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Adding resistor and capacitor in parallel

  1. Nov 14, 2007 #1
    Hi all,

    I am meant to find the total impedance across the circuit but am having trouble with one section of the circuit. This section comprises of 2 resistors and 1 capacitor in parallel.

    The two resistors are 8 ohms each and the capacitor is 3 ohms. So the process is meant to be:

    Z = ( 1/8 + 1/8 + 1/j3) ^(-1)

    and the answer is meant to be: 1.44 + j1.92

    I know it is really simple but I cant wrap my head around it at the moment and it's driving me insane! Any help much appreciated.
     
  2. jcsd
  3. Nov 14, 2007 #2

    rl.bhat

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    Homework Helper

    The secytion can be written as 4 ohms and 3j ohm are in parallel. Then
    1/Z = 1/4 + 1/3j
    Z = 12j /( 4 + 3j) . Now rationalise the denominator and simplify.
     
  4. Nov 14, 2007 #3
    Thanks rl.bhat.

    I am wondering though how did you get 12j for the numerator? I'm sure it relates to 4 x 3j = 12j somehow but I'm not sure why...
     
  5. Nov 14, 2007 #4

    rl.bhat

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    When you add 1/4 + 1/3j you get (3j + 4)/12j. This is 1/Z. therefore
    Z = 12j/(4 + 3j)
     
  6. Nov 14, 2007 #5
    Thanks a lot man!
     
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