1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    (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
    I_1+I_2=I_3.
    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.
     
  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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Kirchoff's Laws
  1. Kirchoff's Laws (Replies: 25)

  2. Kirchoffs law (Replies: 1)

  3. Kirchoff's Law (Replies: 5)

  4. Kirchoff's Laws (Replies: 26)

  5. Kirchoffs Laws (Replies: 6)

Loading...