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Homework Help: Kirchoff's Laws

  1. Mar 9, 2008 #1
    [SOLVED] Kirchoff's Laws

    1. The problem statement, all variables and given/known data
    A link to the problem statement is provided here: http://img231.imageshack.us/img231/5305/kirchoffry1.png [Broken]
    (with one modification: the question should read find the current in the .75 ohm resistor)

    2. Relevant equations
    Ohm's Law: R=delta(V)/I
    loop rule: sum(changes in potentials) = 0 (for closed loop)
    junction rule: because of conservation of charge, the currents follow such that:
    I_1 + I_2 = I_3

    3. The attempt at a solution
    I am supposed to solve a set of simultaneous linear equations. The first one is simply the node equation
    Then by looking at the top loop, I have:
    2.25 - (I_2)*(3) - (I_3) * (4.2) = 0
    The next part is what I am somewhat confused about.
    I can either make one of two loops for the final equations: the bottom one with the middle wire included, or the top and bottom one as a whole.
    For just the bottom loop, I get:
    4.75 - (I_1)*(.75) + 1 - (I_1)*(1.2) + (I_2)*(3) +2.25 = 0
    I am not sure if this is right because I don't know if it is appropriate to add the voltages 1 and 4.75 when they have the resistor between them. Perhaps I am supposed to use their potential difference? But then in that case, the .75 ohms would be internal resistance and I don't have any emf. And, I think it is probably incorrect to assume that the current going through the 1.2 ohm resistor is the same as I_1.
    For the whole loop (the biggest one which does not include the middle wire), I get:
    4.75 - (I_1)*(.75) + 1 - (I_1)*(1.2) - (I_3)*(4.2) = 0
    When I solve this simultaneously with the first two equations, I get a different answer from when I solve the first three simultaneously.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Mar 9, 2008 #2
    EDIT: sorry for double posting, but maybe I should rephrase my question since nobody has offered any assistance.

    Why doesn't Ohm's Law work? The potential difference is (4.74-1) V or 3.75 V, applying Ohm's Law, delta(V)/R= I, so 3.75/.75 = 5 A but that is incorrect. Why?
  4. Mar 9, 2008 #3
    You have the sign of the 2.25 V voltage source wrong.
  5. Mar 9, 2008 #4
    Because your potential difference across the 0.75 ohm resistance is wrong. I can see no reason why it would have that value.
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