Kirchoffs Law(edited)

Discussion in 'Homework Help' started by jiggssmalls, Oct 30, 2013.

  1. jiggssmalls

    Thread Starter New Member

    Oct 21, 2013
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    Hello,can someone help me solve this please?I am supposed to find the currents flowing through all branches using the Kirchhoff's law.Thank you
     
  2. WBahn

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    Mar 31, 2012
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    First you need to post YOUR best effort to solve YOUR homework problem. That will give us a starting point from which we can see what you are doing right and what you are doing wrong and try to make suggestions to help you find your way.
     
  3. jiggssmalls

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    Oct 21, 2013
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    ohh okay..i will then..thank you very much..i will be on it and post it tomorrow morning as its really late here..:):)..thank you
     
  4. WBahn

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    Mar 31, 2012
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    Ah, good. I see that you have (hopefully) given up on trying to get someone to do your work for you via e-mail. Third thread's the charm, I guess. Perhaps now we can get down to business and get you moving in the right direction. Look forward to seeing your work.
     
  5. jiggssmalls

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    Oct 21, 2013
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    hello,I am back with some work..just want to check if u are around,so i can post my workout.:):)
     
  6. WBahn

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    Please, post your work.
     
  7. jiggssmalls

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    Oct 21, 2013
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    well.this is how i have started to my best understanding.it is not complete as i am not sure wht to do next..now i got 2 loops which i think i have to solve it simultenously or something like that.which i intend mutiplying eq 1 by 2 and eq 2 by 1 giving me eq 3 and 4 right?..after which i have to subtract eq 4 from eq 3...but thie thing is I'm even not sure i have started it correctly.:(:(
     
    Last edited: Nov 3, 2013
  8. WBahn

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    In your first equation, you have the current flowing in R3 as (I1-I2). Does that make sense? Aren't R1, V1, and R3 in series? What does that tell you about the current flowing in them?
     
  9. jiggssmalls

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    Oct 21, 2013
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    well,i thought when the current gets to the junction,it spread to both left and right,that's why i made is I1-I2..so does it have to be postive since they are in series?
     
  10. WBahn

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    What do you mean, "does it have to be positive"?

    It is the SAME. So if I1 is flowing in R1 (left to right), then I1 is flowing in V1 (bottom to top) and I1 is flowing in R3 (right to left).

    If the current in R3 (right to left) is I1-I2, and the current (downward) through R2 is I1+I2, then that would require that the current flowing to the left from that bottom junction be 2*I2. Would that seem reasonable?

    Remember, KCL requires that the net current flowing across ANY bounding surface be zero. That means that if you draw a circuit around any portion of the circuit, the currents crossing that circle have to sum to zero.
     
  11. jiggssmalls

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    Oct 21, 2013
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    ohh okay.I understand now...so after i get that..i have to the solve both of them simulteneously?just like i said before?or was my idea wrong?
     
  12. WBahn

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    You've got more than just two currents to worry about.

    Update and post your corrected equations and let's take it from there.

    Once those equations are correct, then it becomes a matter of solving them simultaneously. All of the physics/electronics is in setting up the equations. Everything else is just math. So break it into those two parts -- get the physics taken care of, then deal with the math.
     
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  13. jiggssmalls

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    Oct 21, 2013
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    well said sir...this is the corrected equation of loop 1:)
     
  14. WBahn

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    You have the exact same loop equation, but you have somehow changed the second line so that it no longer agrees with the first.

    And in neither of them do you impose the requirement that the current flowing in R3 has to be the SAME as the current flowing in R1. You second equation asserts that the current flowing in R3 is the same as the current flowing in R2. If that were the case, then how would the current I2 on the top get back to where it came from?

    Just present your loop equations. Don't simplify them. There is no point simplifying them until they are correct. Let's get them correct.

    You still need to identify and annotate on your diagram the third current.
     
  15. jiggssmalls

    Thread Starter New Member

    Oct 21, 2013
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    Please I do not understand what you just explained...does that mean my first and second equations are wrong?I just got to this level and it is the first topic were introduced to.and only an example was given to us and that even was only with two loops...So im really pleading just to make it a bit more simple for me,because I have to hand it in on Thursday.im so confused because I thought the first and second equations were right,but to what I understand now,they are wrong right?
     
  16. WBahn

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    So far I've only seen one equation and manipulations on just that one equation. You did post two equations but we immediately found that the first had problems and I asked you to post ALL THREE of the loop equations. You then just posted the first one.

