other forms of the same equations)


    I actually don't know weather the electrical resistance given in the problem was r, R, VR or Vr
     
  2. jcsd
  3. May 14, 2010 #2

    tiny-tim

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    Hi phy_freak! :smile:
    In this case, it's the inital letter that tells you …

    r or R is for Resistance,

    v or V is for Voltage (= electric potential = emf = driving force) :wink:

    (and i suppose I is for Intensity of current, and W is for energy because it stands for Work, which is a form of energy)
     
  4. May 14, 2010 #3
    so we use the equation: ε= VR + Vr to solve the first question? then it means that VR=6 (the given bulb's resistance) and Vr=2 (battery given resistance)
    so ε=6+2 =8V? is that correct?
     
  5. May 14, 2010 #4

    tiny-tim

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    No, you're misunderstanding what this equation is.

    It's ε= VR + Vr,

    which in words means the emf (driving force) equals the voltage drop (electric potential difference) across the resistance R plus the voltage drop across the resistance r.

    R and r in this equation aren't factors, to be multiplied … they're only labels, to tell you which V you're talking about.

    (so if the resistances were called R1 and R2, the voltage drops would be called V1 and V2, with the labels "1" and "2")

    The equation you need is ε = I(R+r) …

    (because ε= VR + Vr, and VR = IR and Vr = Ir, so altogether ε = IR + Ir = I(R+r):wink:)
     
  6. May 14, 2010 #5
    i used ε = I(R+r), assuming that R=6 (the bulb's resistance) and r=2 (battery resistance), ε equaled 96 V.

    and i used I=ε/R+r (with the ε equaling 96) for the second question, (I) equaled 16A

    are those correct?
     
  7. May 14, 2010 #6

    tiny-tim

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    Yup! :biggrin:
     
  8. May 14, 2010 #7
    thank you very much you were helpful :)
     
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