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Ohm's and Kirchhoff's simple problem

  1. Apr 14, 2007 #1
    Ohm's and Kirchhoff's "simple problem"

    For the figure below solve for:

    I(1) = ? mA V(1) = ? V
    I(2) = ? mA V(2) = ? V
    I(3) = ? mA V(3) = ? V

    I don't seem to be able to apply the correct combination or Ohm's and Kirchhoff's laws to get the correct values? Tips on an approach here?

    - Sav
     

    Attached Files:

  2. jcsd
  3. Apr 14, 2007 #2
    Welcome to the Physics Forum?

    Are you told you need to use K'..s laws here? In other words this circuit can be simplified witout resort to their use. Hint: Find an equivalent resistance for the 20 and 10 ohm resistors.

    In the future, it is best to show some effort on solving the problem using standard format, before expecting help.
     
  4. Apr 14, 2007 #3
    I'm taking electricity and magnetism and I got questions like these, use Kirchoff's laws even if not necessary for practice sake. First write the junction equation or sum of all the currents should equal? Then for each loop(as you can see there are two loops, or two small rectangles on the big one) write a equation, remember if you are going in the direction of the current on any resister it would have a voltage drop or negative and against would be the opposite or positive. Also, on the left loop there is an electromotive force or simply a voltage provider for the whole circuit. You should be aware what direction you are traveling on the loop and the direction of the current, very important!!!!
     
  5. Apr 14, 2007 #4
    Using Kirchoffs loop rule for the parallel loop proves that the voltage drop across both is resistors is the same. Only the current varies between parallel resistors
     
  6. Apr 14, 2007 #5
    Yes, you are correct, but if you use Kirchoff's law, the math will do all the work for you, not much thinking required as long as you know his rules.
     
  7. Apr 15, 2007 #6
    Yes, new here and just starting in circuits.
    Sorry about the nettiquete slip... still learning about this forum. I think I saw a form somewhere when I first started looking through the site, but don't see it now. I'll brush up on that before my next post! :eek:

    I should have said that I'm looking for a way to get started... thanks for the tips!

    TO the poster regarding Ohm's and K's laws, yes that is all he have to work with for this problem. I'm sure them must be a good reason ;)

    If I look at them all at the same time going counterclockwise starting at the bottom I get...

    -I(R3) - I(R2) -I(R1) + 6V = 0 (Kirchhoff)
    -I(20) - I(10) -I(10) + 6V = 0 (sub.)
    I = 0.15 A (solved)

    So I get 0.15 amps for the current for the circuit.

    I(R1) = 1.5 V
    I(R2) = 3.0 V
    I(R3) = 1.5 V

    Which looks ok for the 6 V emf.

    But... shouldn't my current be the same for the two resistors in parrallel (R2 and R3)?

    I'm sure I'm just not looking at the circuit in the right 'chuncks' ;(

    Using just Ohm's law, and Kirchhoffs current and voltage laws are all I have to work with at this point.
     
  8. Apr 15, 2007 #7
    Check out this thread. The methods are the same for your circuit. Also, post pictures using the picture icon so they are easier to read.

    Welcome to PF!
     
  9. Apr 15, 2007 #8
    Thanks! Nice to be here... the thread looks like just what I needed, but unfortunately I can't see any of the graphs (just a big X where they should be)... any way I can get you to repost those? o:)
     
  10. Apr 15, 2007 #9
    Your equations are not set up right:

    I. write a junction equation: ie. In the form of I(net)=I1-I2-I3=0 or you can write I1=I2+I3 which are pretty much the same.

    II. write out equations of the total net voltage of each loop, look there are only two loops.

    You don't have your junction eqation. You have the two equations of the loops, but one of them is incorrect. Remember, you are traveling through one loop for each equation, emf is only on
    one loop, not two!!! Each loop, what is there on each loop is one equation. Read my previous tread, what two important elements you should take into consideration?

    Remember!! You have three unknowns. How many equations do you need to solve the problem? It should be automatic if you set it up right. Junction equation is one, then the two are from the loops(hint hint)
     
    Last edited: Apr 15, 2007
  11. Apr 15, 2007 #10
    OK...
    I. I1 - I2 - I3 = 0 so, I1 = I2 + I3 (Time to find current so on to part II)

    II.
    Loop 1 (I hope):
    +6 V - V1 - V2 = 0
    + 6 = v1 + v2
    6 = I(R1) + I(R2)
    6 = I(10) + I(20)
    I = 0.2 A

    Loop 2 (This one I'm really not sure about)
    +v2 - v3 = 0
    v2 = v3

    I'm still working on the next part... someone stop me if I'm headed the wrong way! ;)
     
  12. Apr 15, 2007 #11
    Label the Is accordingly I1 I2 I3 etc...

    you are trying to solve for values of each I, and you have to distinguish them.
     
