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Kirchoffs law

Discussion in 'Electronics Homework Help' started by tom6123, Jun 27, 2014.

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  1. tom6123

    tom6123

    29
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    May 31, 2014
    hello, ive been trying to figure out how to do kirchoffs law.
    i havent got far into it but i keep getting lost here is what i have so far
    Using simultaneous equations

    R1=10 Ohms
    R2=10 Ohms
    R3= 75 Ohms
    R4 = 33 Ohms
    R5=47 Ohms
    V1= 5v
    V2 = 21V

    what i got so far is:
    Us1= I1(10+33+75)+I2(75)
    US1= I1 118 + I2 75

    Us2=I2 (10+47+75)+ I1 (75)
    Us2=I2 132 +I1 75

    5v=I1 118 + I2 75 (X132)
    21V= I2 132 +I1 75 (X75)
    -------------------------
    660=I1 23760 +I2 9900
    1575= I1 5625 +I2 9900
    -------------------------
    -915 + I1 18135 -915/18135 = -50.5 mA



    5v=I1 118 + I2 75 (X75)
    21V= I2 132 +I1 75 (x180)
    ---------------------------
    375= i1 13500+I2 5625
    3780=I1 13500+ I2 23760
    -----------------------------
    -3405=-18135 -3405/-18135= 187.8mA



    Am i on the right tracks because i think im missing something
    im trying to find the
    Current through R1 UR1=I1(R1)
    Current through R2 UR2=I2(R2)
    Current through R3 UR3=(I1+I1)+R3
    Current through R4 UR4=I1(R4)
    Current through R5 UR5=I2(R5)
    Voltage through R3
    Power Dissipated in R3

    And i dont know the formulae for the last two.but i dont know if my working out is correct right now.
    i think the power dissipated in r3 is PR3=(I1+I2)UR3
     

    Attached Files:

  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,389
    2,774
    Jan 21, 2010
    Firstly, let's make that easier to see:

    [​IMG]

    Kirchoff's laws tell us 2 things:

    1) V1 = VR1 + VR3 + VR4 (and the same for similar loops)
    2) IR3 =IR1 + IR2 and the same for similar nodes

    In both cases the direction of this assumed current and the assumed polarity make a difference because the actual rules are

    1) in a loop all voltages sum to 0
    2) at a node all currents sum to 0

    It's often easier to label the nodes, and if we label them like this:

    KirchhoffCircuit1.jpg

    Then we can more unambiguously say:

    1) VAD = VAB + VBE + VED
    2) IBE = IAB + ICB

    (clearly there are more equations than that)

    Then, since we know the resistances, we can write one in terms of the other

    Then, you solve it :)

    It's described reasonably well here.

    The easiest way to check is to calculate the voltages ACROSS, and the currents THROUGH each component, then see if:

    a) V = IR
    b) the voltage laws apply (try a loop you didn't use in your solving process)
    c) the current law applies for some node (preferably one with more than 2 connections to it).

    Have you done these three things?

    What was the result?

    The most common mistake is to get your sign wrong. This is most obvious with currents at a node having only 2 wires attached. The current should have an opposite sign here. Make sure you don't (for instance) use VAB and IBA (VAB = -VBA -- and the same for current)
     
  3. Ehsan

    Ehsan

    100
    1
    Jun 12, 2014
    Read this book : "Engineering circuit analysis" by William H. Hayt - chapter 3: /voltage and current laws/

    I have the PDF version and I can upload it in Dropbox and put a link to it here but I am not sure if website policy allows me to do that.
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,389
    2,774
    Jan 21, 2010
    That would depend on the situation with regard to copyright...
     
  5. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    chopnhack likes this.
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    I'm happy with a direct link to archive.org. :)
     
  7. Ehsan

    Ehsan

    100
    1
    Jun 12, 2014

    Second edition is completely fine. The technology in that level has not changed for decades. The Ohm law, KVL, KCL and etc.. presented in eighth and second level are the same of course lol The only changes are: things like adding PSpice tutorial and changing chapter problems. That is all.
     
  8. tom6123

    tom6123

    29
    0
    May 31, 2014

    right so regarding the question.
    i can just use the formulas above ( UR1=I1(R1) ) if i just transpose them making the current the subject for all resistors 1-5? but for resistor 3 would i have to include both currents (UR3=(I1+I2)R3?
     
