soo stuck :/ distances in a circuit!

Discussion in 'Homework Help' started by champ01, Aug 22, 2012.

  1. champ01

    Thread Starter New Member

    Aug 22, 2012
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    Hey guys, I have this question that was in a lecture slide.. I have no clue where to start for question a) :/

    [​IMG]

    I know that the r ohms/m is the resistance of the wires per m..thus they give you lengths of x and L.. but where do i go from here? I was thinking super position and neglecting v2 or something then vice versa for v1?

    Thankss heaps!
     
  2. panic mode

    Senior Member

    Oct 10, 2011
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    draw a circuit, label everything and you will get three resistors in series (wire1, load, wire2). then calculate voltage drop on load.
     
  3. champ01

    Thread Starter New Member

    Aug 22, 2012
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    thanks you for that :) i am just wondering if i am heading in the right direction now?

    [​IMG]
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    No you aren't heading in the right direction.

    In the original diagram the "sliding" resistor R forms a bridge across the lines at the point of connection to the lines.

    If the sliding R is a distance x from source V1 then the line resistance in each line conductor from V1 to the connection point is x*r Ω. The line resistance in each line conductor from source V2 to the point of connection is (L-x)*r Ω.

    Redraw the schematic with the line elements replaced with those equivalent resistances and with R bridging the lines at the connection points. You'll end up with a two loop network which you can then analyze using standard DC circuit analysis techniques.
     
  5. champ01

    Thread Starter New Member

    Aug 22, 2012
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    okay.. i think i have done that or i am on the right track now? hopefully :/

    [​IMG]
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    For your current annotations I would think

    I2=I1+I3

    where

    I2=V/R

    Rather than what you have for I2.

    I1 & I3 correct as you have them.
     
  7. champ01

    Thread Starter New Member

    Aug 22, 2012
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    oh okay, thanks! but then I am not including the distance L in? as when I work that equation out after changing I2, I don't get to the equation they want :S just about to go over it again to see if I did any arthimetic errors..
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    If you start with ...

    I_2=I_1+I_3

    or

    \frac{V}{R}=\frac{V_1-V}{2xr}+\frac{V_2-V}{2\(L-x\)r}

    and work through the somewhat complex algebra to make V the subject of the equation then you will arrive at the correct answer.
     
  9. panic mode

    Senior Member

    Oct 10, 2011
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    just multiply last equations by 2xr(L-x) and solve in terms of V and you get the answer to first part. for second part you need to solve dV/dx=0 and rest is just plugging in numbers
     
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