I am not understanding this behavior, and I'm hoping that someone with a lot more experience with wiring than I have will know immediately what's going on.
I have two 1940s-era fabric-clad wires attached to a simple two-pole on/off toggle switch in the wall. There's no ground wire. The switch controls a light on the wall.
I know which breaker in the basement controls it. I can flip the breaker off and then test the wires for power with a voltage tester, and see that it's dead.
So, with the power enabled:
When the light switch is in the OFF position, if I touch the tester prongs to the wires, the tester lights up.
When the switch is in the ON position, if I touch the tester prongs to the wires, the tester does not light up.
That is counterintuitive to me, so I flipped the breaker off, reversed the wires on the switch, and turned power back on.
But the behavior remains the same.
Why does turning the switch ON cut the circuit?
Best Answer
If you consider this simplified diagram, showing a light & switch, with the switch in the off position on the left, and in the on position on the right:
When the switch is OFF, its contacts are open - preventing current from flowing through the circuit to the light bulb.
When the switch is ON, its contacts are closed - allowing current to flow through the light bulb.
Your "voltage tester" is measuring voltage, not current.
When the switch is on the ON/closed position, your "voltage tester" measures 0V across the switch because that is exactly the case - a closed switch is a short-circuit and even though there may well be current flowing through it, there is no voltage across it.
If you were to put your "voltage tester" across the light bulb instead, you would find exactly the opposite case.
With the switch OFF/open there will be 0V across the light bulb and when the switch is ON/closed you will measure a voltage across the light bulb.