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,389
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    Jan 21, 2010
    Essentially, yes. But if you read ahead it might be better to do that with voltages to try to eliminate one of those.

    In this problem you have a certain number of unknowns. To number them, there are voltages across and currents through each resistor plus the currents through the voltage sources.

    That's 12 unknowns. You could try to get 12 equations and solve them, but it would be tedious.

    Kirchoff's laws allow you to simply derive equations for node currents and loop voltages.

    Using ohms law, since you know the resistor values, you can now rewrite the node (current) equations in terms of the loop voltages.

    At this point you'll have m equations in n unknowns. n should be no more than 5, and m should be n or greater.

    now you can solve the equations quite simply, then use ohms law to get the currents.

    At this point you will have all the voltages and all the currents except the currents through the voltage sources. Hopefully it will be obvious how you can derive these.

    As I've said before, the most difficult part is getting the signs of the voltages and currents correct.
     
  10. tom6123

    tom6123

    29
    0
    May 31, 2014
    ok so i tried finding the voltages using these:

    UR1=I1(R1) -50.5 mA X 10ohm= -0.505V
    UR2=I2(R2) 187.8 mA X 10 Ohm 1.878V
    UR3=(I1+I1)R3 ( -50.5 mA + 187.8 mA) 75ohm=10.2975
    UR4=I1(R4) -50.5 mA X 33Ohm= -1.6665V
    UR5=I2(R5) 187.8 mA X 47Ohm= 8.8266V

    Are these correct because if i add the I1 voltages up its more than my voltage sources
     
  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Well, they're clearly wrong because the voltages don't work.

    Why don't you start by showing us the basic equations you wrote for the node currents and the loop voltages.
     
  12. tom6123

    tom6123

    29
    0
    May 31, 2014
    well i was taught to find the current by using simultaneous equations which is at the top of the thread
     
  13. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    I have just done the calculations and I have the right answer.

    Yep, you have the wrong currents (which is obvious I guess).

    start from your basic equations Vs1 = ... and Vs2 = ... then try again, you solved it incorrectly. Did you substitute the values back into an equation to check your result?

    The method you're using is not the basic form derived from Kirchoff's laws -- have you done it the basic way?
     
  14. tom6123

    tom6123

    29
    0
    May 31, 2014
    no i haven't. i was only taught the simultaneous way, that was to find the current.

    i placed the current i calculated back into the equation (-50.5mA x 180) + (187.8mA X 75) =4.995
    and i also did (-50.5mA X 75) + (187.8mA X 132) = 21.0021
     
    Last edited: Jun 30, 2014
  15. tom6123

    tom6123

    29
    0
    May 31, 2014
    does it have to be dead on 5v cause 4.995 is pretty close.
     
  16. tom6123

    tom6123

    29
    0
    May 31, 2014
    ok, i found my problem, it was in the simultaneous equations, i was multiplying by 180 when i was meant to be multiplying by 118, but now my currents are I1=-91.95mA , I2=211.3 mA now these work
     
    Last edited: Jun 30, 2014
  17. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    These are your equations, right?

    VS1 = 118 i1 + 75 i2

    VS2 = 75 i1 + 132 i2

    Can you see your mistake now?

    And, no it doesn't have to be dead on, but both the voltages and the currents need to work.
    You've not (apparently ever) done this using the basic form of Kirchoff's laws. And this is why you've come so badly unstuck (in my opinion).

    And I see you've found the problem. :)
     
  18. tom6123

    tom6123

    29
    0
    May 31, 2014
    Yes i have found my mistake and corrected it... the calculations are so very close. it has been a pain staking, head scratching amount of hours but hopefully i work out the current through the resistors
     
  19. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,389
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    Jan 21, 2010
    You'll find it is time well spent.

    These are mistakes that you either won't make again, or will know where to look if you suspect you've made them.

    Unless you're highly time constrained, these problems have a really useful property of the solution being easily checked.

    If you come up against them in an exam, it's always nice to *know* you've got the right answer :)

    In real life, if you ever have to do these calculations, it will be even nicer to know.
     
  20. tom6123

    tom6123

    29
    0
    May 31, 2014
    is there a simple way of working out the direction of a current in this circuit assuming conventional current flow[​IMG]

    Because i worked out that the current comes from v2 to r5 then r3. where as V1 comes to r1 through to r3, which keeps the calculations positive
     
    Last edited: Jul 1, 2014
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