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How to draw this?

Discussion in 'General Electronics Discussion' started by EFDG, Jun 15, 2016.

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  1. EFDG

    EFDG

    33
    0
    Jun 11, 2016
    Hello,
    How to convert polar form to rectangular with drawing axis.
    Xc =31,847ohm/ 90 degree
    actually i was simulating a Sin wave form source to integrator in op amp.
    the current is 2uA and the impedance of Xc is Xc =31,847ohm/ 90 degree.
    so, i need help for learning it and do calculation fast.
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    you need a protractor and a ruler.

    do you understand that one is an angle and a distance? Do you know how to plot that?
     
  3. EFDG

    EFDG

    33
    0
    Jun 11, 2016
    Hello,
    It is 90 degree right angle triangle what is the 31,847ohm base or hypotenous?
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    I guess the answer is NO.

    Do you have a textbook?
     
  5. EFDG

    EFDG

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    0
    Jun 11, 2016
    drwan in paint i will understand
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    0 degrees (0 radians) is a direction from the origin going in a positive direction on the X axis.

    90 degrees (pi/2 radians) is a direction from the origin going in a positive direction on the Y axis.

    180 degrees (pi radians) is in a direction from the origin going in a negative direction along the X axis.

    and so on...

    A polar coordinate is a pair r and theta where the point is plotted r units along a line at an angle of theta from the X axis as described above.
     
  7. EFDG

    EFDG

    33
    0
    Jun 11, 2016
    ok, how to plot in triangle?
     
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    what triangle do you mean?
     
  9. EFDG

    EFDG

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    0
    Jun 11, 2016

    How to convert polar form to rectangular with drawing axis.
    Xc =31,847ohm/ 90 degree
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    so, do you want the real and imaginary parts of this resistance, or do you want to plot I on a graph?

    and is this homework?
     
  11. EFDG

    EFDG

    33
    0
    Jun 11, 2016
    I have left /forget some basic so, i need how to drawn it both imaginary and real part
     
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Plotting the point gives you the real part along the X axis and the imaginary part along the Y axis.
     
  13. EFDG

    EFDG

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    0
    Jun 11, 2016
    how to drawn sqrt (R²+Xc²)
     
  14. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    I'm afraid I'll have to give up.

    I really don't understand what you are asking for.
     
    EFDG likes this.
  15. EFDG

    EFDG

    33
    0
    Jun 11, 2016
    ok
     
  16. hevans1944

    hevans1944 Hop - AC8NS

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    Jun 21, 2012
    On the Cartesian coordinate system, the vertical axis represents reactance. Positive reactance is inductive reactance. Negative reactance is capacitive reactance. The horizontal axis represents resistance. Positive resistance is on the positive horizontal axis. Negative resistance is normally not a consideration with passive resistor, inductor and capacitor components. You plot a capacitive reactance as a point on the negative vertical axis. There is NO TRIANGLE if there is no resistor involved.

    If you have resistance AND reactance you plot these on the horizontal and vertical axes respectively. Then draw a horizontal line parallel with the horizontal axis through the point (reactance) plotted on the vertical axis. Next, draw a vertical line parallel with the vertical axis through the point (resistance) plotted on the horizontal axis. The intersection of these two lines is a point that represents the (complex) impedance. The magnitude of the impedance is the square root of the sum of the squares of resistance and reactance, which is also the hypotenuse of the right triangle formed by the point at the origin, the point on the horizontal axis, and the point at the intersection of the two lines. This impedance value will be the distance of the intersection of the two lines from the origin. The arc-tangent of the ratio of the reactance divided by the resistance is the angle between the impedance vector and the horizontal axis, measured counter clock-wise from the positive horizontal axis.

    Example: A resistance of +3 ohms and a reactance of +4 ohms (inductive reactance) results in an impedance of 5 ohms, (3 +j4) in rectangular coordinates, making an angle of 53.1 degrees with the horizontal axis. If the reactance is negative, the triangle is below the horizontal axis -53.1 degrees, or 306.9 measured counter clock-wise from the horizontal axis.
     
  17. ElectronicsR

    ElectronicsR

    72
    1
    Mar 23, 2016
    ok, thanks sir this was what i am asking to understand.

    why?
     
  18. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    because it makes the math work.
     
  19. ElectronicsR

    ElectronicsR

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    1
    Mar 23, 2016
    please give example to know it better?
     
  20. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Inductor: v=L*di/dt: current lags voltage (or voltage leads current, which is the same from another perspective)
    Capacitor:iI=C*dv/dt: voltage lags current
    The phase difference between voltage and current shows as a phase angle in the diagram. Depending on the component this angle is either positive or negative.
     
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