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CVT impedance

CVT impedance

CVT impedance

(OP)
Hello, I am a new starting out engineer in the power field and I have a fairly elementary question.  The theory behind a CVT is basically just a capacitive divider with a series inductor to cancel out the phase shift.  So, when a CVT is connected to a phase, there is a path to ground through 2 capacitors (one large and one small).  What I am failing to understand is how there can be a large enough capacitive reactance as to prevent any noticeable fault currents from going through the divider to ground?  I mean if a CVT were connected to a 240kV bus rated at 3kA, would there not have to be smaller than ~100nF to get a 10A current to ground?
 

Any help would be appreciated,


Daniel  

RE: CVT impedance

Typical capacitance range for a 230 kV class CVT is 3000 pF to 12000pF.

Assuming 3000pf (and ignoring the impact of the electromagnetic unit and burden connected across the C2 capacitance), the capacitive reactance would be 884 kohms. At 230 kV L-L (133 kV L-G), that would be approx. 150 mA of current through capacitor divider.

In reality, when the electromagnetic unit (compensating reactor, intermediate transformer, ferro-resonance suppression device), is included in the calculation, the L-G impedance goes up a fair bit.

 

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