Phasor diagrams and RLC curcuits

Discussion in 'Homework Help' started by Silent_64, Dec 19, 2007.

  1. Silent_64

    Thread Starter New Member

    Dec 19, 2007
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    I've been presented with an assignment.

    I've got a series RLC circuit where I've been asked to find the phase angle φ and the generator voltage.

    I know that the voltage over the resistor is 100 V, and over the capacitor it's 200 V and the inductor it is 300 V. The frequency is 50 Hz and the current in the circuit is 2 A.

    I've calculated the generator voltage (Vmax) to be 245 V, unfortunatly I've never dealt with phasor diagrams, and it seems to be the only way to determine the phase angle.

    Any help on how to draw the phasor diagram and/or how to find the phase angle would be much appreciated.

    Thank you
     
  2. agentofdarkness

    Active Member

    Oct 9, 2007
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    Do you know what a Phasor is? I know that I didn't know this when I first did Phasors. Its pretty simple. First start with the current, I. Draw it in the first quadrant at an angle ωt. 90° forward draw VL (voltage of the inductor), this will be in the 2nd quadrant. 90° behind I draw VC (voltage of the capacitor), this will be in the 4th quadrant. You can prove this pretty simply. If I is cos(ωt), then VL = L di/dt = -IωLsin(ωt) = IωLsin(ωt + 90). Similarly, Vc = (I/ω)sin(ωt) = (I/ω)cos(ωt - 90). VR is along I so it will be right next to I Phasor(should actually occupy the same space but you can't draw that). The magnitude was different though. V = cos(ωt + ϕ) so the V phasor will be at an angle ϕ from the I phasor.
     
  3. Silent_64

    Thread Starter New Member

    Dec 19, 2007
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    Well, isn't a phasor is just a vector with a magnitude instead of coordinates? That's at least how i look at them.

    And I'm still having trouble drawing it, I just can't seem to quite get my head around it. It really bugs me.
     
  4. agentofdarkness

    Active Member

    Oct 9, 2007
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    That is correct. The phasor is basically a vector with a magnitude. The magnitude of the phasor is the amplitude of the voltage/current you are drawing. Like I(t) = IoCos(ωt). The phasor will have a length of Io. The angle of the phasor will be ωt with respect to the x axis. Draw the magnitude of I(Io) on the x axis. Turn the line ωt degrees counter clockwise (toward the +Y axis). That will be the phasor for I. You can get the other phasors similarly. I would draw it out but I don't have my scanner.
     
  5. recca02

    Senior Member

    Apr 2, 2007
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    i havent solved it, but will it be 245?
    i wud like to be corrected.
     
  6. Silent_64

    Thread Starter New Member

    Dec 19, 2007
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    Recca you might be correct, I've just typed it in again, and now I'm getting 141.421 V, it might be a parentheses I forgot the first time. Can anyone confirm that?

    Can I assume that the the VL will be -90° and VC +90°, shouldn't it depend on the frequency? Also I'm assuming that I set ωt to be any angle I like.
     
  7. recca02

    Senior Member

    Apr 2, 2007
    1,211
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    141.4 is correct i think.

    for a purely inductive and capacitive components yes, it wont depend on frequency.
    only the voltage magnitude will change with frequency(since reactance changes with frequency).

    as a piece of advice do google terms like 'phasor diagram for rlc' etc to have a better grasp of subject.

    your required phase angle is the angle between the voltage across resistance and the total voltage you calculated.
    find it using V(apparent or total)*cos(phase angle)= V(across resistance)
     
  8. Silent_64

    Thread Starter New Member

    Dec 19, 2007
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    But solving V(apparent or total)*cos(phase angle)= V(across resistance) just leaves me with an unknown constant.

    x=2*@n1* pi + arccos( (100)/(141) ) or x=2*@n1* pi - arccos( (100)/(141) )
     
  9. recca02

    Senior Member

    Apr 2, 2007
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    why bother with the 2npi?
    in vectors/phasors only 0-360 rotation is taken into consideration.
    we dont say V lags I by 370 deg. we say V leads I by 10 deg.
     
  10. Silent_64

    Thread Starter New Member

    Dec 19, 2007
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    That's what I get from typing it into my TI-89, the @n1, is mearly an unknown constant. Which in this case means that the result is no good, as I need an angle and not something with an unknown constant in it. Are you sure that the forumla above the, V(apparent or total)*cos(phase angle)= V(across resistance), is correct?
     
  11. Silent_64

    Thread Starter New Member

    Dec 19, 2007
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    I got ahead of myself, 45 deg. Sorry guys, I'll go off and do something useful now. Thanks a lot.
     
  12. techroomt

    Senior Member

    May 19, 2004
    198
    1
    several phasor diagrams can be plotted. and the nice thing is that once theta is found for one, it will be the same for all. imagine an x y graph: x axis on horizon with positive to the right, and y axis on the vertical with positive to the top. the positive x axis (to the right) is the reference axis and 0 degrees. this is where resistance values (and in-phase values) are placed. angular rotation is counter-clockwise from there. you can plot the capacitive reactance phasor on the positive y axis. calculate the Xc = 1 / 2 pi F C and plot. the inductive reactance is plotted on the negative y axis. calculate Lc = 2 pi F L and plot. subtract the smaller reactance from the larger, and plot the difference as a phasor on the axis that was larger. you now have the resistance and resulatant reactance phasors that can be used in trig (sin, cos) formulaes for calculating the impedance (total opposition of circuit) and theta, which was formed from the reference axis to the resultant impedance phasor. realize that the most angular displacement any reactance can have is 90 degrees from the reference, so all phasors are in quadrants 1 and 4. the same plotting process can be performed with voltage. but not current because the current plot would have all phasors on the reference axis because they are all in phase in a series circuit. if you were plotting a parallel circuit the voltage phasors would all be on the reference axis as they would all be in phase.
     
  13. recca02

    Senior Member

    Apr 2, 2007
    1,211
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    remember for finding power factor.
    use
    cos(phi)= voltage across resistance/total voltage...(resistance has to be total resistance of ckt.)
    or,
    cos(phi)= total impedance/resistance
    or,
    tan (phi)= reactance/resistance
    all this is from simple trigonometry and can easily be found by drawing phasors.
     
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