Which leg from the center tap is it? (Electrical)

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When power is sent to a home it comes in as 240v split by the center tap of the pole transformer, giving 120v on either side. Is there any way to determine which side of the center tap a particular outlet or other 120v device is connected, other than tracing wires and opening the breaker box? I'm mostly referring to using some sort of test gadget.
I'm asking this because I have a device that sends a signal thru the power lines and I discovered that it only operates if the transmitter and receiver are on the same side of the center tap. This is not that hard to determine if it's in the same building, but I'm running to another building on the same electrical system, but the wires are impossible to trace visually, since they are an overhead twisted triplex cable.
Short of physically disconnecting one leg of the mains, there dont seem to be any other method, unless there is some tester which can determine it. I'd prefer having to disconnect the mains, which would require pulling the meter and all of that.....
Anyone???
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On 8/31/2012 5:17 AM, snipped-for-privacy@thecave.com wrote: ...

...
It's one of two -- connect the powerline datalink to one in the far building and it'll either be right or wrong. That determines which is which.
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That was what I was thinking too. From a practical standpoint, the only thing that matters if it works or not, no? So, why not just use the actual gear and see if it works on whatever outlets it's possible to use?
Otherwise I don't know of any way to figure out which is which in two seperate buildings if they are fed from the street via seperate feeds that are not traceable
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On Fri, 31 Aug 2012 05:17:34 -0500, snipped-for-privacy@thecave.com wrote:

Turn off every other breaker (checkerboard) and see if the outlet is on/off?

Oh, you have two entrance panels. Just fix the problem. There are x-10 devices (basically a line-rated capacitor and enclosure) that are made to go into the panel and act as a bridge between the two legs.

Measure the voltage between the outlet hots (0V=same leg. 240V=opposite)? If you do this in one location, along with the "checkerboard" breaker test, above, you can extrapolate map all outlets.
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On Aug 31, 9:41 am, " snipped-for-privacy@att.bizzzzzzzzzzzz"

He has two seperate buildings. That method will work within each building, but not between the buildings. He could run an extension cord between buildings, if that is feasible and then use this method.
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On Fri, 31 Aug 2012 06:45:10 -0700 (PDT), " snipped-for-privacy@optonline.net"

Right, you need a wire between points to measure the voltage. ;-)
If it were only one breaker panel there would be no need to measure anything. Just turn every other breaker off and see what's live/dead.
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On 8/31/2012 9:45 AM, snipped-for-privacy@optonline.net wrote:

The simplest and best way is to install an X-10 *coupler repeater*. That way if the signal works on one it will work on both phase legs and the amplifier will boost the signal. They sell X-10 couplers which are nothing more than a non-polarized capacitor which do not work well.
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On 08/31/2012 11:04 AM, George wrote:
[snip]

I tried just a capacitor. It seemed to have no effect at all on X10. An active coupler helped a little. You really need the repeater.
However, did the OP say he was using X10? Maybe that repeater is repeating the wrong thing.
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On 8/31/2012 3:14 PM, Mark Lloyd wrote:

If you think about how a capacitor works on an AC waveform in this case you can see why.

Maybe, I see the OP never disclosed what the device might be so who knows.
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wrote:

You guys keep talking about an X10. What the hell is an X10 anyhow?
As I said, what I have is an Instajack. It's purpose is to transmit a landline phone signal to another location.
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On Tue, 04 Sep 2012 03:49:13 -0500, snipped-for-privacy@thecave.com wrote:

The most common carrier current appliance found in homes (by far).
http://www.x10.com/homepage.htm
Be careful if you order anything from them, lest you be buried in spam for a lifetime.

It might have helped if you'd specified this in the beginning instead of left people to guess.

