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Complicated Resistance

  1. Feb 7, 2007 #1
    1. The problem statement, all variables and given/known data
    Okay - the picture is attached, but what I need to find is the current and voltage throughout certain points on the circuit. What I need help on is figuring out how to simplify this circuit in order to find resistance.


    2. Relevant equations
    Parallel Req = (1/R1 + 1/R2)^-1
    Series Req = R1 + R2


    3. The attempt at a solution
    I know how to combine equations for series and parallel circuits - but what can I do with R4? I need to simplify the circuit, but I know I can't put it in parallel with R5 or R6 since it has resistance in both loops. Is there a way to combine R5 and R6. Let me know please...
     

    Attached Files:

  2. jcsd
  3. Feb 7, 2007 #2
    I think R5 and R6 can be combined since you have only the one voltage source, then added to R4.
     
  4. Feb 7, 2007 #3
    I tried that before, but I was unsure. The eq of R5 and R6 is (1/(R5) + 1/(R6))^-1 Then that would be added to R4, since R4 is in series. Okay, that makes more sense now - I think I have the rest from here!
     
  5. Feb 7, 2007 #4

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    I believe that tou can join the R5/R6 in parallel more the R4 in series to form a "block" with the R4 in parallel with this block
     
  6. Feb 7, 2007 #5
    I beleive that the circuit would be the same as the attachment from R1, R2, and R3 - but the last branch would have R5 and R6 in parallel in the "block" as you called it, in series with R4.

    Would my total resistance than be:
    Req = R1 + R2 + (1/(R3) + (1/R4) + (1/(1/R5 + R6))

    Specifically, how would I add up the resistance of the R5 and R6 Block?
     
  7. Feb 7, 2007 #6
    Right on.

    But just wait til they start adding batteries all over. :yuck:
     
  8. Feb 7, 2007 #7
    I think thats close call R4+1/(1/R5+1/R6)=Z
    Req=R1 + R2 + 1/(1/r3+1/z) is what I see.
     
  9. Feb 7, 2007 #8

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    R1+{1/R3+[(1/R5+1/R6)+R4]}+R2
    [tex] R1+ \{ \frac{1}{R3} + (( \frac{1}{R5}+ \frac{1}{R6} )+ R4) \} + R2[/tex]
     
    Last edited: Feb 7, 2007
  10. Feb 7, 2007 #9

    Curious3141

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    Teach you a little trick to make these problems easier.

    When two resistors R and r are in parallel, use the notation (R || r) to express the effective resistance. The two parallel lines '||' have an obvious meaning.

    When two resistors R and r are in series, just add them up as usual, i.e. the effective resistance is R + r.

    Now, your objective is to reduce the diagram in stages to a single resistance. Just forget about the mathematics and express everything in this notation at first.

    When you reduce a pair of resistors, immediately redraw the diagram with the effective resistance of the pair as a single resistor. Just stick with the notation.

    So, you'd begin : R5 is in parallel with R6, the effective resistance is (R5 || R6). That is now in series with R4, the effective resistance is now (R5 || R6) + R4, and so forth.

    When you finally finish the reduction, you should get the effective resistance of the entire circuit (R) as :

    R = {[(R5 || R6) + R4] || R3} + R2 + R1

    (Be sure to bracket each pair of resistances as you reduce them, to avoid making a mistake).

    Looks complicated, but it isn't really when you work through it yourself. It becomes a lot easier if you keep drawing the intermediate steps in the reduction with single resistances replacing the pairs you've reduced.

    Now just apply the relation R || r = (1/R + 1/r)^(-1) = (Rr)/(R+r), while taking care with the brackets, and you've got the effective resistance of the circuit.
     
    Last edited: Feb 7, 2007
  11. Feb 7, 2007 #10
    Slick, I like that. Improves legibility, reduces chance for error, avoids the need to make subs as I did....
     
  12. Feb 7, 2007 #11

    Curious3141

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    Wait till you see how legible they make (otherwise) complicated complex impedance problems! :smile:
     
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