    [​IMG]

    This is the same loop equation that you originally posted, although notice that your equation magically changes from the first line to the second. Why?

    Looking at just your initial equation, you are claiming that the current through R1 is I1. You are also claiming that the current through R3 is (I1-I2) -- or, if we use the second line, that it is (Ii+I2). But since R1 and R3 HAVE to have the same current (you do see that, right?), then the only way this is possible is if I2 is identically equal to zero. But that doesn't make any sense, does it?

    I think it might help if you take a step back and come at it from the beginnings.

    [​IMG]

    In the diagram above, I have assigned a current and a voltage for each component. For the resistors, I randomly picked a direction for each one. I then assigned the voltage according to the passign sign convention which says that, through a passive device, current flows from positive to negative.

    Then I assigned the currents for the two supplies according to the passive sign conventions, which says that for sources the current flows through the device from negative to positive.

    Those are all indicated in red. The naming is very simple, Ix is the current flowing in Rx and Vx is the voltage across Rx. For the two supplies, it's Isx and Vsx.

    Keep in mind that the assignments were largely random. I did this because I think it might be beneficial for you to see that it doesn't matter, as long as you are then consistent with the assignments as made.

    In blue, I identified three loops and, again, assigned a random direction to each of them to make the point that it doesn't matter.

    In yellow, I identified the five nodes that exist in the circuit and labeled each with a letter. We can refer to the voltage on a given node, relative to the reference (which we haven't specified yet) by simply using the node letter as the subscript on the voltage, so Va or Vd, for instance. We can talk about the voltage difference between two nodes by using a double subscript in which the value represents the voltage of the first node relative to the second. So, for instance, Vad means Va-Vd.

    The next step is to apply KVL to each of the three loops. To do this, you go around the loop and sum up the voltage drops and set the total equal to zero. You can also sum up the voltage gains, it doesn't matter. You can flip a coin for each loop and decide whether you will sum up the drops or the gains for that loop.

    So I will do the first one and then you do the other two.

    Loop #1: Sum up the voltage gains around the loop, starting in the bottom-left corner.

    Vs1 - V1 - V2 - V3 = 0

    I can relate the voltage across a resistor to the current through that resistor using Ohm's Law, so this becomes:

    Loop #1: Vs1 - I1*R1 - I2*R2 - I3*R3 = 0

    Once we have applied KVL to the loops, we need to apply KCL to the nodes. For each node, we sum up the currents leaving the node and set them equal to zero. Or, we could sum up the currents entering the node and set that equal to zero. Or we could set the sum of the currents entering the node equal to the sum of the currents leaving the node. Doesn't matter and we can randomly pick how we do it for each node -- all three are completely equivalent.

    So I will node Node A and Node B. In the first I will sum up the currents entering the node and in the second I will sum up the currents leaving the node.

    Node A: Is1 - I1 = 0
    Node B: -I1 + I2 + I4 - I5 = 0

    So you do the other three.

    You will now have a total of eight equations at your disposal to solve for the seven unknown currents (one of the node equations -- pick one -- is actually redundant and could be arrived at from the others).

    After you brute force your way through this, we can come back to it and see how we could be more systematic and solve it with only three equations, each of which could be written down by inspection. But I think you need to spend some time with the basics, first.

    Do your best to come up with the other equations and let's get them in place before we try to actually solve for anything.
     
    Last edited: Nov 5, 2013
  17. jiggssmalls

    Thread Starter New Member

    Oct 21, 2013
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    allright...i am on it now.:)..thank you very much...will be back with the correct equations.hopefully
     
  18. jiggssmalls

    Thread Starter New Member

    Oct 21, 2013
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    i have v1 and v2 but i dont see any v3.and in your equation,u added a v3.will v2 be same as v3 since in passing through the same circuit?
     
  19. Richjtf

    New Member

    Oct 12, 2013
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    I remember when i was first learning about Kirchoffs law i found this exact same problem, it drove me crazy because any solution i came up with was different from the answer the book gave me, you are already getting help so i'm just going to say: try your best! =)
     
  20. WBahn

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    Mar 31, 2012
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    Huh?

    Did you look at my diagram? I have labeled all the voltages and currents explicitly.

    V3 is the voltage across R3, just as V1 is the voltage across R1. What YOU called v1 and v2 I am calling Vs1 and Vs2, respectively.
     
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