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Kirchoff's law Question

  1. Oct 25, 2015 #1
    1. The problem statement, all variables and given/known data
    In the circuit shown, what must be the size of the resistor R such that the power delivered to the 8 Ω resistoris 160 Watts?
    Circuit is in the first image
    2. Relevant equations
    • ∑I at Junction= 0
    • ∑ΔV though loop=0
    • P=I2R
    • ΔV=ε-Ir

    3. The attempt at a solution
    The circuit I have drawn is in the second image. The first thing I did was found the current that is going through R1 is 4.47 A. Since R1 and R2 are in series, I thought that they had the same resistance as well. Next, I used the first part of Kirchoff's law to find the junction equations:

    • Junction B: I1= I2+ I3
    • Junction E=I4 +I6=I5
    Next, I created some loops using the second part of Kirchoff's law:
    • Loop AFEBA: -ε1+ R1I5+R2I5+R3I62+rI6+R4I2=0
    • Note, that for the ε2, there is a internal resistance within the battery, such that the ΔV=ε-Ir. I think that the I would be equal to the one in the unknown resistor, I6.
    • Loop EDCBE: -ε3+R5I3-R4I22-I6r -R3I6=0
    • Loop ACDFA: -R5I33-R2I5-R1I51=0
    From the last loop, I was able to find the I3 is 4.83 A. I used this in the other loops and simplified. The problem is that I don't understand how to continue to solve this problem.
    Also, I am not completely sure with the circuit I have drawn the currents are all correct the way I have drawn them.

    Any help will be much appreciated.
     

    Attached Files:

  2. jcsd
  3. Oct 25, 2015 #2

    gneill

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    Staff: Mentor

    Bit of a tricky question here. I agree with your determination of the required current through the 8 Ω resistor in order for it to dissipate the required 160 W. But knowing the power dissipated by a resistor doesn't tell you which direction the current flows through it, so you have to test both cases to see which will produce plausible results.

    What I suggest is that you simplify the circuit a bit by gathering up known resistors where possible to begin with, and add a convenient reference node:
    Fig1.png

    Then use the given information about the branch with the "known" current to determine the potential at the V1 node with respect to the reference node One value will be plausible when you consider the constituents of the other branches, the other not so much (without invoking some exotic characteristics for R). What value will you choose for V1?

    Hint: The highest potential source in the circuit is the 36 V source. Being the highest source, you can be sure that any current from it must flow out of its positive terminal (again, unless exotic characteristics are invoked for R).
     
  4. Oct 25, 2015 #3
    Thanks for the comment! I see that simplyifiing it makes it easier to read and such. However, on the 1 ohm resistor, I am curious as to its placement. Shouldn't it be on the other side of the 36 V elctromotive? The way its been changed doesn't agree with the problem does it?

    Thanks again!!!
     
  5. Oct 25, 2015 #4
    And another question, is the I3 going in that direction because of the 36 V electromotive? Because I also thought that it was going in that direction originally, but didn't think it was plausible. I said it because the 36 V seemed it would have the greater influence.
     
  6. Oct 25, 2015 #5

    gneill

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    Staff: Mentor

    The order of the components in a given branch doesn't matter. I chose to "move" the resistor for purely aesthetic reasons, making the image easier on the eye. Adding up potential drops in series is much like adding up several numbers; the order that you sum them doesn't affect the total.
    Yes, if you want to choose directions for your currents then you can make a judgement call about about what influence you think the circuit's emfs will have. But I should point out that there's really no "wrong" choice; the math will eventually take care of the sign of the current regardless.

    I pointed out that the current direction in the bottom branch is constrained by it containing the largest voltage source inn the circuit, and you should be able to work out what the direction of the current must be in the top branch from the resulting potential it forces at V1. That leaves the middle branch as an unknown, but should become obvious once the other two currents are calculated.
     
  7. Oct 25, 2015 #6
    Ok, thanks for the help! Really, this has been an eye opening experience here! So I did the problem using I1 going both ways. With the I1 going towards, I got a positive R, but its kind of big. With the I1 going away, I got a negative R, and an I that is over 100 A, so I think that is wrong.
     
  8. Oct 25, 2015 #7

    gneill

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    Staff: Mentor

    It's great if you've learned a trick or two :smile:
     
  9. Oct 25, 2015 #8

    Mister T

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    What did you get? Mine is not so large.

    By the way, in your original drawing you've got way too many I's. You need only three, one for each branch. Use the idea that resistors in the same branch have the same current.
     
  10. Oct 27, 2015 #9
    The resistance I got was like in the 42.25 ohms.
     
  11. Oct 27, 2015 #10

    Mister T

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    My notes are not with me, but that sounds about right.
     
  12. Oct 27, 2015 #11

    SammyS

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    I get about 46.4 Ω .
     
  13. Oct 27, 2015 #12

    Mister T

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    Me, too. The node at V1 is 31.2 volts above the reference level (ground).
     
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