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High Xc with stray capacitance

  1. Jun 29, 2011 #1
    Hello.

    I know the formula for capacitive reactance is 1/2PIfC. The function says that if frequency is constant, the Xc increases as C decreases. Question is, in a 60 hz system, stray capacitance in wire does not turn a jumper into an infinite impedance. A piece of wire 1 foot long had capacitance of 30 nF (according to my passive component tester). In this formula it creates 88.5K ohms impedance. But if I bend it around and stick it in an outlet there will be a big ball of fire! What am I missing here?

    Thanks,
    J
     
  2. jcsd
  3. Jun 29, 2011 #2

    vk6kro

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    Try measuring the resistance of your piece of wire from end to end.

    Now, work out the current that will flow if you put it across your 120 volt or 230 volt mains supply.

    The stray capacitance is to ground and this still exists, but the very low resistance of such a piece of wire is a lot more important if you use it to short circuit a source of voltage.

    Welcome to Physics Forums.
     
  4. Jun 30, 2011 #3
    Thanks for the reply, but I am still a little confused.

    If the impedance is the sq.rt. of the sum of the squares of the resistance (very low, like .01 ohms) and the reactance (very high), the impedance is almost equal to the reactance itself. In most of my work experience, we use these formulas when dealing with a range of frequencies, usually high ones. Now I don't understand what's happening mathematically when I have to deal with a low frequency and a range of capacitances.

    I appreciate your answer, hope I'll be able to get it :)

    J
     
  5. Jun 30, 2011 #4

    f95toli

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    The stray capacitance is the capacitance to ground, but the resistance is (obviously) a series resistance.
    Hence, your 88.5kOhm stray capacitance is sort of on connected in parallel with a 0.1 ohm resistor. Do you see now why the current is so high?
     
  6. Jun 30, 2011 #5

    f95toli

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    The stray capacitance is the capacitance to ground, but the resistance is (obviously) a series resistance.
    Hence, your 88.5kOhm stray capacitance is sort of on connected in parallel with a 0.1 ohm resistor. Do you see now why the current is so high?
     
  7. Jun 30, 2011 #6
    I think I see it now.

    If I have 2 wires going out to a light bulb, with the stray capacitance between the wires (88.5Kohm in Xc), then this impedance is in parallel with the resistance of the wire (0.5ohm). The formula for parallel resistances says the total impedance will be less than the smaller of these two. So the very-low resistance with the lightbulb in the circuit. Then, if the bulb blows out, the 88.5Kohm impedance would be the capacitive coupling when the DC resistance through the bulb goes to infinity.

    For some reason I got it in my head that the Xc was series impedance, but really it's parallel.

    Does this sound right?
     
  8. Jun 30, 2011 #7

    vk6kro

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    Yes, that is right.

    Incidentally, that capacitance of 30 nF in your first post seems very unlikely.
    It does agree with the 88.5K reactance figure you calculated, but the stray capacitance of such a short piece of wire would be more like 50 pF than 30 nF.
     
  9. Jun 30, 2011 #8
    Thank you everyone for the help.

    I'll check my tester again, or maybe I got the nF/pF mixed up in my memory. I appreciate everyone's willingness to help me.

    J
     
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