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Kirchoffs law

  1. Aug 13, 2008 #1
    In the following circuit what does the current source mean?
    Am I supposed to use 3A to determine how much voltage is running across the 10 ohm resistor? If not, how do i determine how much voltage is running across that resistor? i need to know so i can determine the final equation of my set of simultaneous equations so i can determine what current is running through each loop.

    [​IMG]
     
  2. jcsd
  3. Aug 13, 2008 #2
    If that really is an ammeter (with the 3A by it), then it effectively bypasses the 10 ohm resistor since they are in parallel. You can essentially replace the 10 ohm resistor with a straight wire that has 3A going through it (and thus you know I_2).
     
  4. Aug 13, 2008 #3

    Defennder

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    I don't think that is an ammeter. It is more likely a constant current source of 3A and it just means that 3A of current is flowing through that part of the circuit. It appears you have to solve this by mesh analysis, judging from the clockwise loops. Use Kirchoff voltage law to write the linear equations. Also note that the potential drop across any current source is not zero. There's one voltage source that's not labeled, was that omitted?
     
  5. Aug 13, 2008 #4
    assume that the other voltage source is 40v.
    im not sure if the 3a has 3 different paths it can take, or 2. But assuming it takes 2
    first find how much of the 3a goes where
    (50/60)x3=2.5a
    (10/60)x=.5a
    so is this equation correct for the bottom right loop?
    0=(I1+I2)x15 + (I3+(I2-2.5))x50 + ((I2+.5)x10)
     
  6. Aug 13, 2008 #5
    The easiest thing to do is to use Thevenin's theorem to replace the 10 ohm resistor and 3 amp current source with a voltage source of 30 volts in series with a 10 ohm resistor. This combination will be in place of the 10 ohm resistor. Then you can easily write the 3 loop equations.
     
  7. Aug 13, 2008 #6
    so in this circuit the 3a in parallel is equivalent to a voltage source in series?
    how do i determine the output of the voltage source?
     
  8. Aug 13, 2008 #7
    The 10 ohm resistor in parallel with a 3 amp current source is equivalent to a 30 volt voltage source in series with a 10 ohm resistor. If you consider only the 10 ohm resistor in parallel with a 3 amp current source, convert it to a Thevenin equivalent by first determining the open circuit voltage of the combination (30 volts in this case) and then determining the short circuit current of the combination (3 amps in this case).

    You can always find a Thevenin equivalent to a current source in parallel with a resistance.

    See:

    http://en.wikipedia.org/wiki/Thevenin's_theorem

    and:

    http://en.wikipedia.org/wiki/Norton's_theorem
     
  9. Aug 16, 2008 #8
    So would these equations be correct?
    0=-40+(25)(I1)+15(I1-I2)
    0=-30+10(I2)+50(I2-I3)+15(I1-I2)
    0=-35+(I3-I2)(50)+20(I3)
     
  10. Aug 16, 2008 #9
    I think the 2nd equation should be:

    0=-30+10(I2)+50(I2-I3)+15(I2-I1)

    and the 3rd should be:

    0=+35+(I3-I2)(50)+20(I3)
     
  11. Aug 19, 2008 #10
    After knowing the three currents, how would i calculate the voltages at nodes a and b with respect to the ground node?
     
  12. Aug 19, 2008 #11
    The current in the 15 ohm resistor is I1-I2, so the voltage drop across that resistor, which is the voltage at node A, is 15*(I1-I2).

    Similarly, the current in the 50 ohm resistor is I2-I3, so the voltage at node B is the voltage across the 15 ohm resistor minus the voltage across the 50 ohm resistor (the assumed direction for I2 means that the voltage across the 50 ohm resistor must be subtracted), or 15*(I1-I2)-50*(I2-I3). If any currents actually have the opposite of the directions assumed, the polarities of the voltages across some of the resistors may be reversed.
     
  13. Aug 19, 2008 #12
    But doesn't the I3 current contribute to the potential difference at node A as well?
     
  14. Aug 19, 2008 #13
    I suppose one could say that all 3 currents are involved in one way or another in the voltages at all of the nodes. You can probably derive a calculation that involves all of them, but it's not necessary. For the method I gave you, only I1 and I2 are needed to calculate the voltage at node A.

    The current I3 doesn't pass through the 15 ohm resistor, so it doesn't enter in to the calculation I gave you. You only need multiply the current in a resistor by the resistance to get the voltage drop across a resistor. Currents that don't pass through the resistor don't have any effect.

    Another way to calculate the voltage at node A would be this:

    The current in the 25 ohm resistor is only I1, so the voltage at node A is equal to the -40 volt source, minus the voltage drop across the 25 ohm resistor. In other words, the voltage at node A is -40-I1*25.

    You could calculate the voltage at node A by several paths. They must all give the same answer, or you haven't solved the network correctly.
     
  15. Aug 20, 2008 #14
    Those two different methods for calculating the voltage at node A give different answers.
    I think you need to consider the 50ohm resistor when determining node A's voltage.
     
  16. Aug 20, 2008 #15
    I made a mistake in post #9; I got some signs backwards. You were supposed to notice this! :-(

    The equations should be:

    0=+40+(25)(I1)+15(I1-I2)
    0=+30+10(I2)+50(I2-I3)+15(I2-I1)
    0=-35+(I3-I2)(50)+20(I3)

    When I solve these, I get:

    I1 = -1.22281167109
    I2 = -.594164456233
    I3 = +.0755968169761

    Then the voltage at node A is -9.42970822286 and the voltage at node b is 24.0583554377.

    If you have the right currents, the methods for finding the node voltages will work as I explained them. The 50 ohm resistor need not be considered in finding the voltage at node A; it's just 15*(I1-I2).The method of calculating -40-I1*25 gives the same answer.
     
  17. Aug 20, 2008 #16
    those current values are very different from what i got.
    And I double checked mine with circuit simulation software
    I got:
    I1=1.6604
    I2=.7612
    I3=.0437
     
  18. Aug 20, 2008 #17
    I get numbers like that if the voltage source on the far left is 55 volts. In post #4, you said it was 40 volts. Check your simulation and see if you haven't used 55 volts for that source.

    Also, I think you have some signs wrong. Given the magnitude of the leftmost voltage source and its polarity, there's no way I1 is flowing in the assumed direction (clockwise), which it must do if I1 is not a negative number.
     
  19. Aug 20, 2008 #18
    sorry my mistake i gave you the wrong value for one of the voltages.
    sorry if i wasted any of your time
     
  20. Aug 20, 2008 #19
    You didn't waste my time. I hope I helped you. Have you been able to get a solution to the network that makes sense?
     
  21. Aug 20, 2008 #20
    I just want to clarify some things in regards to how you calculate the voltage at a particular node.
    Do you only consider the currents/voltages that are present on one particular branch coming off the node being calculated?
    If the voltage source to the left was flipped around would you still be able to use your second method of calculating the voltage at node A?

    Also, I have to find the thevinin equivalents of each of the sub-circuits A-G B-G and A-B then put them back together and show that the voltages at those nodes remain the same.

    For the left sub-circuit, A-G, immediately i can see that if i replace the 20ohm resistor with Rth, which would be 40 ohms the voltage at node A would be different from in the original circuit. So how do i go about doing this?
     
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