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Power to a resister in circuit

  1. Oct 31, 2007 #1
    All resistors, R1 through R6, have the same resistance R. What is the power delivered by the batteries to R6?


    I worked it like this first...

    I saw two junctions, the nodes at the left and right where all three connections meet. I figured I1, I2 and I3 would meet here, combine, then split back up along the connections in the other direction as I = ((I1 + I2 + I3)/3). Power = I^2R, so I solved as (I3 - ((I1 + I2 + I3)/3))^2*R6... but mainly I have only a lot of pictures to prove that, and it's not right at all...

    I got some help and redrew it like this. Someone explained kirchoff's rules for loops to me briefly, and that I have to add up all the power contributions along each loop, then set them to 0. I simplify and solve simultaneous equations, also using I3 = I1 + I2. I will update with my progress... any advice would be expressly appreciated.
    Last edited: Oct 31, 2007
  2. jcsd
  3. Oct 31, 2007 #2
    So when the current is going 'forwards' across an element, delV = -IR. The opposite is true if its going backwards. Across the batteries (E1-E3), delV is positive when going from the negative to positive side of the battery, and vice-versa.

    Accordingly, I've set up a couple equations which start at E1 and E2 and form two loops.

    E1 - I1R2 + I3R6 - E3 + I3R5 - I1R1 = 0
    E2 - I2R3 + I3R6 - E3 + I3R5 - I2R4 = 0
    I3 = I1 + I2

    Assuming this is correct... I can simplify and substitute.

    But if I set these equal to one another, my term with R6 just cancels out... obviously I'm doing something wrong here... back to the drawing board.
    Last edited: Oct 31, 2007
  4. Oct 31, 2007 #3
    I am suggesting just another approach.

    As the resistances are equal, symmetry considerations might reduce the effort. (It would have been of a great help had all emfs also equal.)

    Anyways, still I think, principle of superposition would reduce the effort to a great deal!
    Considering two emf sources to be 'off' (replace by a short-circuit) and one 'on':
    Equivalent resistance = R + (2R*2R/(2R + 2R)) + R = 3R.
    Therefore, current (through R6) =
    E3/3R (if the emf source which is 'on' is E3)
    -E1/6R or -E2/6R (if the emf source which is 'on' is E1 or E2) [Note the opposite direction!]
    Therefore, total current current = (2E3 - E1 - E2)/6R
    I hope now you can get the power.
  5. Oct 31, 2007 #4
    (You are always travelling in the direction of current, so how come +ve sign with I3??)

    I am referring to the original figure of the question. Call the left node A and the right node B.
    Then the current in the segement joining A & B through E3. You have a problem here. In this segment, the current will have same value, call it I3 (as you have named).
    The first two equations that you have written is assuming the direction of I3 reverse to what you have shown, while the last equation is according to your diagram.
    Either, reverse the direction of I3 shown and change the last equation as I1 + I2 + I3 = 0 and solve; Or, else.. your equations must look like:
    E1 - I1R2 - I3R6 - E3 - I3R5 - I1R1 = 0
    E2 - I2R3 - I3R6 - E3 - I3R5 - I2R4 = 0
    I3 = I1 + I2
  6. Oct 31, 2007 #5
    Okay, tryin' to follow you...

    For E3, R + R refers to the resistors on either side of the battery which are in series, and can just be added together. (2R*2R/2R +2R) refers to the parallel equivalent resistors on the other two branches. That makes sense...

    I'm not so sure how you're getting the 6R for the other two loops, though...
  7. Oct 31, 2007 #6
    When we consider other two loops, note that, the R6 is in one of the branches of two similar parallel branches. So, current is divided by 2 (Resistance is NOT multiplied by 2).
  8. Oct 31, 2007 #7
    Ah, damn it... you're right. I got jumbled up in my professors notes and my own. It's been a long day... went ahead and changed the drawing i made to have the right direction for the current.
    Last edited: Oct 31, 2007
  9. Nov 1, 2007 #8
    I'm not actually certain how to continue with the equations I was using (having corrected them). Even considering my mistake, I've ended up with the same thing which is

    E1 - I1R2 - I1R1 = E2 - I2R3 - I2R4

    From there I'm not sure what the point of substitution is... since the R6 term canceled out. Maybe that's still messed up.

    Your method makes a lot of sense, though, and it doesn't require the variables for current, which the problem doesn't provide. Additionally it uses the general R values, which the question hints at. I think I will stick with your method, though I'm still curious how to do it the other way...
  10. Nov 1, 2007 #9
    Go ahead, you will get it. Three equations, three unknowns, solve it.

    Basically, you havent messed up, it is just that you have opted for a 'long-cut'!

