Why Doesn't Stray Capacitance Create Infinite Impedance in a 60 Hz System?

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Discussion Overview

The discussion revolves around the behavior of stray capacitance in a 60 Hz electrical system, particularly why it does not result in infinite impedance when a piece of wire is used in a circuit. Participants explore the implications of capacitive reactance and resistance in this context.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant notes that the formula for capacitive reactance indicates that as capacitance decreases, reactance increases, yet stray capacitance does not lead to infinite impedance in practical scenarios.
  • Another participant suggests measuring the resistance of the wire and calculating the current when connected to a mains supply, emphasizing the importance of the wire's low resistance compared to the stray capacitance.
  • A participant expresses confusion about the mathematical relationship between resistance and reactance at low frequencies, particularly regarding the implications of using these formulas across different frequency ranges.
  • One participant clarifies that the stray capacitance is in parallel with the resistance of the wire, which explains the high current observed in the circuit.
  • A later reply confirms the understanding that the impedance of the stray capacitance is indeed in parallel with the wire's resistance, affecting the overall circuit behavior.
  • Another participant questions the initial capacitance value provided, suggesting that it is likely incorrect and should be lower than stated.

Areas of Agreement / Disagreement

Participants generally agree on the relationship between stray capacitance and resistance, but there is some disagreement regarding the initial capacitance value and its implications. The discussion remains unresolved regarding the exact nature of the stray capacitance measurement.

Contextual Notes

There are limitations regarding the assumptions made about the stray capacitance values and the specific conditions under which the measurements were taken. The discussion also highlights the dependence on definitions of capacitance in practical applications.

JimAlexander
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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
 
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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.
 
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
 
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?
 
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?
 
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?
 
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.
 
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