  13. Apr 15, 2007 #12
    Yea! I take it then that my two loops look good!!

    As for labling the Is... I thought I did that in part 1 above!? based on the figure in my first post.
     
  14. Apr 15, 2007 #13
    Using I1=0.2A for 'loop 1', I get 2 volts for R1. Where go I go from there?
     
  15. Apr 15, 2007 #14
    labels the Is for each loop equation

    then you can solve for each I by deduction
     
    Last edited: Apr 15, 2007
  16. Apr 15, 2007 #15
    Well, I just can't seem to 'get' this... sorry.

    I keep getting the same wrong answers, but thanks for the help. If you have the solution for me that would be great for next time. SOrry to be so dense on this one!
     
  17. Apr 15, 2007 #16
    This is where I ended up (just we can all see just how confused I am)...

    From my figure (in post one);
    I1 = 0.6 A; I2 = 0.2 A; I3 = 0.4 A
    V1 = 2 v; V2 = 4 v; V3 = 4 v

    The reasoning goes like this...
    For the first I1 current, its just Ohms, 6v/10 ohm = 0.6 A

    starting off on the first loop with Kirchhoff gets me,
    +6v -V1 - V2 = 0
    6 = IR1 + IR2
    I = 0.2 A (for the 'first loop' circuit)
    so, in V1 = IR1 = 0.2(10) = 2v

    V2=I(R2)=0.2(20)=4v

    Kirchhoff for the second loop gets me, V2 = V3 = 4v
    Ohm's law gets me, I3 = V3/R3 = 4v/10 ohm = 0.4 A

    This seems reasonable... my loop volts seem to zero out AND my current nodes seem to work (what goes in, comes out).

    Anyway... thats what I'm turning in, we'll see how it goes.
    Thanks! Savoy
     
    Last edited: Apr 15, 2007
  18. Apr 15, 2007 #17
    Okay, I give up here is how you should solve it
    Junction equation: I1=I2+I3

    Loop1(going clockwise): 6-10I1-20I2=0

    Loop2(going clockwise): -10I3+20I2=0

    Then solve for I1, I2, I3
     
    Last edited: Apr 15, 2007
  19. Apr 15, 2007 #18
    I used those (see above)... where did my logic break down?

    From your loop one, I = 0.2 A (at the junction)
    so I know 2A flow into the junction.
    From your loop two, I'm left with 10(I) which is also 0.2A. (whichis nice for proving conservation, but how do I 'split' it?)
     
  20. Apr 16, 2007 #19
    Yes the current get spit at the junction, hence I1=I2+I3. But, also it can be written as a equation of net Junction equation which is I1-I2-I3=0, is this not true? Can you relate this to the other two equations? When you solve circuits with mulitpul loops involving more than one emf you may appreciate Kirchoff's method.
     
    Last edited: Apr 16, 2007
  21. Apr 17, 2007 #20
    OK, here I go:...
    Junction: I1 = I2 + I3
    Loop 1: +6 -v1 + v2 = 0
    Loop 2: -v3 +v2 = 0 | v2 = v3

    Solving for I1, I2, I3

    Junction: I1 = I2 + I3 | v1/r1 = v2/r2 + v3/r3
    v1/10 = v2/20 + v3/10
    2 v1 = v2 + 3 v3 (v2 = v3) | 2 v1 = 3 v(2-3)
    v1 = 3/2 v2

    Loop 1:
    6 -v1 + v2 = 0
    6 -3/2 v2 + v2 = 0 (subbed)
    6 - 5/2 v2 = 0
    6 = 5/2 v2

    v2 = 2.4 v
    v3 = 2.4 v
    v1 = 3/2 v2 = 3.6 v

    I3 = V3/R3 = 0.24 A = 240 mA
    I2 = v2/R3 = 0.12 A = 120 mA
    I1 = v1/R1 = 0.36 A = 360 mA

    I'm sure there was an easier way, but this is where I ended up. Thanks! Savoy
     
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