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Circuit Problem electrical engineering

  1. Sep 20, 2012 #1
    1. The problem statement, all variables and given/known data

    Here's the question:
    http://puu.sh/171ls [Broken]


    3. The attempt at a solution

    I have no idea where to begin. It says R(AB) = R(L) = 301 ohms. What does that mean? Does R(AB) mean the resistance from A to B? Is that the total resistance? The total resistance is equal to 301 ohms which is also equal to R(L)? I don't know. Any help would be appreciated.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 20, 2012 #2

    gneill

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    Staff: Mentor

    Yes, RAB is the equivalent resistance of the entire network as "seen" from the terminals AB.
     
    Last edited by a moderator: May 6, 2017
  4. Sep 20, 2012 #3
    Ok, I think I got it. R should equal roughly 174 ohms.
     
  5. Sep 20, 2012 #4

    gneill

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    Staff: Mentor

    Looks good :smile:
     
  6. Sep 20, 2012 #5
    Hi! Note that the circuit can be drawn as a resistance R in series with the parallel combination of: (a) a resistance of R and (b) a series combination of the resistances R and R(L).

    Then, the circuit can be drawn as a resistance of R in series with the parallel combination of R and (R + R(L)).

    First, determine the equivalent resistance of the parallel portion of the circuit, R and R + R(L).

    [tex]\frac{1}{R(Parallel)} = \frac{1}{R + R(L)} + \frac{1}{R}[/tex]
    [tex]\Rightarrow R(Parallel) = \frac{1}{\frac{1}{R + R(L)} + \frac{1}{R}}[/tex]

    Multiply the numerator and denominator of the right side of the equation by [itex]\frac{R(R + R(L))}{R(R + R(L))}[/itex] to get the following:

    [tex]R(Parallel) = \frac{R(R + R(L))}{R + R + R(L)} = \frac{R^2 + R(L)}{2R + R(L)}[/tex]

    You are then told that R(AB) must equal R(L). This becomes the following:

    [tex]R + \frac{R^2 + R(L)}{2R + R(L)} = R(L)[/tex]

    Then, multiply both sides of the equation by (2R + R(L)) to get the following:

    [tex]R(2R + R(L)) + R^2 + R(L) = (2R + R(L))R(L) = 2RR(L) + (R(L))^2[/tex]
    [tex]\Rightarrow 2R^2 + RR(L) + R^2 + R(L) = 2RR(L) + (R(L))^2[/tex]
    [tex]\Rightarrow 3R^2 - RR(L) + R(L) - (R(L))^2 = 0[/tex]
    [tex]\Rightarrow 3R^2 - 301R + (301) - (301)^2 = 0[/tex]
    [tex]\Rightarrow 3R^2 -301R + 301 - 90601 = 0[/tex]
    [tex]\Rightarrow 3R^2 - 301R - 90300 = 0[/tex]

    Applying the quadratic equation (I'll leave the details up to you) yields the following:

    [tex]R = 230.7675926 ohms ≈ 231 ohms[/tex]

    You will want to double check my result; but I think that I did everything okay.
     
  7. Sep 20, 2012 #6
    ^If I plug 231 in for R, Rab comes out to be about 390 ohms.

    174 for R gives me 301 ohms for Rab though.
     
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