Voltage drops across RL series/parallel circuit

Discussion in 'Homework Help' started by wildnixon, Jun 4, 2008.

  1. wildnixon

    Thread Starter Member

    May 1, 2008
    12
    0
    I have a homework question regarding the total voltage drop across the resistance in the attached circuit. There are two parallel inductors that come before the resistors, and so I must figure out the voltage drop across this before I can determine the drop across the resistors. I am given the source voltage (5 VAC) and the frequency of 10 KHz. Do I use tau to get to the drop across L? I know that Vs=VL +VR; but how do I get VL? I know it's probably staring me right in the face; just not sure.
    Any advice is appreciated.
    Thanks,
    Deanna
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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  3. Ratch

    New Member

    Mar 20, 2007
    1,068
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    wildnixon,

    Tisk, tisk. You are calculating the impedance of the 3 inductors as if all of them were in parallel. Only two of them are. Same for the resistors. The impedance of L3 (63 ohms) gets added to the parallel combination of the L1 and L2 impedances. Same for R1 and R2||R3. Once you get the total impedance of the series circuit, divide it into the voltage and obtain the current. From the current, you can get the voltage across any of the circuit elements such as the resistors. Ratch
     
  4. Ken89wh

    New Member

    May 2, 2008
    8
    0
    wildnixon,

    Ratch is correct!

    And since VR=(I)*RT:
    VR=(VT/((L1||L2)+R1+(R2||R3)+L3)) * (R1+(R2||R3))

    This is because I = VT/Z
     
  5. wildnixon

    Thread Starter Member

    May 1, 2008
    12
    0
    Yes, I realized that fortunately before I turned it in last night! Thanks for all of your help!
     
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