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Electricity: Basic circuit question with a switch

  1. Jul 10, 2011 #1

    Femme_physics

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    This is a question that was on my test

    1. The problem statement, all variables and given/known data



    http://img269.imageshack.us/img269/5125/circuitthingy.jpg [Broken]


    In the circuit in the drawing switch S is open. Both resistors are idential, both having resistance R each. Given the source is E, and R = r (the resistance of the voltage source). The ammeter resistance is negligible.

    Closing the switch

    A) Will the ammeter's reading go up, down, or won't change? Base your answer on a physical consideration or prove it by calculation.

    ((There were several questions but I'll just start with A for now))

    3. The attempt at a solution


    The ammeter's reading decreases since there is more resistance.

    Proof

    http://img543.imageshack.us/img543/1949/ianswer.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 10, 2011 #2

    I like Serena

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    Hi again Fp! :smile:

    Do you have a physical consideration for your answer?

    As for your calculation, I'm afraid you made a simple conceptual mistake.
    But before I tell you what it is, I'd like you to think about the physical consideration, and I kinda hope it will *click* then! :smile:
     
  4. Jul 10, 2011 #3

    Femme_physics

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    Actually, I just copied it from my test (it's the question I skipped and ended up deleting)
    , but I see what you mean now
     
    Last edited: Jul 10, 2011
  5. Jul 10, 2011 #4

    I like Serena

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    You do?
    Well, don't keep me in suspense... what is it? :confused:
     
  6. Jul 10, 2011 #5

    Femme_physics

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    I can't believe I would've been wrong at THIS question if I handed it on the test. Why am I so careless?!? Here we go


    http://img15.imageshack.us/img15/2825/fixxooros.jpg [Broken]

    So clearly I was wrong.

    The current INCREASES when the switch is closed.

    As for the physical explaination, let me think about it.
     
    Last edited by a moderator: May 5, 2017
  7. Jul 10, 2011 #6

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    Aaaaaaaaaaaaaaaah! Now I get it! :smile:

    As for being careless, that's what we train and exercise for.
    The reason I'm asking for the physical consideration, is because it will give you a visual (or rather conceptual) cue about what you're doing.

    The point being that when your calculated results don't match your visual cues, you'll try and find out what you did wrong and fix it.
     
  8. Jul 10, 2011 #7

    Femme_physics

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    True, I learned that in mechanics, I think I got it in me after passing both the internal and external tests successfully. Sadly, I did the electronics test before them...before I became a super-test-taking-ninja!

    I agree. Then, for the physical explanation, I'd say the current has more flowing options. Does that work? :smile:
     
  9. Jul 10, 2011 #8

    Femme_physics

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    Question 2: Does the voltage go up, down, or doesn't change? Base the answer on a physical consideration or prove it ia calculation


    Physical Explanation:


    It doesn't change. The voltage remains constant regardless of the alteration in the construct of the circuit since V is the source, while R and I are determined by the construct of the circuit.
     
  10. Jul 10, 2011 #9

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    Yieaah!
    Can you catch a speeding bullet's kinetic energy now, when such a question is fired at you?


    Yes, that works.
    I can't believe how far you've come!!! :!!)


    Uh oh, I take that back.
    The idealized voltage source indeed keeps a constant voltage, but......
     
  11. Jul 10, 2011 #10
    for physical explanations for increase in current when a parallel resistance is added, think of the first wire getting fatter thus allowing more current to pass and since the voltage is same, desired result follows.
     
  12. Jul 10, 2011 #11

    Femme_physics

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    What? But what?

    Yep! The water analogy is also great :smile:
     
  13. Jul 10, 2011 #12

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    If you look at your problem, you'll see that an internal resistance was specified.
    This means that the terminal voltage is not constant.
    (Hehe, I managed to insert "terminal voltage" into a sentence! :smile:)

    Now what would the consequences be that the voltage source is not perfect? :confused:
     
    Last edited: Jul 10, 2011
  14. Jul 10, 2011 #13

    Femme_physics

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    I thought that the terminal voltage is constant. It's just

    Vf = Vi - I1r

    I don't see anything about rates of change or "time" in this formula

    Sorry for the mechanics notation. Force of habit! :biggrin:
     
    Last edited: Jul 10, 2011
  15. Jul 10, 2011 #14

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    Mechanics and electronics are interwoven (didn't you study mechatronics?)
    So it's sort of fine to use the same symbols. :wink:
    Although..... in mechanics you have a velocity v (lower case letter), whereas in electronics you have a voltage V (capital letter).

    Nope. It depends on the current I1.
    The more current flows the lower the terminal voltage.

    Just like in real life.....
    If you make a battery do too much, at some point it won't be able to provide voltage anymore to make a current flow. :smile:
     
  16. Jul 10, 2011 #15

    Femme_physics

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    Am "studying", not studied! But yes, I do realize they're intrervoven, but having the mathmatical introspection as to how, well-- that I didn't fully grasp yet.

    Fair enough, but I don't see "time" in ohm's law. Ohm's law doesn't care about time!
     
  17. Jul 10, 2011 #16

    Femme_physics

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    Come to think about it, there was a word I wasn't sure how to translate



    in hebrew it's

    מתח ההדקים

    I just figured it means voltage according to my dictionaries and translators

    Well, one of the words

    מתח

    Does mean voltage. The other words

    הדקים

    I'm not so sure about, but I just figured it's a fancy way of writing voltage. Since we're out of school semester, I can't get access to ask my teacher. I tried emailing him but he hasn't emailed me back.

    Maybe Quabache can help us here.. :smile:
     
  18. Jul 10, 2011 #17

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    All in good time. You'll get there. And it will be amazing! :smile:



    No, "time" is not in Ohm's law.
    However, voltage and current are.

    Remember that a voltage source is modelled as an ideal voltage source in series with an internal resistor.
    When the current increases, so does the voltage drop over the internal resistor.
    In effect, the voltage source delivers less voltage on its "terminals".
    (There! I got to insert "terminals" again in a sentence! :smile:)
     
  19. Jul 10, 2011 #18

    Femme_physics

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    If I calculate V across all the resistors will it be sufficient proof?


    PS made a reply earlier you might have missed
     
  20. Jul 10, 2011 #19

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    According to google translate, it's "power terminals".
    That makes sense, since it would indicate the "power" available at the "terminals" of the battery.
    A "terminal" would be a point on the outside of a device where you can connect something.


    Right.


    According to google translate it mean "thin", but I have no clue to what that might mean.


    Sure. Let's see if it will match with the physical consideration that the voltage should have dropped. :)
     
  21. Jul 10, 2011 #20

    Femme_physics

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    Hmm....then I think I understand. They're really asking for the voltage of the internal resistance, right?


    Power terminals?


    But the word

    מתח

    means voltage.

    How did we get to "power"?!?
     
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