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The clock stops


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I'll explain the symptom then I'm hoping someone can explain the circuit.

1. Electric 120VAC clock

2. 3 ceiling fans controlled by old fashioned rheostat

3. rheostat wired with nm; line-in black, line-out white, no ground

When I turn the rheostat to run the fans on high speed, the clock stops.

As I slow the fan speed down, the clock runs progressively faster as the fans run slower.

WTF

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I think I know what's happening but let me clarify it for others to understand:

A rheostat is simply a variable resistor. It has only two wires. In this circuit, it's connected between the 120 V supply and the fans. At minimum resistance setting, the fans run fast because the least resistance supplies the highest voltage to the fans.

The two wires of the clock are connected not to neutral and hot but to the rheostat output (which is also white) and hot. So, when the fans are getting the most voltage, so is the neutral on the clock. It stops because both its two wires are at the same potential - 120 V. There's no voltage differental across it to power it.

Marc

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I think I know what's happening but let me clarify it for others to understand:

A rheostat is simply a variable resistor. It has only two wires. In this circuit, it's connected between the 120 V supply and the fans. At minimum resistance setting, the fans run fast because the least resistance supplies the highest voltage to the fans.

The two wires of the clock are connected not to neutral and hot but to the rheostat output (which is also white) and hot. So, when the fans are getting the most voltage, so is the neutral on the clock. It stops because both its two wires are at the same potential - 120 V. There's no voltage differental across it to power it.

Marc

Ok, I get what you're saying except the part that the clock runs when the rheostat is off. What is the clock using for the neutral when the rheostat is in the off position?

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Perhaps the 'off' position on the rheostat is actually a bypass and closes the circuit.

I haven't examined a rheostat in decades but it makes sense that the progression of the dial towards points of lower resistance would end with zero resistance, which is a switch in the closed position.

Marc

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What about this:

The clock neutral uses the fan neutral in series in a different junction box. Both the clock and the fan are powered by the same pole. When the fan is turned on the current for the fan motors seeks a path through the clock receptacle and through its neutral wiring ( all paths) and the potential between the clock hot and the neutral is eliminated. As the rheostat decreases current to the fans, the clock motor starts spinning at a rate proportionate to the voltage difference between the hot and neutral.

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When the rheostat is in the off position, the white wire on the clock remains connected to hot wire on the fans. The fans have no power and do not run but the fan windings have a much lower resistance than the motor windings that turns the clock hands so with the rheostat off, the neutral wire on the clock is able to find the neutral bus via the fan motor windings. With the clock motor so much smaller than the fan motor, nearly all of the voltage is upon the clock, at least enough to run it.

In other words Chad, you're right, the clock is sourcing its neutral through the fan motor windings when the rheostat is in the off position.

Somebody hand this guy a cigar.

Move the clock neutral wire from the rheostat 'line out' to a connection that leads straight to the panel neutral bus.

Marc

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What about this:

The clock neutral uses the fan neutral in series in a different junction box.

That's one way to say it.
Both the clock and the fan are powered by the same pole.

Yes.

When the fan is turned on the current for the fan motors seeks a path through the clock receptacle and through its neutral wiring ( all paths) and the potential between the clock hot and the neutral is eliminated.

I had to read this several times but yes, it's one way to say it.

As the rheostat decreases current to the fans, the clock motor starts spinning at a rate proportionate to the voltage difference between the hot and neutral.

You got it.

Marc

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