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Find the total Impedance Z

  1. Jan 5, 2012 #1
    1. The problem statement, all variables and given/known data

    We have a resistor R1 in series with a capacitor C1, then we have a capacitor C2 in parallel with a resistor R2

    I dont want to achieve the whole question, just the start as it is confusing me.
    2. Relevant equations

    1/Rz = 1/r1+1/r2 (parallel)

    3. The attempt at a solution

    Total Impedance of the circuit= (R1+Jwc1)+ (R2//C2)

    Zt = (R1+jwc1)+(1/(1/R2+1/jwc)

    I am presuming 1/zt = 1/r1+1/jwc

    zt = (R1+jwc1)+(1/(1/R2+1/jwc)

    zt = (R1+jwc1)+(R2/(1+jwc2R2)

    This is as far as I want to go. I believe all is right so far. If not please let me know.

    My question starts with zt = (R1+jwc1) I thought (R1-jwc1) would be correct. What I mean is -Jwc or +1/jwc. Could anyone enlighten me?

    See below to for more information if necessary from a previous question.

     
  2. jcsd
  3. Jan 5, 2012 #2

    NascentOxygen

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    Why presume? If you can't confirm it, then to proceed further is likely to be a waste of time and effort.

    So start with what you do know. What is XC for a capacitor? If you haven't yet memorized it, look it up. Is there more than one way to write it?
     
  4. Jan 5, 2012 #3
    Thanks for looking at it, but rather than presume, I was meaning that this is what has been done to get to the next stage.

    The start is o.k, the problem is as mentioned. The left hand side does not make sense to me and I am trying to figure out why this has been done. If you have any ideas that would be great. Also see the other questions comments if necessary.

    XC = 1/wC
     
  5. Jan 5, 2012 #4

    NascentOxygen

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    Shouldn't there be a j in that expression for Xc?
     
  6. Jan 5, 2012 #5

    tiny-tim

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    hi lubo! happy new year! :smile:

    nooo, that would be the impedance for an inductor circuit …

    Z = (R1+jωL1)+(1/(1/R2+1/jωL2)

    for a capacitor, you must always use ZC = 1/jωC (= -j/ωC)

    (as opposed to ZL = jωL for an inductor)

    so, inside that big bracket, you should have 1/(1/jωC), = jωC :wink:

    (which incidentally is why two or more capacitors "add" when they're in parallel)

    try again! :smile:

    (btw, what is the "t" in zt ? :confused:)
    mmm … they're not the same, are they? :wink:
     
  7. Jan 5, 2012 #6
    Yes, thanks,

    ZC = 1/jωC (= -j/ωC)

    happy new year, to you also "Tiny"-Tim with "big" answers :smile:

    That is what I thought, the answer must be wrong. It goes like that for a long time, so the autor definately thinks it is right?

    If the full question is: show that the I(current) out for the cct "as above" is given by... Iout =V/Zt would that make any difference?

    I think I have got what you mean for the right hand side (RHS). Thanks.

    Ztotal = (R1+Jwc1)+ (R2//C2)

    Zt = (R1+jwc1)+(1/(1/R2+1/1/jwc)

    Only taking the RHS:

    I am presuming 1/zt = (1/R2)+(1/(1/jwc))

    zt = (R1+jwc1)+(1/(1/R2+jwc2) I didnt mean to put the 1/jwc2 here

    zt = (R1+jwc1)+(R2/(1+jwc2R2)


    Impedance "Total(t)" I should be more clear, its easy to forget when you are caught up in the question.

    mmm … they're not the same, are they? :wink:[/QUOTE]

    … What I mean is -J(1/wc) or +1/jwc …

    Thanks.
     
  8. Jan 5, 2012 #7

    tiny-tim

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    hi lubo! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    i can't see what difference that makes … Itotal is always equal to V/Ztotal :confused:
    shouldn't it be 1/jwC1 ?
     
  9. Jan 5, 2012 #8
    I have found the X2 tag, thanks.

    Sorry about that, yes it should be on the Left hand side (LHS) "(R1+(1/jwc1))", I hope this is what you meant. All I wanted was confirmation that the answer I had in my book was not correct. It is frustrating looking at something that should be correct and it is not. It is another to guarantee that without some help! No need go any further at this time.
    Thanks again.
     
  10. Jan 5, 2012 #9

    NascentOxygen

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    Glad you are sorted. But I can't refrain from pointing out where you have been typing 1/jwC there are many who contend you should be in the good habit of typing 1/(jwc) since that is what you mean here.

    We have all found there comes a time when learned sloppiness has been our undoing. :grumpy:
     
  11. Jan 6, 2012 #10

    tiny-tim

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    aren't they the same? :confused:
     
  12. Jan 6, 2012 #11

    NascentOxygen

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    1/2mv2 is not the same as 1/(2mv2)
     
  13. Jan 6, 2012 #12

    tiny-tim

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    erm … yes it is! :redface:

    if you want to write the characters in that order, you should put a bracket round the fraction, as (1/2)mv2

    (alternatively you can type the fraction as a single character, as ½mv2,

    or you can leave a gap, as 1/2 mv2)
     
  14. Jan 8, 2012 #13

    NascentOxygen

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    No need for brackets here, because multiplication does not have precedence over division.

    Gaps are not recognized mathematical operators. They mean nothing, fortunately.
     
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