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Kirchoff's Rule Problem

  1. Jun 19, 2011 #1
    1. The problem statement, all variables and given/known data
    http://img585.imageshack.us/img585/5079/scan0007v.jpg [Broken]


    2. Relevant equations
    I_1=I_2+I_3

    3. The attempt at a solution

    I got the following equations by splitting the problem up into the left loop and right loop

    Right Loop starting at the negative terminal of the middle battery and going CCW-> 6-I_2(10)-8-I_3(3)

    Left loop starting at the positive terminal of the left battery going CW -> 12-5I_1+6-10I_2

    I then got the following equations (including the first equation)

    10I_2 = -3I_3 -2
    5I_1 = 18 - 10I_2

    I then solved them but my answers are not the same as in the problem.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jun 19, 2011 #2

    ojs

    User Avatar

    You have two signs wrong in the equations, one in each loop.
    What happens to the voltage if the current is going from plus to minus but the loop is going in the opposite direction?
    Also, what happens to the voltage over a resistor (the 3 ohm resistor) when the current is going CW but the loop is going CCW?
     
  4. Jun 19, 2011 #3
    I knew there was something wrong with my signs.

    1. If you are talking about the voltages of the 2 batteries, then they should be added from what I understand (from positive to negative)
    2. That I'm not sure about, from what i understood is that the only difference would be the sign. If the loop direction and current direction are opposite then they should be added if they are the same then they should be subtracted. Correct?

    Let me recalculate my stuff. I think the problem was that I was adding 6 to 12 and subtracting the 3 ohm resistor

    EDIT:
    Still no luck:
    10I_2 = 3I_3 - 2
    5I_1 = 6 - 10I_2
     
    Last edited: Jun 19, 2011
  5. Jun 19, 2011 #4

    gneill

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    Staff: Mentor

    Let's take a walking tour through the circuit. Starting with the left hand loop at it's bottom left hand corner (where the "I1" label is located). Walking clockwise we first encounter a 12V voltage supply, and have to "walk up" the increasing voltage (- to +). So we mark that down:

    +12V

    Next we encounter a 5Ω resistor. Since we're walking in the same direction as the current that's been indicated (I1), we'll be "walking downhill" with the current, and there will be a voltage drop of I1*5Ω. So mark that down:

    +12V - I1*5Ω

    Next we turn the corner into the middle branch (where the current is now I2 in the direction of our walk), and encounter first another battery. This time we first meet the + terminal and will therefore be "walking downhill" as we traverse it. So a voltage drop of 6V is indicated:

    +12V - I1*5Ω - 6V

    Next in line is the 10Ω resistor. We are "walking" in the same direction as the indicated current I2, so there will be a voltage drop traversing the resistance of I2*10Ω. Mark it down:

    +12V - I1*5Ω - 6V - I2*10Ω

    Since we're now back where we started from, KVL says we should have a net "elevation" change of zero, so the equation for the first loop is:

    +12V - I1*5Ω - 6V - I2*10Ω = 0

    Combining the voltage terms:

    6V - I1*5Ω - I2*10Ω = 0

    Remember to keep the chosen current directions in mind; you're either walking with the current or against it. If you're walking with the current, traversing a current means a voltage drop. If you're walking against the current, then you are climbing up the slope that current's running down and there is a voltage rise across the resistance.

    Can you produce a similar "walking tour" for the second loop?
     
  6. Jun 19, 2011 #5
    Yeah I got that for the left loop!

    For the right loop I did the following. Go CCW and starting at the negative pole of the 6V battery. So we have 6 - 10I_2 - 8 + 3I_3

    But now I am kind of confused since there are actually 2 currents in this loop. So i'm not quite sure which current to take at the 3 Ohm resistor. Heck I don't even know if i am supposed to add the 8 V to the 6 or subtract it since the 8V battery's current is flowing in the other direction.
     
  7. Jun 19, 2011 #6

    gneill

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    Staff: Mentor

    If you begin your tour at the negative pole of the 6V battery, you haven't yet crossed that battery, so there's no 6V to begin. You start with the drop on the 10Ω resistor caused by I2 flowing through it (the indicated direction for I2 is in the direction that you're walking).

    When you reach the junction at the bottom and proceed into the rightmost branch, the current flowing in that branch is I3, and you'll be walking against the flow -- upstream as it were. The first thing you encounter is the 8V supply, and it's a climb up 8V (voltage sources don't care about what direction the current is flowing -- they always present a voltage change of their indicated value, - to + or + to - depending upon which way you walk over them). Then you encounter the 3Ω resistor, and you're waling against the flow so it's a voltage rise (you're walking against the flow, so "uphill" for that resistance).

    Finally you encounter the 6V battery, and it's a downhill grade, + to -. You should be able to write the equation from that description.
     
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