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Capacitor to be connected to increase power factor!

  1. Aug 28, 2011 #1
    A resistance 'R' ohm and inductance of 'L' henry are connected across 240V, 50Hz supply. Power distributed in the circuit is 300W, and the voltage across R is 100V. In order to improve the power factor to unity, the capacitor to be connected in series should have a value of __.
    a) 43.7 uF b) 4.37 uF c) 437 uF d) 0.437 uF

    Step 1) Using P=V/I, got I=1.25A
    Step 2) Using R=VR/I, got R=80 ohm
    Step 3) Now since total voltage is 240V, VL is 240-100=140V
    Step 4) Using XL=VL/I, got XL=112 ohm.
    Step 5) For power factor to be unity, XL=XC, which means that XC should also be equal to 112 ohm, which leads me to the answer that C is 28.4uF. (since XC=1/2πfC)

    Where did I go wrong? Any help would be appreciated. :wink:
    Last edited: Aug 28, 2011
  2. jcsd
  3. Aug 28, 2011 #2


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    First line (ignoring the typo. P=V/I), you used P=VI to find I. This is incorrect, VI is the apparent power if you use the total voltage. If however you just use the voltage across the resistor ...
  4. Aug 28, 2011 #3


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    Last edited by a moderator: Apr 26, 2017
  5. Aug 28, 2011 #4


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    BTW. It is a poor exercise to refer to this a merely "power factor correction". Series resonating out a reactive element like this does a lot more than merely correcting power factor. In this case for example it increases the total power by a multiple of about 6. If the load was designed to run at 300W then it would likely be destroyed.

    This is why PFC is normally done with a parallel capacitor.
  6. Aug 28, 2011 #5
    Why is it necessary to calculate the impedance?
    Btw, are all my other steps right?
  7. Aug 28, 2011 #6


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    Because you want to find the reactance "X". So find the resistance "R", find the impedance magnitude |Z| and use R^2 + X^2 = |Z|^2 to solve for X.
  8. Aug 28, 2011 #7
    Thanks for your help, uart and tim! :)
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