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Voltage divider using capacitor question

  1. Oct 11, 2009 #1
    I'm calculating the size of a capacitance I need to drop a certain voltage at a certain current for a tube filament, and doing it two different ways gives two different answers. For example:

    Tube needs 13 volts across it at 0.725 amps current. I pretend that the tube's heater is a resistance with an equivalent resistance of 17.93 Ohms. I then do a calculation to see what size capacitor I need to give the required current from the 120 volt line.

    [tex]120 V = 0.725 A * 165.5 Ohms [/tex]

    I want the vector sum of the capacitive reactance and resistance of the tube heater to equal this..

    [tex] \sqrt{17.93^2 + X_c^2} = 165.5[/tex]

    [tex]X_c = 164.5[/tex]

    [tex]\frac{1}{2\pi (60)(C)} = 164.5[/tex]

    So C is equal to about 16 uF.

    Now I do it this way: I assume the resistance of the filament is the same at 17.93 Ohms, and I calculate the capacitance I need for a voltage divider to give 13 volts from the 120 volt line across the tube.

    [tex]\frac{17.93}{X_c + 17.93}*120 = 13[/tex]

    When I do it this way, the capacitive reactance is equal to about 148 ohms, and I get about 18 microfarads as the size of the capacitor needed. I'm wondering if anyone can tell me what the reason for the difference is and where I've gone wrong in my thinking? Or maybe there's just roundoff error?
     
    Last edited: Oct 11, 2009
  2. jcsd
  3. Oct 11, 2009 #2

    vk6kro

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    Science Advisor

    Yes. The second method is invalid. You can't add vectors like that unless they at the same phase angle and these are at right angles to each other.

    Your first method shows you know how to do it. Much better :)
     
  4. Oct 11, 2009 #3
    I see my mistake now. I think the second method would work also if I made sure to treat the reactance of the capacitor as the imaginary part of a complex number and then took the magnitude of the equation to get the voltage division ratio. I should know better! :D
     
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