Voltage at X1/X2 terminals

[David Hopkins]

Hopefully this isn't foolish, but I am curious if current transformers use the same math as I learned for voltage transformers - basically, that the product of the primary voltage and current ideally will equal the product of the secondary voltage and current. If that is the case, would a 100:0.1A current transformer at full scale, 120V, really generate 120,000 Volts at X1/X2? (120 * 100) = (120,000 * 0.1)

 

[Tim Law]

It's a valid question. I think the attached document will help to explain how these CTs work. Following is an excerpt with a summary of the math used with these CTs.

The principle of operation of a CT is the sensing of the magnetomotive force around a current-carrying conductor. The CT contains a high permeability magnetic core, and a multiple turn secondary winding. This secondary winding links all the magnetic flux generated in the core by the magnetomotive force (mmf) caused by the current in the primary conductor. The CT therefore resembles other transformers. The goal in designing a CT is to approach the behavior of an ideal transformer, where the permeability of the core can be considered infinite; the resistance of the windings is zero, and both windings link exactly the same magnetic flux. To the extent these conditions are met, the transformer will have two properties:

1. The voltage per turn will be equal on both the primary and the secondary windings, so that the ratio V (sec)/V (pri) = N (sec)/N (pri).
2. The net magnetizing current is zero, so I (pri) x N(pri) + I (sec) x N (sec)=0, or I (sec)/N(pri)= -I (pri)/N(sec). It can be seen that an ideal transformer can be described by one number, the ratio N (sec)/N (pri). For the CT, N (pri) is usually 1, so that I (sec) = I (pri}/N (sec), or, as is usually written, I(pri)/N. (the negative sign can usually be ignored, unless the phase relationship between I(pri) and I(sec) is of importance in the measurement). For most measurements, the CT secondary winding is permanently connected to a low value resistor, called the burden resistor. The voltage across this resistor is then V= [I (pri)/N] x R. The value of R is determined by the maximum value of I (pri) to be measured, the number of secondary turns (N), and the full scale voltage of the measuring or recording device.
   The value of R is then equal to (V x N) / I (pri). As an example, if V = 0.333 volts.
   N = 5000, and I (pri) = 100 Amps, R will be16.65 ohms, and its full scale power dissipation will be 0.0067 watts.