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Kirchhoff's Current law

  1. May 23, 2009 #1
    Please see question at http://www.flickr.com/photos/38702264@N06/3556612295/sizes/o/ [Broken]

    Which one of these solutions is correct?
    1) V / R = I
    6 / 1.5 = 4

    i1 = 4a

    3 / 2 = 1.5

    i2 = 1.5a

    i1 + i2 = i3

    i3 = 5.5 amps

    2)V / R = I
    6 / 11.5 = 0.5217

    i1 = 0.5217a

    3 / 12 = 0.25

    i2 = 0.25a

    i1 + i2 = i3

    i3 = 0.7717amps


    Thanks for the help
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 23, 2009 #2
    definitely not the first, as the voltage across the 1.5 ohm resistor isn't 6 volts.

    and the second one doesn't take into account that both currents flow through R3.

    i could be wrong, but i think you have to consider it like this:

    kirchoff said the vectorial sum of the voltages in a loop is 0.

    but the voltage across the ten ohm resistor is the sum of both currents. ie, 10xI3. this can be written as 10(I1 + I2)

    so for your first loop, I1,

    6 - 1.5(I1) - 10(I1 + I2) = 0

    do that for the second loop I2, and you'll get 2 equations, 2 unknowns.
     
  4. May 24, 2009 #3
    Can anyone confirm this??

    Thanks
     
  5. May 24, 2009 #4
    Easiest way to confirm a solution is check that the answers work. Solve two equations for two unknowns. Using those answers, do the voltages around very loop add up to zero as required, and do currents at every node add up to zero as required? If so, you have checked your math.
     
  6. May 24, 2009 #5
    I get an interesting result: I2 comes out negative, which is okay -- that's the way the math informs us that it flows in the direction opposite to the assumed arrow. So conventional current is flowing into the positive terminal of the 3 V voltage source.
     
  7. May 25, 2009 #6
    Thats the problem i dont know how to check the answers by using two equations,

    can anyone just tell me if either solution is correct or if they are both wrong?

    Thanks
     
  8. May 25, 2009 #7
    Both wrong.
    You started out using Ohm's Law right away, which isn't allowed here.
    If you know a resistor's current, Ohm's Law will tell you its voltage.
    If you know a resistor's voltage, Ohm's Law will tell you its current.
    But we have no way in the beginning to know either one, so we can't use Ohm's Law to find the other.

    There is no other way except the simultaneous solution of two equations.

    Using a positive term wherever a voltage source produces a "voltage rise",
    and a negative term wherever a resistor produces a "voltage drop" ...

    Sum of voltages around the first loop = 0.
    6 - 1.5 I1 - 10 (I1+I2) = 0
    Sum of voltages around the second loop = 0
    3 - 2 I2 - 10 (I1+I2) = 0

    Observation - I now have two equations and two unknowns. The number of equations is greater than or equal to the number of unknowns. That tells me that a solution is possible.

    Then I personally use Cramer's Rule to solve them, but it's not the only way.
    The other method is to substitute one equation into the other one.

    Getting rid of the parentheses (the distributive property of algebra):
    6 - 1.5 I1 -10 I1 -10 I2 = 0
    3 - 2 I2 - 10 I1 - 10 I2 = 0

    Putting them into the general form "a I1 + b I2 = c":
    11.5 I1 + 10 I2 = 6
    10 I1 + 12 I2 = 3
    Now it's in the form to use Cramer's Rule immediately.

    Do you know how solve the rest of it?
     
    Last edited: May 25, 2009
  9. May 25, 2009 #8
    By the way, about the phrase you typed as the header of the forum thread, I'm using Kirchhoff's voltage law here to write my two equations. I'm not using Kirchhoff's current law to write my two equations.
     
  10. May 25, 2009 #9
    you are using KCL when you define I3 = I1 + I2. but thats as far as it can go to my knowledge...

    the rest is KVL
     
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