Rl circ

Discussion in 'Homework Help' started by caique221, Nov 11, 2013.

  1. caique221

    Thread Starter New Member

    Nov 10, 2013
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    Sorry about the title. I couldn't post, so I read somewhere that a big title could be the cause. I had to make it that short.


    I'm having a hard time with a basic AC RL circuit. Since only the voltage drops are given, I don't have a clue where to start from.

    They are asking for:

    1) Source voltage in polar notation;
    2) Power factor of the circuit;
    3) Phasor diagram with source voltage, voltage drops and current.

    The source voltage frequency can be adopted as 60Hz.


    I would like to know just what should I do with those voltage drops. I can't think of anything.
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    Since all of the components are in series, what you do know about the current in the circuit?

    What do you know about the phase angle for the voltage across a resistor relative to the current through it?

    What do you know about the phase angle for the voltage across an inductor relative to the current through it?
     
  3. adeboy

    New Member

    Dec 12, 2012
    17
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    you can use the face that the circuit is in series to draw the phasor diagram since there is a common current passing through the elements and
    use CIVIL acronym for reactive elements
     
  4. caique221

    Thread Starter New Member

    Nov 10, 2013
    7
    0

    The current is the same along the circuit; the voltage across the resistor is in phase with the current and the voltage across an inductor is 90 degrees ahead of the current going through him. Still, I don't know how to get the magnitude of the current with that.
     
  5. caique221

    Thread Starter New Member

    Nov 10, 2013
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    0

    Sorry, but what did you mean by civil acronym for the elements?
     
  6. WBahn

    Moderator

    Mar 31, 2012
    17,737
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    So let's use the current as our reference and say that it is I@0° (magnitude I at an angle of 0°).

    What is the phasor for the voltage across the resistor?

    What is the phasor for the voltage across the capacitor?

    What does KVL require that the voltage across the source be?
     
  7. WBahn

    Moderator

    Mar 31, 2012
    17,737
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    It is one of a handful of acronyms that people have come up with to help them remember things they don't understand.

    Or, that they use to help them recall basic conclusions about things they do understand and could figure out but don't want to waste time over and over figuring it out from scratch each time they need it.

    So it's either good or bad, depending on why it is being used.

    In this case, you think of CIVIL as being two peices: CIV and VIL

    CIV - In a C, the I leads the V
    VIL - The V leads the I in an L

    Another is ELI the ICE man.

    ELI - The E leads the I in an inductor
    ICE - The I leads the E in a capacitor
     
    Last edited: Nov 13, 2013
  8. LvW

    Active Member

    Jun 13, 2013
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    100
    As the forum member #12 has stated in another thread:
    "Still learning something almost every day on this site."
     
  9. tbinder3

    Member

    Jun 30, 2013
    30
    1
    I can't view your schematic, but adding voltage drops will give you your total voltage. From there, find total current.

    Polar Notation is a complex number, a Magnitude and Vector angle.
    Example: 12v<0° The 12v being your magnitude, and 0° being your vector angle.

    When solving for total current, you must divide two complex numbers. I=V/Z
    Hoping that you know how to solve for impedance.

    In my example i'm using the voltage source as a reference, at 0°
    Continuing my example: Giving us 12v<0° / 12Ω<30°

    Now, when dividing vectors, you divide the magnitudes (12v / 12Ω =1A) AND subtract the angles (0° - 30° =-30°). Giving us I total= 1A<-30°.

    Just want to point out that, when subtracting angles, it is common to subtract by a negative number, example: (0° - -30°) resulting in a positive angle.

    If you were to add or subtract a vector, you would have to put it in Rectangular Notation first. Rectangular notation has 2 parts as well, Real and "Imaginary" . It will look something like this (-1 + J6) The "-1" being the real part, and "J6" being the so-called "imaginary" part.

    When converting a Polar Notation number to Rectangular Notation, I memorized this example: C(magnitude)<θ°(angle) (polar notation)
    The Real part= C * cos(θ) or Magnitude * cos(Angle)
    Imaginary= C * sin(θ) or Magnitude * sin(Angle)

    Example: 1A<-30°
    Real = 1 * cos(-30) = .866
    Imaginary = 1 * sin(-30) = -.5

    Answer= .866<-.5°

    Of course there is a lot going on, and a lot more to understand, and i'm sure i missed basic concepts, but this was put together quickly as an example.
     
  10. tbinder3

    Member

    Jun 30, 2013
    30
    1
    Very nice example. Same that was used when I was in school.
     
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