It is known that:
- AC current can flow through capacitors
- A wire has some inherent capacitance
- A capacitor is the same as an open circuit with plates at either end, and the size of the plates corresponds to the capacitance
So if you have a circuit like this:
then it is equivalent to a circuit like this:
where C1 is miniscule, and equivalent to the capacity of the wires. If current can flow through circuit 2, why can't it flow through circuit 1 using the same logic? Does this imply that an open switch doesn't prevent the flow of electricity?
Edit
Andy Aka said AC current does flow through an open switch. So can it be said that:
a) it is impossible to turn a lightbulb off
and
b) it is impossible to turn a lightbulb on
My explanation for statement a) is that, if an open switch is identical to a closed one, in that current can flow through it, then it should be impossible for an open switch to stop electricity from flowing through a lightbulb.
For statement b), given that the following two circuits are equivalent, that current should flow over the gap in circuit 3 instead of through the lightbulb, since that open switch has no resistance to the AC current. Therefore the lightbulb would never be powered.
Since these two statements are obviously false, the only explanation I can think of is that either Andy Aka's answer is false, or that an open circuit still has a finite resistance. (ignoring sparking)
Best Answer
This is your question: -
However, you have made a contradiction in saying this: -
If both wires have capacitance then an AC current will flow because it's no longer a true open-circuit.
Correct. An open circuit (real) switch will have open-contact capacitance and, although it might only be 0.1 pF, an AC current will flow if there's an AC voltage across the switch contacts.
So, how much current might flow? If we take a UK household supply of 230 volts, 50 Hz and, we assume the open-circuit capacitance is 0.1 pF then, the current that flows is this: -
$$I_{RMS} = 230 \times 2\pi\times 50\times 0.1\times 10^{-12} = 7.23\text{ nA}$$
As one would expect it's not a lot of current because, a 0.1 pF capacitor creates an impedance of around 32 GΩ. If it were a bigger capacitor, the current would be proportionately bigger.