RLC phasor diagram

Discussion in 'Homework Help' started by xx123321, Apr 27, 2011.

  1. xx123321

    Thread Starter New Member

    Apr 27, 2011
    15
    0
    it is a series RLC circuit.

    VR(voltage of resistor) is given by 10sin(ωt-53.13°)
    VL(voltage of inductor) is given by 20sin(ωt+36.87°)
    VC(voltage of capacitor) is given by 10sin(ωt-143.13°)

    How about the phasor diagram of E,VR,VL,VC?
     
  2. bluemarvin

    New Member

    Apr 19, 2011
    8
    2
    I'm not completely sure about this, but I think you draw 3 arrows head-to tail. The first arrow, V(R) starts at the origin of a cartesian graph, is 10 units long, and is at an angle 53.10 deg below the X-axis. The next V(L) arrow is 20 units long, is attached to the R arrowhead, and points up 36.87 deg above horizontal. The last arrow (for V(C)) is attached to the head of the L arrow, is 10 units long, and points 143.13 deg clockwise (down) from horizontal. Your E(t) arrow starts at the origin and points to the tip of the C arrowhead...it's voltage has to equal the sum of the 3 series components. You will have to do the trigonometry to figure out the equation for E(t). Also keep in mind it is helpful to set up a spreadsheet with these equations across 3 or 4 sine wave cycles ---you can plot the sum E(t) against the constituent voltages and see how it all works.
     
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  3. xx123321

    Thread Starter New Member

    Apr 27, 2011
    15
    0
    one more question, if e(in) is given by 100sinωt.
    I need to calculate the total power,and I wonder the answer whether is VR^2/3?
     
  4. bluemarvin

    New Member

    Apr 19, 2011
    8
    2
    Average power would be .707 * 100 *R since the L and C give back all the power they soak up, therefore power would be dissipated only by the R. Instantaneous power will be dependent on the phase of the cycle, but would be [(100*sin(ωt))^2]/[R+jωL+(1/jωC)]. Ever looked through 'The Art of Electronics' by Horowitz and Hill? It's got lots of good info.
     
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