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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?

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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:

schematic

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.

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V = I . R

This is Ohm's law where

V is the voltage across the switch
I is the current flowing through the switch
R is the resistance of the switch (~0 when on, ~infinite when off)

enter image description here

Explanation 1

When the switch is off, the resistance of the air gap is very high, so the voltage dropped across that air gap is nearly full supply voltage

When the switch is on, the resistance of the switch is nearly zero, so the voltage is almost zero.


Explanation 2

Your voltage tester probably has a high resistance - typical multimeters have a resistance of 10000000 ohms, an incandescent lightbulb in a room-light has a resistance of maybe 10 ohms when cold, maybe 100 when hot. This means the bulb in your tester is designed to work with very small currents (e.g 10 mA) when compared with the currents a normal 120V 60W incandescent bulb needs to light up fully (e.g. 500 mA).

With the switch open, electric current flows through your tester, then through the room-light and back to the supply through the neutral wires. The current is enough to light the small bulb in your tester but not enough to light the bulb in the room light.

With the switch closed, all the current flows through the switch because it has a very low resistance and almost none flows through the high-resistance tester - not enough to light the tester's bulb. The "on" switch bypasses the tester.


Explanation 3

Try the old water/pipe analogy. voltage is like pressure (e.g. PSI), current is like flow-rate (e.g. gallons/second) a switch is like a valve. If a valve is off, the full water supply pressure is exerted across the valve. If the valve is fully open there is very little difference in pressure across the valve. Pressure drops across a partly closed valve because it constricts the flow.

  • Might want to extrapolate the symbols there. "I" in particular is not intuitive. "V" is better than "E", though. :) – isherwood Feb 9 '17 at 22:07
  • I'm not following what you mean with "the voltage dropped across". – TRomano Feb 9 '17 at 22:14
  • @TRomano. Try the old water/pipe analogy. voltage is like pressure (e.g. PSI), current is like flow-rate (e.g. gallons/second) a switch is like a valve. Pressure drops across a partly closed valve because it constricts the flow. If a valve is off, the full pressure is exerted across the valve. If the valve is fully open there is very little difference in pressure across the valve – RedGrittyBrick Feb 9 '17 at 22:15
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Perfectly reasonable, actually.

All you are measuring is the voltage between the switch terminals.

When the switch is on, it directly connects both sides of the switch to each other. The voltage between them must be zero since they are connected.

One of the terminals is always hot. Therefore the other terminal must also be hot if the switch is on.

You are not measuring the voltage compared to earth or neutral, since neither one is present in the box.

So with the switch off, you are looking at hot on one side, and the bulb on the other, which is connected to neutral and not connected to anything else. 120V wants to go through the switch, but can't, which is why there's a 120V difference.

It's like if a water valve is shut, there is 0 presssure outside and full pressure inside the pipe.

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