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

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Stray capacitance in a 60 Hz system does not create infinite impedance due to the low resistance of the wire, which significantly influences the overall current flow. The calculated capacitive reactance of 88.5K ohms is in parallel with the wire's low resistance, resulting in a total impedance that is much lower than the reactance alone. When a light bulb is connected, the low resistance dominates, and if the bulb fails, the capacitive coupling becomes more relevant. The initial assumption of stray capacitance being 30 nF is likely incorrect, as it should be closer to 50 pF for a short piece of wire. Understanding the relationship between resistance and reactance clarifies why high current can occur despite the presence of capacitance.
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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
 

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