[auondicz]

I got some strange voltage measurement when I was repairing an electrical outlet and hope someone can explain this to me.

I have a electrical outlet with hot wire connected and neutral wire disconnected. A lamp with a simple rotary switch at the light bulb socket is plugged in to the socket. Breaker is on.

When I switch the lamp to on (it does not actually turn on becase neutral is disconnected) and measure the voltage between the neutral terminal and ground I got 110V. This is what I expected. As there is no current flowing the voltage drop across the bulb is zero and I am just measuring the voltage of the hot wire.

When I switch the lamp to off I expected the voltage between the neutral terminal and ground to be zero, since the neutral side is effectively disconnected from the hot side through the open switch. But instead I measured about 55V from neutral to ground. I try it again with switch - on and unscrewed the bulb and again got the same thing. For some reason there is a 55V drop across open switch or the open bulb socket.

Can anyone tell me what is happening here?

[tkrussell]

Since the switch is open ,and your testing across the neutral and ground, you must be reading volts across the neutral and ground from another source.

Can you test other outlets? All you need to do is check any unused outlet.

May be a clue to a hidden problem. I don't think the lamp is the issue.

[auondicz]

Actually I found out my digital voltmeter is causing the problem. When I test the above with a cheap analog meter I got the expected results.

I stick one lead from the digital voltmeter into a (normal) hot outlet (the other lead just dangling in mid air) and it shows between 20 to 50V. Is this how digital voltmeter is supposed to behave? Or is mine just busted? Weird!

[bhayne]

Not wierd- expected. The analog meter has a low resistance compared to the higher quality of the digital meter.

The neutral is open, so the voltage is floating. A floating voltage will not drive an analog meter but a digital meter will measure the voltage as half of the supply voltage; 110V/ 2=55V.

[auondicz]

Thanks for clearing this up!