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Homework Help: Direct Current Circuit with Galvanometer.

  1. Jan 26, 2008 #1
    [SOLVED] Direct Current Circuit with Galvanometer.

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

    http://img299.imageshack.us/img299/6357/document002aa5.jpg [Broken]

    2. Relevant equations

    V = IR

    3. The attempt at a solution

    http://img263.imageshack.us/img263/6684/document003kl5.jpg [Broken]

    I did the following but the answer, to me, makes absolutely no sense at all. From my understanding, the galvanometer has no reading because there is no current fluctuation but I no real idea on how to do this.

    I thank you in advance.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 26, 2008 #2

    Tom Mattson

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    You're going about this all wrong. You need to to KVL around the left loop, and then again around the outside loop. You will obtain 2 equations in 2 unknowns: the unknown resistance X, and the current i in the left loop.
  4. Jan 26, 2008 #3
    Consider the problem conditions given. You are told that the Galvanometer reads zero. This menas that the current flow from the 2 volt battery must be zero. In order for this to happen there must be a voltage developed accross the "X" resistor to match the voltage supplied from the 2 volt battery. What must this be?

    Another way to look at this would be that since the current flow from the 2 volt battery must be zero, you could choose to ignore it completely, assume it does not exist. Now solve for "X". Once you get this answer, place the resistor value you came up with back into the circuit and attempt to solve this with two KVL loops. You will find one current will slove to the current from the 8 volt battery and the other will solve to the current from the 2 volt battery and will be zero.

    Hope this helps.
  5. Jan 26, 2008 #4
    What is KVL?

    Alright, so from what you are saying, the potential difference across the resistor X, must be equivalent to the electromotive force of the battery on the left hand side of the circuit which is 2.

    From this, I can gather that potential difference across X is equivalent to:
    [tex](\frac{X}{X+75})[/tex][tex](2)[/tex] = 2V

    and that for the right hand loop,
    [tex](\frac{X}{X+150})[/tex][tex](8)[/tex] = 2V

    You guys mentioned something about finding current, then X. Can I just use potential dividers to solve it? That way now I can sue either equation to get X.

    Please let me know if this makes any sense at all. :-X
  6. Jan 27, 2008 #5
    Your first equation will not solve for "X". Your second equation solves for "X=50 ohms" which is the correct answer for this problem. The suggestion was for you to use two possible solution paths to get this answer. One by simple inspection of the problem recognizing that the current flow given a zero from the 2 volt battery circuit meant you could ignore it's affect and just solve for "X" looking at only the 8 volt battery circuit. There are a number of ways to solve this simple circuit and using the current divider or voltage divider. A second method would be to use the KVL method of current loops solving for two unkonwns with two equations. This method would be more work but it introduces you to a method that may be used to solve much harder circuits.

    Hope this helps.
  7. Jan 27, 2008 #6
    Hmmm...thanks fir the insight. Again, what does KVL stand for?

    Also, I am not getting your method of using the current decider. I am not sure I have learnt that yet (or maybe I did and I forgot). If it's not too much hassle, can you please show me this method? I am rather interested now.

    Once again, thanks for your reply.
  8. Jan 29, 2008 #7

    Tom Mattson

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    Sorry, KVL=Kirchhoff's Voltage Law. I thought it was a common abbreviation, but maybe not.
  9. Feb 2, 2008 #8
    Ah. Haha..I did learn that after all. Including the current one but I suppose I just kind of forgot how to use and apply them in this question effectively.

    Anyway, thanks for the help. Question solved. (:
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