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Kirchhoff's Current Law

  1. Jan 28, 2010 #1
    Problem i am stuck on:
    In 10 seconds, an observer - Willy Nilly - notices that 35 coulombs of charge leaves node "n" in this circuit, heading for node "x". (Vn is the voltage at node "n", etc.) In the same ten seconds, 22 coulombs of charge leaves node "n" heading for node "z". Determine the current, Iy.


    I have said that

    so Iy=13

  2. jcsd
  3. Jan 28, 2010 #2


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    The current into a junction must equal the current flowing out of it.

    Does your solution show this?
  4. Jan 28, 2010 #3
    the problem is i don't have solutions.
  5. Jan 29, 2010 #4
    Current entering node n = current leaving node n? sum the currents leaving this node and you'll get the current entering this node. Right now you have a net loss of current from a single node.

    think of it as a zero sum equation:

    0 = (Inx + Iny + Inz)
    0 = (-35 + x + -22)
  6. Jan 29, 2010 #5
  7. Jan 29, 2010 #6
    Now you need to get the signs right.
    If charge moves in the direction of the arrows then you have a positive current.
    Charge leaving node n for node x flows in the same direction as the arrow next
    to I_x, so I_x is positive. Same for I_y.

    Also current is in Ampere wich is coulomb/second so 35 coulombs in 10 seconds is only 3.5 A
  8. Jan 29, 2010 #7
    sign is very important, but in this case charge moving is comprised of electrons, and electrical current is defined to be opposite their flow. 35 [C] flowing from n to x over 10 seconds means that a current of 3.5 A flows FROM x TO n. Similarly, 57 [C] from y to n means 5.7 [A] FROM n TO y. Iy is therefore negative, Ix is positive, Iz is positive, but I personally find it easiest to think first in terms of electrons moving and then to realize that current flow is defined (originally by ben franklin) to be opposite that of electrons

    yes current = Ampere = [C/s], assuming you're not supposed to sum current over the 10 second time period
  9. Jan 29, 2010 #8
    ok thanks guys for answering the question i understand now.
    Last edited: Jan 29, 2010
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