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Really dumb Ohm's Law question

  1. Oct 1, 2004 #1
    A 12.0 V battery has an internal resistance r. The measured voltage (emf of the battery) is 12.0 V. When connected to a resistor R the terminal voltage is 11.7 V and the current is 0.10 Amperes. What is the value of the external resistor R in ohm?

    Ok I thought the problem was simple but for some reason I keep getting it wrong. (E denotes electromotive force). I set up conservation of energy like this:

    E - V(resistor) = 11.7

    E = 12 in this case and V = i*R = 0.1*R
    so..

    12 - 0.1*R = 11.7

    R equals 3 from my equation but it's wrong. Dead wrong. Any ideas? It's the language that seems most unclear to me.
     
  2. jcsd
  3. Oct 1, 2004 #2
    Try this:
    [tex]I_{T}=\frac{V_{applied}}{R_{T}}[/tex]
    and remember what the total resistance is in this case.
     
  4. Oct 1, 2004 #3
    Maybe my problem is more how the question is worded.. here is what I interpreted from what you said. A simple yea or nay will either send me back to the books or confirm I understood something:

    It = Vapplied/Rt

    since the resistors (r and R) are in parallel, Rt = R + r. Current is constant in the system so It = 0.1a but I'm a little confused as to what Vapplied is. Is that the emf? So I'm left with

    0.1 = 12V / (r + R) but now I have two unknowns. ugh. Undoubtedly little r was given to me in the question but I'm having trouble wading through the semantics. I see that I haven't used that 11.7 yet.

    Edit wait: emf = 0.1 * r right? So r must be 1.2..

    0.1 = 12V / (1.2 + R)

    R = 118.8? Edit again: Got it right. Thanks I can be so nearsighted sometimes. That was discussed in the lecture I came from not two days ago!
     
    Last edited by a moderator: Oct 2, 2004
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