Re: Need Information about a Current Transformer


"ehsjr" <ehsjr@xxxxxxxxxxxxxxxx> wrote in message
news:_0yyf.1625$8r1.1054@xxxxxxxxxxx
> scada wrote:
> > "DaveM" <masondg4499@xxxxxxxxxxxxx> wrote in message
> > news:uIqdnZZR0KHyMlTenZ2dnUVZ_v6dnZ2d@xxxxxxxxxxxxxx
> >
> >>"scada" <scada@xxxxxxxxxxxxx> wrote in message
> >>news:qzfyf.63$5u5.7@xxxxxxxxxxx
> >>
> >>>"DaveM" <masondg4499@xxxxxxxxxxxxx> wrote in message
> >>>news:G4-dnQdIv62K_lTenZ2dnUVZ_tKdnZ2d@xxxxxxxxxxxxxx
> >>>
> >>>><slc@xxxxxxxxxx> wrote in message
> >>>>news:1137270022.893690.213610@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> >>>>
> >>>>>Ok, I put a 6 amp load thought the primary and measureed the
> >
> > secondary.
> >
> >>>>>The meter was set on "AC volts" and it read .020
> >>>>>So what would the rateing be, 100 amps to what?
> >>>>>
> >>>>>Thnaks
> >>>>>Kevin
> >>>>>
> >>>>>
> >>>>>Rich Grise wrote:
> >>>>>
> >>>>>>On Mon, 09 Jan 2006 18:19:23 -0800, slc wrote:
> >>>>>>
> >>>>>>
> >>>>>>>I took the CT apart and there is a 1 ohm resistor (about ¼ watt)
> >>>>>>>across the secondary.
> >>>>>>>If I have 6 amps on the primary and I connect a volt meter across
> >>>>>>>the
> >>>>>>>secondary.
> >>>>>>>What should the voltage be?
> >>>>>>
> >>>>>>It depends on the turns ratio.
> >>>>>>
> >>>>>>Don't you have a voltmeter with an "AC Volts" range? Just measure
> >>>>>>it! If it's a 10:1 CT, with your 6A through the primary hole, then
> >>>>>>there will be .6A flowing in the secondary, which, with the 1 ohm
> >>>>>>resistor, would drop .6 V, which is .6 * .6 = .36 watts, but since
> >>>>>>it's a 1/4 watt resistor, it's probably not 10:1. Probably a lot
> >>>>>>closer to 1000:1 - that would give you .006A, or 6 mA, with 6A
> >>>>>>through the primary, which would drop 6 mV across your 1 ohm
> >
> > resistor.
> >
> >>>>>>So it's very probably something in between those two - time to get
> >>>>>>out the meter!
> >>>>>>
> >>>>>>Good Luck!
> >>>>>>Rich
> >>>>
> >>>>
> >>>>
> >>>>Since the burden resistor is 1 ohm, the current ratio would be 6 / .02
0
> >
> > =
> >
> >>>>300:1
> >>>>So, if you put 100A through a single turn primary, you would expect to
> >>>
> >>>read
> >>>
> >>>>0.3333 volts on the secondary. The current through the 1 ohm burden
is
> >>>>0.333A, and the burden dissipation is 0.1111W.
> >>>>
> >>>>Hth
> >>>>--
> >>>>Dave M
> >>>>MasonDG44 at comcast dot net (Just substitute the appropriate
> >
> > characters
> >
> >>>in
> >>>
> >>>>the address)
> >>>>
> >>>>Never take a laxative and a sleeping pill at the same time!!
> >>>>
> >>>>
> >>>
> >>>I don't think the 1 ohm resistor is the actual burden. I believe it is
> >>>there
> >>>for protection, should you run the CT open circuited. Try the test with
> >
> > a
> >
> >>>shorted CT secondary using a clamp on ammeter, or a wired AC Ammeter
> >>>directly connected to the secondary.
> >>>
> >>
> >>No, the 1-ohm resistor is the burden. It's across the secondary. CT
> >>secondary protection would likely be in the form of a series opposed
pair
> >
> > of
> >
> >>zeners. A CT secondary is not normally short-circuited, although it
could
> >>be. The CT's used in power metering equipment are usually run directly
> >
> > into
> >
> >>a 5A AC current meter, but even those are not a direct short circuit to
> >
> > the
> >
> >>CT. Those types of CT are designated for 5A current meter connection...
> >>and the ratio will be expressed as something like 1000:5, meaning a
> >
> > primary
> >
> >>current of 1000A will produce a 5A secondary current when connected to a
> >
> > 5A
> >
> >>current meter.
> >>Cheers!!!
> >>--
> >>Dave M
> >>MasonDG44 at comcast dot net (Just substitute the appropriate
characters
> >
> > in
> >
> >>the address)
> >>
> >>Never take a laxative and a sleeping pill at the same time!!
> >>
> >>
> >
> >
> > If the 1 ohm is the burdon, then when a secondary load (Ammeter, relay,
> > etc...) is connected, then some of the current would flow through the
> > resistor and some through the load, wouldn't it! Or am I wrong in my
> > thinking? Or does one have to remove the resistor when the CT is put
into
> > service? I never worked with a CT that could be energized without a load
> > connected to it's secondary.
> >
> > Thanks...
> >
> >
>
>
> A CT with *nothing* connected to its output other than the
> burden resistor would be almost useless. (I suppose you could
> measure the temperature in the burden and do something with
> that.) So, for it to be included in a circuit and serve a
> useful purpose, there would have to be some use of the
> current on the secondary, other than through the burden.
>
> In the case we are discussing, connect an 11 megohm volt meter
> across the CT secondary. It will introduce an undetectable
> error - the meter won't be able to display a difference
> that small. (Do the math - 11000000/11000001 )
>
> Or get a worse indicator - say a 1 mA meter movement with 87
> ohms resistance. It will introduce an error of less that
> 2 percent, indicating 296.59 amps when 300 amps is the
> primary current.
>
> If the CT is used to switch a relay, there would be a high
> impedance connected to the CT, so the additional load
> wouldn't matter.
>
> Ed

CT secondaries drive current devices, not to measure potential. A PT relay
is used to monitor potential (voltage). A relay wired in series with a CT's
secondary is a "Current Relay" and is of extremely low impeadance! That
impeadance must be within the design of the CT burden. Such relays are used
typically as overcurrent protection for equipment.

Thanks...


.

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