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Equivalent Resistances

  1. Feb 16, 2004 #1
    To begin with, I need to find the R(eq) of the following (beg my endeavor with paint). What I don't seem to get at all is what to do with the diagonal 6 ohm resistor. I don't see the grouping at all..
    [​IMG]
     

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    Last edited: Feb 16, 2004
  2. jcsd
  3. Feb 16, 2004 #2
    I think you are letting the diagonal position of it throw you off.
    That R is simply in parallel with the 9 ohm between points d and f.

    Try redrawing the diagram, but with the 6 ohm R running vertically on your paper, and kick the 9 ohm to the right and turn it vertical as well. Just be sure to keep all your connections and labeled points in the correct relation. I think that will aid your visualization of the problem.

    Let us know how this works for you.
     
  4. Feb 16, 2004 #3
    So... the de 6 ohm is parallel to the 9 ohm, but the 2.4 ohm isn't parallel to those two (b/c of the cd resistor?)?
     
  5. Feb 17, 2004 #4

    Integral

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    Draw your picture as above, you will see that the 2.4 ohm res. is in series with the 6 and 9 ohm parallel pair.
     
  6. Feb 17, 2004 #5
    6 ohm + 9 ohm = 3.6 ohms
    Then the rest is series?
    3.6ohms + (3)6 ohms + 2.4 ohms = 24 ohms?
     
  7. Feb 17, 2004 #6
    Wait, or is it, the 6ohm and 9 ohm are parallel.
    The 2.4 ohm is series to the (6 & 9).
    The 6 ohm in the middle is parallel to the combined (2.4 ohm series to (6 &9)).
    Then all series? To get 15 ohms?
     
  8. Feb 17, 2004 #7
    I believe you have it.
     
  9. Feb 17, 2004 #8
    Atleast I got that part , so I can traverse around the circuit
    a. I need to find I1, I2, .. I5
    This seems to be a bit of a problem for me. I know I = V/R
    So the absolute total would be 1 amp for each way, but where next?

    b. Find potential difference across each resistor.
    So V=IR
    For the 6 Ohm on the a-c, that would be 6V? (same for the bottom b-d) Then I use the remaining 3V and break into the 2nd half, correct?

    c. Find power dissipated by each resistor.
    P=IV, Use the V's figured from b. to get the wattage for each?
     
  10. Feb 18, 2004 #9
    So I1 = 1 amp is confirmed,
    someone said I2 & I3 = .5 amps each. (I4 & I4 = .25 amps each?)

    Now if those are what they are, I'm all confused on the current to use for part C & D. Anyone with any insight? (Someone got 2.5v from the a to c 6ohm resistor and the bottom resistor too, from 15v/6ohm)
     
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