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Change in ammeter reading due to change in resistance

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  1. Aug 19, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-8-19_22-25-11.png

    2. Relevant equations
    1/R= 1/R1 + 1/R2 +...
    I1= R2/R1 x I

    3. The attempt at a solution
    I chose B because I thought that since total resistance in the branch of variable resistor has increased, the branch with ammeter will get a greater share of current. Since I'm not too sure about it because the overall increase in resistance does cause a drop in main current, I put in some random values of R and V , and I still get an increase in ammeter reading. So why is the answer C?
     
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  3. Aug 19, 2015 #2

    phinds

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    Your ammeter, with its internal resistance, are shown in parallel to the battery, so

    1) the rest of the circuit is irrelevant to the ammeter reading
    2) the ammeter will likely burn up rather quickly, or if it is strong enough not to, your battery will melt down due to the short circuit.
     
  4. Aug 19, 2015 #3
    Huh? Isn't that symbol a resistor? It's not stated in the question that it's the internal resistance of ammeter...
     
  5. Aug 19, 2015 #4

    phinds

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    Oh. I just assume that's what it was. Perhaps a bad assumption. In that case, ignore my point 2. Point 1 still holds.
     
  6. Aug 19, 2015 #5
    Why? Isn't ammeter connected to circuit too?
     
  7. Aug 19, 2015 #6

    phinds

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    Sure. And you could connect it to lots of other resistors too, but all would be irrelevant. The resistor/ammeter pair are connected directly across the battery and no amount of other circuitry is going to change that.
     
  8. Aug 19, 2015 #7
    :confused: I still don't understand. If I increase the resistance in variable resistor, this will cause an overall drop in current.
    I don't know... I just put in some random values...into the components and I do get an increase in ammeter reading.

    So...I can't see why it's irrelevant.:smile:
     
  9. Aug 19, 2015 #8
    Hint : Use Kirchoff's loop law in one inner , and one outer loop , both of which include the battery .
     
  10. Aug 19, 2015 #9
    Ah I see now. Since potential drop is always the same across resistor in outer loop, and its resistance is constant, I also must be constant.
    Thanks for this 'light-bulb' moment.:smile:
     
  11. Aug 19, 2015 #10

    phinds

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    Then your drawing does not represent what you think it represents.

    EDIT: OOPS. I was responding to a different post (#7)
     
  12. Aug 19, 2015 #11
    Yes. I see where I'm wrong now. Got the ratios of currents wrong.:-p
     
  13. Aug 19, 2015 #12

    phinds

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    Ratios of currents in this case have nothing to do with the ammeter reading.
     
  14. Aug 19, 2015 #13
    Uhhm...ratios of resistors?
     
  15. Aug 19, 2015 #14

    phinds

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    Has nothing to do with the ammeter reading in this circuit.
     
  16. Aug 19, 2015 #15
    is it because of what I've said in post #9?
    Then, my previous approach cannot be used at all is it?
     
  17. Aug 19, 2015 #16

    phinds

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    yes
    Not sure what your previous approach was but if it gives anything but a constant reading on the ammeter it's wrong.
     
  18. Aug 19, 2015 #17
    my previous approach was calculating new total resistance of circuit, hence finding new main current and from the ratios of resistors in two branches calcuate the current in each small branch.
     
  19. Aug 19, 2015 #18
    my previous approach was calculating new total resistance of circuit, hence finding new main current and from the ratios of resistors in two branches calcuate the current in each small branch.
     
  20. Aug 19, 2015 #19

    phinds

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    New main current has nothing to do with what flows through the ammeter. At this point I'm not sure whether you're just trolling to see how long you can string me along trying to explain this trivial fact. Draw the circuit without the middle vertical part there at all. That's what the ammeter sees. Nothing else is relevant to the ammeter.
     
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