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Kirchhhoffs rule

  1. Feb 16, 2007 #1


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

    2. The attempt at a solution

    I followed Kirchoff's Rules for part A of this question by outlining three formulas. Each starts at point A on the diagram
    -(6)I1 + 18 - (6)I1 + 8 - (22)I3
    I3 = [-26 +(12)I1]/-22 ******formula 1

    -(6)I1 +18 -(6)I1 - (14)I2 = 0
    I2 = [-30 + (12)I1]/-14 *****formula 2

    I1 = I2 + I3 *******formula 13

    By combining these three formulas I solved for I1 and used that to find I2 by using formula #2

    My final answer is 0.96 amp

    Because this is positive it means I chose the right direction for my current with my original formulas. Therefore the current is flowing counter clockwise in the circuit... Is that right?

    does this look right. I have never done a question with three batteries of which one is facing the opposite way to the others...

    for part b do I just add each emf to find the voltage across ab


    12V - 8V + 18v = 22V

    I minus because it faces the opposite direction as the 12 V emf

    I don't think this looks right at all but I am unsure where to go.
    Last edited: Feb 16, 2007
  2. jcsd
  3. Feb 17, 2007 #2
    Whatever sign convention you chose for your original equations is the direction your current will flow in, which I am not sure if that is what you said because it seems like you flipped the direction for some reason. The electrons will flow in the opposite direction, but don't get that confused with the flow of current.

    I am not really sure what you did for part b), it looks like you just added up the voltages from the batteries. This is almost right (maybe it even is, I didn't do the calculations), but I don't know what your reasoning behind it is, and if it is right it would probably be for the wrong reason. Regardless, for part b) you ought to use the current you found in part a). Notice that the potential between a and b has all three of the parts in parallel. Find one part, you found them all.
  4. Feb 17, 2007 #3


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    does this make more sense

    You're right my formulas went around the loop in a clockwise direction not counter clockwise. Either way I should come out with the same answer right??? Just one would be negative????

    Any how for part B does this make more sense?

    V = IR
    I = I1 = 1.38 A
    R = Req = the total resistance
    V = ?

    Resistor 22 and resistor 14 are in parallel therefore

    R (22&14) = [(22)(14)]/[(22)+(14)]
    R (22&14) = 8.55555

    Resistor 6 and 6 are in series therefore

    R (6&6) = 6+6
    R (6&6) = 12

    Resistor R (6&6) and R (22&14) are in parallel therefore

    Req = [(8.5555)(12)]/[(8.5555)+(12)]
    Req = 4.995 ohms

    V = I1 * Req
    V = (1.38 A) (4.995 ohms)
    V = 6.9 V

    Please tell me this looks right
  5. Feb 18, 2007 #4
    Yes, the magnitude of your current should be right, and you would just change the sign (i.e. direction).

    I suppose that circuit reduction could have worked, but I don't agree that resistor 6 and 6 are in series. Yes it looks like they are since the same current goes through them, but remember that for series the two elements must share a common node, or you could also look at it like they have to share the same wire with nothing between them. The battery is between them so they don't share the same node.

    Also the resistors you put in parallel aren't actually in parallel. The way I always look at parallel, which works most of the time (sometimes if a ground is defined in a funny place this won't work), is that I see if their ends connect together.

    Try this out. Calculate the potential across 22 ohm resistor, and add it to the potential of the 8V battery - store this number. Now calculate the potential across the 14 ohm resistor, and add it to the potential of the 12V battery - store this number too. Finally, figure out the potential across the two six ohm resistors and add those potentials to the potential of the 18V battery. You should notice something about all the numbers (be careful about getting your signs right!).

    *By the way, if you want to check your answers you can try to learn PSpice. It has a free student (lite) edition from Orcad (I think) that you can download. There are a lot of strange niches about it, and I personally hate it, but the option is there.
    Last edited: Feb 18, 2007
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