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#55229 08/21/05 08:08 AM
Joined: Jul 2005
Posts: 1
fungguy Offline OP
Junior Member
Can someone please explain what the use of CT's are in relation to electrical metering

#55230 08/21/05 08:19 AM
Joined: Jan 2003
Posts: 4,391
Cts are current transformers.

When used for metering it is simply to keep electric meters reasonably sized and priced.

In my area any service larger than 400 amps uses CTs, some areas go up to 600 amps before CTs are used.

A standard meter runs the entire current directly through the meter, once you get to 400 600 amps this starts to become very impractical.

So at this point we use a remote meter driven by the CTs.

The entire buildings current is run through the CTs which in turn put out a voltage in proportion to the current running through them.

Connected to the correct type of meter this variable voltage along with voltage leads connected directly to the service conductors can be converted and recorded as KWH.

Bob Badger
Construction & Maintenance Electrician
#55231 08/21/05 02:15 PM
Joined: Feb 2005
Posts: 693
If I may elaborate...

A CT (current transformer) is a type of instrument transformer that sends a representative current through the meter circuit, rather than the full load current.

A typical CT may have a 400:5 ratio, which translates to an 80:1 ratio. This means that, for every 80 amps that flow through the service, 1 amp flows in the meter circuit.

Most CT's are a simple coil on a core that the load conductor passes through (often called a "doughnut"), or a similar core with a bar that the conductors bolt to.

While considered a current transformer, if the secondary circuit is not closed either by the meter or shorting bars, it will function as a voltage transformer at a 1:80 ratio.

This can step up a typical service voltage to the point that the insulation within the coil can be damaged, or worse, damage nearby equipment. After all, 240 x 80 = 19,200!

Larry Fine
Fine Electric Co.
#55232 08/22/05 12:08 PM
Joined: May 2005
Posts: 178
Never one to let a nit escape picking...

After all, 240 x 80 = 19,200!

How do you achieve 240 volts across the primary half-turn of the CT, in order to obtain that secondary voltage?

#55233 08/22/05 03:29 PM
Joined: Oct 2004
Posts: 806
To add another nit: [Linked Image]

Who says a transformer has to have more than one turn in either winding?

Since it is possible to induce a voltage/current flow in conductors laid side by each, they still act as a transformer.

The voltage across the primary is equal to the service voltage regardless of the number of turns, is it not?

So the example shown is valid and illustrates why you must never open-circuit a CT!!

To take Larry's example to the next level, CT metering is common at the 480 volt level. 480 x 80 = 38,400!!

Break out the hotsticks!

Stupid should be painful.
#55234 08/22/05 03:55 PM
Joined: Jul 2004
Posts: 9,796
Likes: 15
This will be a fairly low current but I suppose it could still be dangerous.

Greg Fretwell
#55235 08/22/05 04:59 PM
Joined: Sep 2003
Posts: 650
Well, um, not quite.

The voltage actually dropped by the primary of a transformer is set by the number of primary turns and the magnetic flux developed in the core. The lower the magnetic flux in the core, the lower the voltage on the primary side.

With normal voltage transformer, the core is sized so that with the full primary voltage across the primary coil, some relatively small magnetizing current will flow (relative is a relative term; this magnetizing current will be 10% to 50% of the full load current of the device!). But the point is that with the core not saturated, sufficient magnetic flux is produced to significantly impede the supply voltage.

As secondary current is drawn, it has the effect of 'balancing' some of the primary current, reducing the core flux. More primary current flows to keep this balanced, and we get the happy result that as we draw secondary current we see a corresponding increase in primary current flow.

In a current transformer, the core is sized rather differently. Remember that the primary is in series with the load, so that even if the primary impedance was _zero_ the current would still be reasonably limited. In fact, it is desirable to have a low primary side impedance for a CT, so that it doesn't waste energy that would otherwise go to the load.

In normal operation, there is very little secondary impedance, so current flows in the secondary to match the current flowing in the primary. Rather than seeing greater current flow in the primary, the result is to reduce the primary side impedance. Current flows in the primary, current flows in the secondary, and very little magnetic flux is actually produced.

If you open the secondary, then the current flow in the secondary ceases. Now, if this happened to be an extremely large transformer with enough core to produce lots of magnetic flux, then the open secondary would increase the primary side impedance. A large fraction of the supply voltage would be dropped across the primary, and the voltage delivered to the load would drop quite a bit.

But this is a small current transformer, which means that there isn't much core area, and thus not too much magnetic flux. For most applications, the load in series with the primary would still be the dominant control of current. Based on the current and core area, you could calculate the magnetic flux in the donut, and from this you could calculate the primary and secondary voltage drops. The primary voltage drop will be much less than the full supply voltage, but the secondary voltage drop will still be 80x the primary drop.


#55236 08/22/05 05:57 PM
Joined: May 2005
Posts: 178
Well said, Jon.

I thought iwire answered fungguy's original question very well, but it's always educational to delve a little deeper into theory.

If I may summarize:

1. The secondary current in a CT is a fixed fraction of the primary current. As Larry pointed out, 1/80 is common.

2. Ideally, the meter and wiring on the secondary side have minimal impedance. That way, even though several amps may be flowing, the voltage is negligible and the wasted power is minimized.

3. The minimized impedance on the secondary is also "reflected" into the primary, so the voltage drop is negligible.

We return now to the discussion of minutiae:

Remember that the primary is in series with the load, so that even if the primary impedance was _zero_ the current would still be reasonably limited.

I didn't understand that statement, but I agree with everything else.

As to the voltage present at the open secondary of a CT, I just don't believe the "hot-stick" levels being suggested. That voltage will be a function of the secondary turns count and the magnetic flux magnitude, and nothing else. As Jon mentioned, CTs have relatively small cores that are easily saturated, so the flux magnitude won't be nearly high enough to induce multi-kV voltages on the secondary.

Just how much voltage can be dropped across that big wire or bus passing through the CT with the open secondary? I'd be surprised to see more than a few tenths of a volt, so the secondary voltage can only be 80 times that, on the order of a few tens of volts.

What will happen is that the saturated core will dissipate a lot of heat, which certainly could damage the windings. That's probably the real reason never to leave the secondary open.

#55237 08/22/05 06:10 PM
Joined: Nov 2001
Posts: 745
My thanks to all who have posted on this thread. I have always been interested in CT's and their purpose, but it's nice to understand HOW and WHY it works.

Mike (mamills)

#55238 08/22/05 06:35 PM
Joined: Jul 2002
Posts: 680
In school we were taught to never open circuit a CT, always short them before working on them or controls. Do we agree on that??

I'll never work on one but I'll always remember the dire warnings given in class [Linked Image]

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