    Your variables are I1, I2, & I3. Note resistances are known, so why eliminate them? Furthermore, to calculate power through R6, you need to know I3. It doesnt matter if you dont know I1 and/or I2.
    So, try to eliminate I1 & I2, instead of I3!!
  11. Nov 1, 2007 #10
    okay, so... I simplified the dual equations to

    E1 - 2R(I3 - I2) - 2R*I3 - E3 = E2 - 2R(I3 - I1) - 2R*I3 - E3

    2R*I3 and E3 cancel out. I expanded to get

    E1 - 2R*I3 - 2R*I2 = E2 - 2R*I3 - 2R*I1

    Again, 2R*I3 cancels out to get

    E1 - 2R*I2 = E2 - 2R*I1

    so I1 = I2 - (E1 +E2)/2R. If I sub this for I1 considering I3 = I1 + I2 I get

    I3 = 2*I2 - (E1 + E2)/2R

    hahah... this is junior-high algebra, yet it is tormenting me still.
  12. Nov 1, 2007 #11
    lol.. so having fun solving 'junior-high algebra'? You are still having two variables in one equation!

    Why don't you try this way: (I am sure, there are other ways.)
    We want to get I3. So, eliminate I1 & I2. One way could be obtaining one of these in terms of remaining two from one equation and substituting back into the remaining two. Then, we would be left with two equations, two variables!
    Now, as we can see, simplest of all is the 3rd equation. Let's use that. I2 = I3 - I1. Substitute in remaining two. Start from here. You should be able to get the result very easily. Just remember, when you will be solving 'two equations, two variables', you have to eliminate I1 and NOT I3. This way, you can avoid some time & effort.
    Have fun! :wink:
    Last edited: Nov 1, 2007
  13. Nov 1, 2007 #12
    okay... i hope this is closer...

    I2 = I3 - I1

    E2 - R(I2) - R(I3) - E3 - R(I3) - R(I2) = 0

    Plug the first into the second and add it to the first equation...

    E2 - 2R(I3) + 2R(I1) - E3 = 0
    + E1 - 2R(I1) - 2R(I3) - E3 = 0

    so E1 + E2 - 2E3 - 4R(I3) = 0
    and I3 = -E3/2R + E1/4R + E2/4R

    but... that's doesn't comply with your original answer. but i guess its a little closer. i messed up something trivial... geez. i'm so tired.
  14. Nov 1, 2007 #13
    Have you made some changes to the figure, or was I sleeping when I looked at it? :surprised
    Last edited: Nov 1, 2007
  15. Nov 1, 2007 #14
    Okay, to avoid any further confusion, I will write down the description of the figure to which I am referring: (I am describing the figure to which I wrote those three equations!)
    Referring to the original figure of the question,
    I1 is the current from node A (the left node) to B (the right node) through the branch having E1.
    I2 is the current from node A to B through the branch having E2.
    I3 is the current from node B to A through the branch having E3.

    Now, the equations are:
    1: E1 - I1R2 - I3R6 - E3 - I3R5 - I1R1 = 0
    2: E2 - I2R3 - I3R6 - E3 - I3R5 - I2R4 = 0
    3: I3 = I1 + I2

    Using equation 3 to eliminate I2, we are left with
    4: E1 - E3 = I3(R5 + R6) + I1(R1 + R2)
    5: E2 - E3 = I3(R3 + R4 + R5 + R6) - I1(R3 + R4)

    Now multiplying 4 by (R3 + R4) and 5 by this (R1 + R2), we get:
    6: (E1 - E3)(R3 + R4) = I3(R5 + R6)(R3 + R4) + I1(R1 + R2)(R3 + R4)
    7: (E2 - E3)(R1 + R2) = I3(R3 + R4 + R5 + R6)(R1 + R2) - I1(R3 + R4)(R1 + R2)

    Adding 6 and 7 and rearranging we get,
    I3 = {(E1 - E3)(R3 + R4) + (E2 - E3)(R1 + R2)} / {(R5 + R6)(R3 + R4) + (R3 + R4 + R5 + R6)(R1 + R2)}

    Oh my God! It's so much more painful to type than write with pen. :grumpy:

    Well, now set R1 = R2 = R3 = R4 = R5 = R6 = R
    and you can easily see that the result by superposition theorem and by this method. {The negative sign implies that sign, that I have chosen in this method is opposite to which I chose by superposition theorem.}

    P.S. : Bold letters refer to equation numbers.
    Last edited: Nov 1, 2007
  16. Nov 1, 2007 #15
    I am quite amazed that worked... and also that superposition predicted the same thing. wow.

    Thanks very much!!! I can see clearly now... :)
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