current transformer voltage

[domnic]

Is the voltage in the same as voltage out? ( if i put a current transformer around a wire with 120 volts feeding the wire . will i get 120 volts out of the leads on the current transformer?

 

[infinity]

What is the turns ratio of the CT?

 

[LarryFine]

No. The voltage in a CT secondary circuit is almost zero; it's all current.

That's why a CT can, and must be shorted when the meter is removed from the circuit.

 

[infinity]

No. The voltage in a CT secondary circuit is almost zero; it's all current.

That's why a CT can, and must be shorted when the meter is removed from the circuit.

What if the shunt is removed and the CT secondary is open?

 

[electrofelon]

A CT is actually a step up transformer, but in normal use it is shorted and the current is measured.

 

[Barbqranch]

Also remember that the current transformer is put in series with some regular load, and the voltage drop across it is very small.

 

[LarryFine]

What if the shunt is removed and the CT secondary is open?

Then it acts like a PT and steps the voltage up at the inverse ratio. Imagine the OC voltage from a heavily-loaded 800:5 CT.

That's why they must be shunted; otherwise, the internal windings burn as the insulation is breached from the high voltage.

 

[infinity]

Then it acts like a PT and steps the voltage up at the inverse ratio. Imagine the OC voltage from a heavily-loaded 800:5 CT.

That's why they must be shunted; otherwise, the internal windings burn as the insulation is breached from the high voltage.

I kind of thought that is what the question was in the OP. Obviously you don't want to run it with an open secondary.

 

[hillbilly1]

This is what happens when the meter circuit goes open.

 

[Carultch]

Is the voltage in the same as voltage out? ( if i put a current transformer around a wire with 120 volts feeding the wire . will i get 120 volts out of the leads on the current transformer?

A CT has a voltage ratio, just like any other transformer, which neglecting internal losses, would be the inverse of the current ratio. The voltage across the primary circuit that runs through the CT, is not the nominal voltage of the circuit it is measuring. Rather we're talking the millivolts that would drop across the circuit in that distance of a couple inches from one side of the CT to the other.

A CT is designed to cause as little voltage as possible on the primary circuit, due to the current-sensing circuit inside the meter being as close to as practical to a short circuit (in terms of its effective impedance). Just like an inline ammeter which is also as close to a short circuit as practical. The CT is there to bring hundreds/thousands of amps, down to the scale of individual amps or milliamps, so it is practical for the meter to measure. You will get a voltage across the measured circuit, if you were to measure both sides of the circuit that runs through its wire window, but it is probably too small to be detected by most voltmeters.

This is why you have to short-circuit the CT when removing the meter, rather than opening the circuit like you ordinarily do to de-energize a circuit. It is one of the rare instances where a short circuit is intentional. Because connecting it to an open circuit, will make it generate extremely high voltages on the output terminals, that will damage the CT. Not all CT's require shorting, as some are specifically built to prevent this issue. Assume shorting the CT is required, unless you know otherwise from its documentation.

 

[LarryFine]

To add:

A CT send a representational current to the meter. The CT and meter must have the same ratio for the reading to be correct.

The supply voltage must also be measured, just like any watt-meter, either directly or through PTs if the voltage is very high.