Parallel resistors - quick question.

Discussion in 'General Electronics Chat' started by Kango, Oct 9, 2007.

  1. Kango

    Thread Starter Active Member

    Sep 22, 2007
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    I have 2 circuits where the resistors are parallel connected but if you look at the circuit (Fig 2) it´s getting all blurry too me.

    Fig 1:

    R1 // R2

    Fig 2:

    In my opinion this should be

    R1//R2//R3 but the answer is

    (R1//R2) + R3 (How can this be so?)
     
  2. recca02

    Senior Member

    Apr 2, 2007
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    i cant see the fig 1 and 2.
    are they posted or is there a specific page u are talking about.
     
  3. Kango

    Thread Starter Active Member

    Sep 22, 2007
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    Wops, forgot to post the figures.;)
     
  4. nomurphy

    AAC Fanatic!

    Aug 8, 2005
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    From a DC perspective, you effectively have no parallel resistors. R1 and R2 are in series for both Fig 1 and Fig 2.

    Were you to replace C in Fig 2 with a short, then you would have:
    (R1||R3) + R2.

    From an AC perspective of Fig 2, the math becomes a bit more complicated, because the series circuit (R3)C is in parallel with R1, which together are in series with R2.
     
  5. cumesoftware

    Senior Member

    Apr 27, 2007
    1,330
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    R1 and (R3 + C) are in parallel with each other, and R2 is in series with the resulting association. It is important to not forget about the capacitors, unless the circuit is under an AC regime under a almost infinite frequency.

    I cannot reply with more detail since I'm out of the context here. What do you wan't to know exacly. Are those figures taken from an exercise? If so, what is asked?
     
  6. recca02

    Senior Member

    Apr 2, 2007
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    until that circuit is complete it is hard to say whether even r1 will parallel r2
    in fig 1 until then it is possible that the last terminals may remain open and if dc is applied they r in series.
    same goes for fig 2. from what i see it is r1//r3 + r2 for DC supply.
    it depends on how the circuit is closed and where is the source connected.
     
  7. Kango

    Thread Starter Active Member

    Sep 22, 2007
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    Well, the question was to figure out the Frequency by "taken away "the condensator. And then using

    f = 1 /2PI*Rtot*C

    It seems to me that i was confusing toghter all things when taken away the capacitator, this thing is fairly basic but I mess it all up with the parallel connection.

    Yes, that seems confusing to me, how come?

    But wouldn´t that be (R1||R2||R3)
     
  8. recca02

    Senior Member

    Apr 2, 2007
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    it wud be r1//r2//r3 only if the other terminals of r1 and r2 were connected.
    remember fro being in parallel the components must be connected across same two points.
     
  9. Kango

    Thread Starter Active Member

    Sep 22, 2007
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    0

    Thanks, I was thinking all wrong :confused:
     
  10. nomurphy

    AAC Fanatic!

    Aug 8, 2005
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    Of course, this all assumes conventional I/O with input on left and output on right. Otherwise, it's all a guessing game.
     
  11. billbehen

    Active Member

    May 10, 2006
    39
    1
    The assumption seems to be that one is operating from a low impedance source (on the left) to a high impedance load (on the right.) Thus, the output pins can be left open, but the input pins are effectively shorted! Then, from the standpoint of the cap, you have R3 in series with the other two in parallel.
     
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