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On 09/04/2012 07:35 AM, snipped-for-privacy@att.bizzzzzzzzzzzz wrote:
[snip]

That site has problems, but X10 is not limited to that site or supplier. It you want X10, look elsewhere.
[snip]
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responding to http://www.homeownershub.com/maintenance/which-leg-from-the-center-tap-is-it-electrical-711865-.htm DA wrote: snipped-for-privacy@att.bizzzzzzzzzzzz wrote:

Ah, if only they worked ...
I have one like you describe and it's not really doing much, communication between the two legs is still impossible. There may be other factors, of course, like switched power supplies, and there's plenty of that in the house. Perhaps an active amplifying phase bridge will do better but I don't think the passive capacitor-based is worth the time to install it (which is, admittedly, not much either).
There are also plug-n-play ones that go right into a 240V outlet, such as one for an electric dryer, and bridge the two hots in that spot. I just don't think they're any more useful than the hardwired passive ones. Maybe just easier to try them out tho.
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On 8/31/2012 12:00 PM, DA wrote:

If you consider how a capacitor does what it does you will will realize it can only work poorly in this application. However there are X-10 active repeater couplers that do work.

between the two legs is still impossible. There may be other factors, of course, like switched power supplies, and there's plenty of that in the house. Perhaps an active amplifying phase bridge will do better but I don't think the passive capacitor-based is worth the time to install it (which is, admittedly, not much either).

for an electric dryer, and bridge the two hots in that spot. I just don't think they're any more useful than the hardwired passive ones. Maybe just easier to try them out tho.

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Wrong.
Which only work for X-10.

between the two legs is still impossible. There may be other factors, of course, like switched power supplies, and there's plenty of that in the house. Perhaps an active amplifying phase bridge will do better but I don't think the passive capacitor-based is worth the time to install it (which is, admittedly, not much either).

for an electric dryer, and bridge the two hots in that spot. I just don't think they're any more useful than the hardwired passive ones. Maybe just easier to try them out tho.

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On Aug 31, 1:16 pm, " snipped-for-privacy@att.bizzzzzzzzzzzz"

Yeah, I'm also wondering what about how a cap works makes it work poorly if you use it to bridge the phases. Cap should pass those high frequencies just fine.
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On 8/31/2012 8:57 PM, snipped-for-privacy@optonline.net wrote:

Capacitive reactance will cause loss. Since the objective is to couple as much of the low level signal as possible you can easily cancel that loss by adding an inductive element in series to form a classic series LC circuit.
http://en.wikipedia.org/wiki/LC_circuit
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You need a tuned circuit only if you want a bandpass filter to pass primarily frequencies within the range the filter is tuned for. A cap will pass high frequencies. An inductor passes low frequencies. Since what we're interested in is passing freq up in the mhz range, a cap wll do that and adding an inductor is not going to get any more of the X10 signal one leg to the other. What you have with just the cap is a high pass filter.
Having a tuned circuit would filter out freq both above the desired target and below it. If there is interference involved that is at other frequencies, then I can see an advantage to using the tuned circuit. But for just getting the X10 from one leg to the other, that LC circuit has no advantage over just using a cap.
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On 9/2/2012 10:00 AM, snipped-for-privacy@optonline.net wrote:

Exactly, X10 operates at 121 kHz and we want to couple as much of that signal as possible. By using a tuned circuit we have the lowest possible loss at the desired frequency. Its the absolute best you can get with passive circuitry.

Sure it will. Its classic AC circuit theory and it works in practice.

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Not exactly. You only need a tuned circuit if you also seek to *not* pass the low frequencies. Otherwise a capacitor passes the high frequency X10 signal just as well as a tuned circuit. Since there is nothing to suggest that there is a need to reject low frequencies, the cap is all that is needed.

In my world, AC circuit theory says that as the frequency increases, the impedance of a capacitor drops. Do a plot of the signal that makes it across the capacitor that is used as a high pass filter and as the frequency goes up, the signal asymptotically approaches 100%. You can't get more than 100%.
With a high pass filter, eg a cap, you have a curve that rises and asymptotically approaches 100% as the freq rises. With a low pass filter, eg an inductor, you have a curve where it passes 100% at low freq and asymptotically declines to zero as the freq increases..
With a tuned circuit, you have a combination of the two, with a bell shaped curve that passes frequencies that it's tuned for, with the curve asymptotically appoaching zero at very high and low freq, away from the tuning point.
None of that says that the tuned circuit is going to pass any more of the X10 signal than a simple cap.
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