Electronics RLC Ciruit question

Discussion in 'Homework Help' started by jeffrey1027, Nov 9, 2014.

  1. jeffrey1027

    Thread Starter New Member

    Nov 9, 2014
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    [​IMG]
    I started by finding
    ZL = jwL which gives j(672 Ohms)

    Don't know where to go from here.

    TIA
     
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Then you find voltage across your 672j Ohm "resistor".

    Also. RLC circuits are not considered electronics.
     
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  3. jeffrey1027

    Thread Starter New Member

    Nov 9, 2014
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    Would it be
    VL = IL
    =( - 65 x 10^-3 A) (672j)
    = -j 43.68V

    Would I just add that to the Given Vs?
    Vx = Vs + VL
     
  4. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Yes.
    But. Now you need to watch out your polarity.

    VX=VS+VL That is correct. But if your polarity is wrong, you will use VL with wrong sign and get wrong answer.
     
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  5. jeffrey1027

    Thread Starter New Member

    Nov 9, 2014
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    Would it not just be :
    Vx = (-55 + j22 V) + ( 0 - j 43.68 V)
    = -55 - 21.7j V
    ?
     
  6. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Replace the inductor with a voltage source. Place the polarity symbols into the voltage source. Now you have two voltage soruces in series. Do you know how to "add" two voltage sources in series?
     
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  7. jeffrey1027

    Thread Starter New Member

    Nov 9, 2014
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    I think i figured out where the mistake I made
    Vx = -VL + (-Vs)
    = - ( 0 - j 43.68 V) + [- (-55 + j22 V)]
    = + 55 + 21.7j V

    Isn't adding voltage sources the same as adding resistors in series? Req = R1 + R2 +...+ Rn
     
  8. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    The current enters the "resistor" from the bottom and moves up. The terminal where current enters is assumed to be positive. The terminal where current exits the "resistor" is assumed to be negative. The current flows from higher voltage potential to lower voltage potentia.

    So the bottom terminal is positive, upper terminal is negative. But the voltage across the "resistor" is -43.68j volts. What does that mean? It means that our earlier assumptions about which terminal is at higher electrical potential (positive) were wrong. So we will do a simple thing. We will make upper terminal of the "resistor" positive, bottom terminal will be negative, and now instead of -43.68j volts we have +43.68j volts.

    Also now we have two voltages connected from positive pole to negative pole, now we really can add two series voltages: Vx=-55+22j+43.68j
     
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  9. jeffrey1027

    Thread Starter New Member

    Nov 9, 2014
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    Thanks a lot for the help!
     
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