[donf]

Hey guys, bare with me on this, I'm down with pneumonia and really should be toes up in bed. Actually that is my next destination.

At my Nice's home, I was asked to look at the wall switch because the light circuit failed.

What was actually there is aluminum wire (12 AWG). The wall switch had supply into the bottom of the switch and out at the top using two black conductors, one of which was broken.

One conductor feeds a fan/light fixture.

I fixed the broken conductor and replaced the switch with a Al/CO switch. I replaced the switch because with the switch in the off position I only got 25 VAC between the terminals of the switch.

I also measured between the supply side of the switch and ground. Again I saw only 25 VAC.. With the switch in the closed position, I read 0 VAC, just like I would expect. I opened the breaker for the circuit and with the fed conductors disconnected the was 0 Ohms with the switch in the closed position and infinite in the open. The old switch seemed to be fine but I changed it out anyway. When I reset the breaker, I got 120 VAC between Supply and ground with the switch open. When tested across the switch terminals, 25 VAC.

I also pulled the fan down to the outlet box and found several loose wire nuts and one wire nut that was completely. These were within the fan/light housing.

At the outlet box, there was 25 VAC between hot and neutral with the switch open and 118 VAC with the switch closed.

I do not understand why the "Open" switch shows 25 VAC. It should show 120 VAC.

Right now my plan is to do a pigtail at the switch and only use one conductor to the switch. I also plan to do a pigtail on the exposed ground conductors and run the new ground conductor to the ground screw on the new switch yoke, cap them with a antioxidant wire nut (both pigtails).

I am also planning on installing a AFCI Breaker since this circuit feeds a living room and two bedrooms.

1) Can anyone explain the 25 VAC on the open switch?
2) Does my plan to clean up the circuit connection points make sense?

P.S. I did not check the second switched conductor against ground. I supposed there is a possibility of it being hot and affecting the reading across an open switch? I also have no idea where this conductor terminates. The receptacle outlets in this room are not switched. Nor did I find any device that also worked off the wall switch. It seems to feed only the fan.

[donf]

Any creative ideas. Can I put a tone on the unknown lead and trace it back to a device or is that impractical in this situation?

[tkrussell]

Measuring across a switch is not useful information, and not normally done by a troubleshooter, as you are measuring a parallel circuit across two points in series.

Your test meter is acting as a device in series with the load.

[donf]

TK,

If I measure across an open switch, I should see 120 VAC, because there is a difference of potential between the terminals.

With the switch closed, I should see 0 VAC, because there is no difference of potential is zero.

I'm real hard pressed to believe that a volt meter can simulate a load in a circuit. If it did, then any time you use the meter, you would have to zero out the AC circuits much like you do with the old analog meters when testing for resistance.

However, I do trust you implicitly so I'll keep looking, when I get out and about again.

Question, do you think toning the unidentified conductor on the switched side, to find out where it goes is an effort worth doing?