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IB physics, potentiometers!

  1. May 8, 2012 #1
    1. The problem statement, all variables and given/known data
    HI! IB physics exams coming up and I'm having real trouble with potentiometers. The two questions I'm posing are in the attachment along with a picture of the circuit they include.


    3. The attempt at a solution
    What I'm having trouble with is the 2nd question, What is the maximum reading of the ammeter? Assuming that I rearrange the voltmeter and ammeter so they're in the right position, I haven't got much of a clue of how to figure this one out. I think this ties in with my poor understanding of potentiometers. I've looked at the answers to find it's 12V / 15, when the resistance of the variable resistor (or potentiometer?) is at its maximum resistance. I would've thought that the voltage across the lamp would be greatest at this setting. But i don't understand how the current= 12V/15. Since the filament lamp and the variable resistor have different resistances won't the current be different in each?

    If any kind soul could shed some/any light on this problem or perhaps help me a little with potentiometers and why at max. resistance= max current through lamp I'd be eternally grateful. Thanks!
     

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  2. jcsd
  3. May 8, 2012 #2

    George Jones

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    Does the question say to do this?
     
  4. May 8, 2012 #3
    Hi! It doesn't explicitly say it but it asks you to draw a diagram of the circuit with the ammeter and voltmeter in correct positions in c). Here's how that diagram looks in the answers! + the answer to b).
     

    Attached Files:

  5. May 8, 2012 #4

    George Jones

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    Because it doesn't ask you to do this until c), you are supposed to analyze the original circuit in b).
     
  6. May 8, 2012 #5
    Hmm. Even so, I'm still not very clear at all on how this would work! So if I leave the circuit intact how would the equation for the current on the ammeter be derived? What does this have to do with the maximum resistance of the variable resistor? (15 ohms)
     
  7. May 8, 2012 #6

    cepheid

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    With the circuit in the present configuration, you have the battery in series with three resistors: 1. the portion of the variable resistor that is above the slider, 2. the voltmeter, and 3. the lamp filament.

    In order for the voltage across the filament alone to be 6.0 V (half the source voltage) what would have to be true about the ratio of the filament resitance to the rest of the resitance in the circuit?

    Hint: how would the voltage be divided amongst them?
     
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