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Capacitive Reactance

  1. Oct 6, 2013 #1
    Hello again, I am trying to knock out some challenge questions in my text (to prepare for my midterms) and I came across this question.

    A 10V, 2.5kHz, sine wave is applied to a series combination of a 0.01µF capacitor and a 10kΩ resistor. What would the total current be?

    The answer in the back of the book says 0.844mA

    I am not getting anything anywhere near that. Can someone please tell me where I am thinking incorrectly? Thanks.

    Here is my thought process.

    First, calculate the capacitive reactance: 1/2πfC = 1/(2πx2500x0.1x10^-6) = Xc = 6366.2Ω

    This is where I think I am going wrong.
    since the capacitor and the resistor are in series, so i added the capacitive reactance and resistance together: 6366.2 + 10,000 = 16,366.2Ω

    I then applied ohm's law to get 10/16366.2 = 0.611mA which is way off

    Using backwards design, I figured 10/0.844mA = 11,848Ω
    which means that in some way shape or form, my total resistance in the circuit should be 11848Ω

    I cannot figure out how to get my Xc of 6366.2Ω and resistor 10k to combine in a manner that would be 11848Ω

    Any insight would be great!

    Thanks!
     
  2. jcsd
  3. Oct 6, 2013 #2

    Drakkith

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    Staff: Mentor

    You don't need to multiply Xc by the square root of -1 or something do you?
     
  4. Oct 6, 2013 #3
    I don't think so? Is there an equation or reason that I should? I don't think this chapter has had any mentioning of imaginary numbers. What was your idea?
     
  5. Oct 6, 2013 #4

    Drakkith

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    Staff: Mentor

    Nothing really, I just looked up Capacitive reactance on wiki and saw an equation using j, which is the square root of -1. But that thing was mentioning phasors and stuff, so I don't know.
     
  6. Oct 6, 2013 #5

    meBigGuy

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    Gold Member

    Since the capacitive reactance and the resistance are at right angles, you need to use the squareroot of the sum of the squares to determine the magnitude of the sum. What is the magnitude of Z = R + jw/C.
     
  7. Oct 6, 2013 #6
    Interesting. I just tried that out and you are absolutely correct! What worries me about this, is that the chapter section on capacitive reactance has NO MENTION of this and there are no sample problems like this. I will have to do some research to figure out why this is...

    I can't even picture why they would be perpendicular to one another.

    Maybe ill ask my TA again next week.
     
  8. Oct 6, 2013 #7

    meBigGuy

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    Gold Member

    That's actually pretty easy. Reactance is modeled by (I hate this term) imaginary numbers, but in simple terms what happens is that there is a phase shift of 90 degrees between the voltage and current caused by the capacitor. The current through the capacitor leads the voltage (essentially it takes time to charge the capacitor) It is described nicely in this tutorial: http://www.electronics-tutorials.ws/accircuits/AC-capacitance.html and there is a nice diagram to demonstrate the effect.

    Inductors have the same effect except the voltage leads the current.

    BTW, for a series LC circuit when the lead in the inductor = the lag in the capacitor you get resonance and the impedance is zero.
     
  9. Oct 6, 2013 #8
    That is perfect! I wish you wrote my textbook! Thank you so much!
     
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