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Finding Currents Using Kirchhoff's Laws

  1. Apr 25, 2015 #1
    1. The problem statement, all variables and given/known data
    P26_34.jpg a. Determine the currents I1, I2, and I3 in the above figure. Assume the intermal resistance of each battery is r = 2.48Ω.

    2. Relevant equations
    Kirchhoff's Laws (Junction Rule and Loop Rule)

    3. The attempt at a solution
    Using Kirchhoff's Laws,
    I ended up with three equations:
    i1-i2-i3=0
    -12i1-2.48i2=4.48
    2.48i2-2.48i3=29

    Using matrices to find i1, i2 and i3 I come up with answers but they are wrong. I am wondering if there is a mistake that I made and am not seeing it?
     
  2. jcsd
  3. Apr 25, 2015 #2

    gneill

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    Staff: Mentor

    Your KCL equation looks okay, but your KVL equations don't look right. Can you give more detail about how you arrived at them? Perhaps write them out showing all contributing terms as you do your "KVL walk" around the loop.
     
  4. Apr 25, 2015 #3

    Doc Al

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    This looks good.

    How did you get these two equations?
     
  5. Apr 25, 2015 #4
    The Loop Rule Equations:

    (Starting from the top, to the left of the 12V battery and going counterclockwise) -2.48-8+12-2.48i2-10-12i+12=0

    (Starting from the junction of the right) 10+2.48i2-12-15+6-2.48i3-18=0
     
  6. Apr 25, 2015 #5

    Doc Al

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    For one thing, you forgot to multiply the internal resistance by the current to get the voltage drop.
     
  7. Apr 25, 2015 #6

    gneill

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    I'm seeing a mix of voltage and resistance terms, some resistances multiplied by currents and some not. So as written those equations cannot be correct. You are summing potential changes around the loop, so any resistor value must be multiplied by a current in order to realize its potential change (Ohm's Law). Perhaps you could group the resistance terms for given currents in parentheses?
     
  8. Apr 25, 2015 #7
     
  9. Apr 25, 2015 #8

    gneill

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    Aren't 2.48 and 8 and 10 all resistances? What currents multiply them? I see only two voltage sources in the loop (both 12 V).

    Again, I don't see currents multiplying each resistance.
     
  10. Apr 25, 2015 #9
     
  11. Apr 25, 2015 #10

    gneill

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    That looks better! What do you find for your currents?
     
  12. Apr 25, 2015 #11
    Got it, thanks! I just needed to be consistent with labeling which current goes with which resistor.
     
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