# Current Transformers



## Rikki (Mar 23, 2007)

Has anyone ever mounted a 18'' round interposing CT in a GFP system for equipment .
First question-CT came with no mounting hardware(or instructions) and I don't feel comfortable just hanging it on the feeds. How are these supposed to be mounted?
second question-How can you hook the leads on the secondary side together on the CT without frying the transformer(is the impedance just that high?)
And if you can how is it possible to ground the secondary?
I'm just thinking about a conventional transformer and hooking the secondary leads together is a short circuit so having a hard time with this type of connection.

Thanks for any help


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## micromind (Aug 11, 2007)

This type of GFP is pretty common. I usually support the CT using snap-ties or something like that. You need to get creative sometimes. Don't use anything metal that goes through the center. 

Though I don't like it, I've seen a lot of CTs that simply rest on the wries.

The secondary leads of all CTs absolutely MUST be shorted, or very nearly shorted. Never open, not even for an instant. 

When you connect the secondary side of the CT into whatever it feeds, connect one of the leads, then connect an ohmmeter between the other lead and the terminal it's going to connect to. The idea here is to insure that the CT operates at a very low resistance.

Rob


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## Jlarson (Jun 28, 2009)

Support the CT with tyraps. I also like to put in a shorting block(its a terminal block with a metal bar that can be connected across the terminals if you have to disconnect the CT) when ever possible.


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## mattsilkwood (Sep 21, 2008)

The ones I have done, we just used ty-raps to hold them in place. 
I have been told that a open secondary can reach as high as 10 times the primary. 
Maybe one of the guys that know a little more than me can chime in.


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## Jlarson (Jun 28, 2009)

mattsilkwood said:


> The ones I have done, we just used ty-raps to hold them in place.
> I have been told that a open secondary can reach as high as 10 times the primary.
> Maybe one of the guys that know a little more than me can chime in.


I will give it a shot it will probably be wrong but who cares. If the ct is 400:1 so 400a on the primary gives 1a on the secondary that means you are dividing the current by 400. Because you are doing that it means you multiply the voltage by 400. The secondary load is based on the primary. When the secondary impedance goes up that is reflected in the primary allowing higher voltage drop and bla, bla(highly technical terms don't ya think) you would end up with like 192kv on a 480 volt system. 

It made sense in my head but I know that dosn't mean anything.


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## etb (Sep 8, 2010)

Jlarson said:


> I will give it a shot it will probably be wrong but who cares. If the ct is 400:1 so 400a on the primary gives 1a on the secondary that means you are dividing the current by 400. Because you are doing that it means you multiply the voltage by 400. The secondary load is based on the primary. When the secondary impedance goes up that is reflected in the primary allowing higher voltage drop and bla, bla(highly technical terms don't ya think) you would end up with like 192kv on a 480 volt system.


Or in other words, it's a step up xfmr. You wasn't kiddin the other day when you said your description of inductive reactance in an xfmr woulda been long and boring.:laughing:



Jlarson said:


> It made sense in my head but I know that dosn't mean anything.


No, I think you got it right; at least until the voltage drops as it arcs. That's about the only easy calc one can make on them though, cause the leakage impedance ans saturable excitation reactance come in to play in real (loaded) operation.


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## etb (Sep 8, 2010)

Rikki said:


> second question-How can you hook the leads on the secondary side together on the CT without frying the transformer(is the impedance just that high?)


Exactly. Anytime you have a signal or power source you want a low output impedance so you don't waste power or loose output voltage inside the source; it's a source, after all. So power xfmrs have low output impedance, as do stereo amps, etc.

Anytime you have an input (PLC input, guitar amp, DMM, scope) you want a high input impedance so you don't draw a lot of current and load down whatever source or whatever it is that you're measuring.

On a PT you have a step down xfmr; takes high voltage to a lower voltage that electronics like. Lots of turns on the primary; few on the secondary; connected in shunt/parallel. The output impedance of this xfmr is low and that feeds into the high input impedance of the poco's meter or your instrumentation (or if you're on a transmission line, the protection relay). Thus the meter doesn't load down the PT.

On a CT you have a step up xfmr; take high current to low current, but the reason is different. The primary has only one turn (the feed conductor passing through it) and the secondary has a lot of turns; connected in series. We have a current that is a scaled-down replica of the real current. But electronics don't "read" current, they read voltage, so you need to convert the current to a voltage. The oldest and simplest way (which is still in wide use) is to use a resistor and ohm's law. This is why you want a small current, so you're not burning up a lot of power in this current-to-voltage resistor (called a burden resistor). In this case, the output impedance of the CT secondary IS high but there's nothing that can be done about that. Since the impedance is high and the output of the CT is an actual current, the burden will be a low resistance itself. (Likewise when you wire them, you don't want as low as resistance as possible.) So...that's why shorting it out doesn't hurt it. I guess it was my turn to be long and boring....oh well.


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## Jlarson (Jun 28, 2009)

etb said:


> Or in other words, it's a step up xfmr.


That works too. :laughing:


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