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Why do high voltages develop across the leads of an open CT secondary, and not on an open VT secondary?
Electric Power Engineering
Answer: The combination of opposite turns ratios and differing system connections causes the difference.
Explanation: The simplest model of an ideal transformer shows this very well. This model steps voltage along with the turns ratio, and current with the inverse of the turns ratio. There are no series impedances or shunt admittances and therefore no losses.
Begin with the VT. The VT primary is connected in parallel with both the source and the system load. With an open circuit, zero current flows on the VT secondary and therefore the primary also has zero current. System voltage is across the VT primary, and the secondary has the same voltage stepped down according to the turns ratio.
Now the CT. The CT primary is connected in series between the source and load. With the secondary shorted, voltage across the secondary is zero. Adjusted by the turns ratio, voltage across the CT primary is also zero. The load, therefore, sees full system voltage and normal current. The same load current flows in the CT primary which steps it down by the turns ratio to the shorted secondary. When the secondary is open circuited, assuming no arcing or tracking occurs, there will be zero current in the secondary. When adjusted by the turns ratio, there will also be zero current in the primary. Zero current in the primary means zero current through the load. Zero current through the load means no voltage drop across the load. Since the sum of the voltages around the loop must equal zero, full system voltage must drop across the CT primary. This voltage is then stepped up by the turns ratio to the voltage that appears across the open CT secondary. (Remember that if the current steps down, the voltage must step up)
Now before I get loads of mail complaining that no way can a few turns of wire and iron around a conductor cause it to open circuit the load..: The above analysis uses the most idealized transformer model. Real transformers have shunt admittances. The next level of analysis would include these elements in the model. We cannot use a CT as a switch because the shunt admittance continues to carry the load. It does create a small voltage drop, however. Even a small voltage drop across the primary is enough to cause very high voltages to develop across the open CT secondary after being stepped up by the turns ratio.
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http://files.engineering.com/getfile.aspx?folder=3edb7d8c-6623-4ec7-978d-d2ef30ebb5fe&file=VT-CT.pdf
Explanation: The simplest model of an ideal transformer shows this very well. This model steps voltage along with the turns ratio, and current with the inverse of the turns ratio. There are no series impedances or shunt admittances and therefore no losses.
Begin with the VT. The VT primary is connected in parallel with both the source and the system load. With an open circuit, zero current flows on the VT secondary and therefore the primary also has zero current. System voltage is across the VT primary, and the secondary has the same voltage stepped down according to the turns ratio.
Now the CT. The CT primary is connected in series between the source and load. With the secondary shorted, voltage across the secondary is zero. Adjusted by the turns ratio, voltage across the CT primary is also zero. The load, therefore, sees full system voltage and normal current. The same load current flows in the CT primary which steps it down by the turns ratio to the shorted secondary. When the secondary is open circuited, assuming no arcing or tracking occurs, there will be zero current in the secondary. When adjusted by the turns ratio, there will also be zero current in the primary. Zero current in the primary means zero current through the load. Zero current through the load means no voltage drop across the load. Since the sum of the voltages around the loop must equal zero, full system voltage must drop across the CT primary. This voltage is then stepped up by the turns ratio to the voltage that appears across the open CT secondary. (Remember that if the current steps down, the voltage must step up)
Now before I get loads of mail complaining that no way can a few turns of wire and iron around a conductor cause it to open circuit the load..: The above analysis uses the most idealized transformer model. Real transformers have shunt admittances. The next level of analysis would include these elements in the model. We cannot use a CT as a switch because the shunt admittance continues to carry the load. It does create a small voltage drop, however. Even a small voltage drop across the primary is enough to cause very high voltages to develop across the open CT secondary after being stepped up by the turns ratio.
If you can't rate this FAQ highly, please offer your comments first. I promise to consider them.
http://files.engineering.com/getfile.aspx?folder=3edb7d8c-6623-4ec7-978d-d2ef30ebb5fe&file=VT-CT.pdf