kirchoff's current law

Discussion in 'Homework Help' started by electronicstech07, Nov 16, 2007.

  1. electronicstech07

    Thread Starter Member

    Nov 16, 2007
    I'm somewhat confused about this law, but am afraid to ask in class. I'm hoping someone can help me out some.

    I understand that current entering a point is positive, and current directed away from that point is negative. The algebraic sum of currents entering and leaving a point must equal to zero.

    Where I am somewhat confused is when I have two voltage sources, both are 10V for example. Each in series with a 1 ohm resistor, with a 1 ohm resistor in parallel with both voltage sources.

    To find IR voltage drops plus each branch I, I must know current total. Do I have it correct that voltage total is Vs1 + Vs2, Rt = R1 + R2 + R3, then It = 20V/3 ohm? Then Itotal from the assumed current flow is positive and I flowing from the negative terminal of the other voltage source is negative? So VR1 = I1R1, VR2 = I2R2, VR3 = I3R3, and I1 = VR1/R1, I2 = VR2/R2, I3 = VR3/R3.

    I am actually very confused. Hopefully someone can clear this up for me. Attached is the diagram of the circuit. Thanks in advance!
  2. recca02

    Senior Member

    Apr 2, 2007
    there is no fixed sign convention assigned for a direction of current u can take whatever sign u want just attach it properly for every direction.
    for the above problem do post a rough dia of the ckt.
    frm what i understand of the problem.
    u are having two loops,
    assume any direction of currents in both loops.
    u'll get two equation just by applying kvl on two loops.
    while forming eqns u can traverse the ckt in any desired direction, preferably in the direction of current.
    while traversing for any resistor take IR for current in direction of travel as +ve and -ve for current in reverse direction.remember the resistor common to two loops will have two currents(just for solving sake it is seen as such in fact only one current flows thru it) flowing and hence for that very resistance u'll have to take IR two times. or u can take the effective current thru that element as (i1-i2)(depends on direction u may get diff current eqn) and multiply with resistance.
    solve the eqn thu obtained for two unknown variables.

    hope this helps.
  3. Distort10n

    Active Member

    Dec 25, 2006
    It does not have to be. You can arbitrarily chose what direction and polarity from a given node. Although, you must remain consistent throughout the analysis of the circuit.
    This means that you cannot say that one node in a circuit current is leaving and is positive, then turn around and say another node in the circuit that current is leaving and is negative. This is Nodal Analysis, and it utilizes KCL.

    In Mesh Analysis you are utilizing KVL. You pick a reference node where your loop currents will be referenced to. You then pick a direction of current flow in each loop which can be clockwise or counter clockwise. It does not matter what direction you choose as long as you are consistent.


    Hmmm. Do you really mean this:


    A 1 ohm resistor that is actually in parallel with both voltage sources would only be an academic exercise, and would not be practical. Although, this whole circuit is not practical.

    Anyway, you have two loops if you are going to use mesh analysis. Call them I1 and I2. Pick the reference node to be GND, and the direction of current flow. Say it is clockwise. That means I1 flows through R1 and R3 clockwise, and I2 flows through R3 and R2 clockwise.
    If I1 flows clockwise through R3 as well as I2, then that means you have two opposing currents through R3: I1 - I2. Using KVL like Recca suggested should help you along the way.
  4. chesart1

    Senior Member

    Jan 23, 2006
    To convince yourself that the current could be assumed in any direction in a given loop, try solving the problem with an assumed curent direction in each loop. Then change the assumed current direction in one of the loops and solve the problem again. Both solutions should yield the same answer.

    If you assume that in one loop the current flows from the positive terminal of V1 to the negative terminal of V1 and in the other loop you assume current flows from negative terminal of V2 to the positive terminal of V2, then you must remember that in one loop the current is positive and in the other loop the current is negative.

    For simplicity, I always assume that current flows in the conventional direction [from positive to negative] and draw my current loops accordingly.
  5. electronicstech07

    Thread Starter Member

    Nov 16, 2007
    Thanks for the replies. Just wondering what you used to create the diagram? Thanks!