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
Any help would be appreciated,
Daniel






RE: CVT impedance
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.