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#129361 - 02/01/05 09:31 AM open circuited ct's
DJT Offline
Junior Member
Registered: 01/25/05
Posts: 2
Loc: Indianapolis, In. 46221
Hello everyone,
I would appreciate a detailed theoretical explanation of why a meter/relay class CT develops such a high open circuited voltage. Aside from the ammount of iron, what other deciding factors determine the voltage level?

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#129362 - 02/01/05 09:52 AM Re: open circuited ct's
Trumpy Offline

Registered: 07/05/02
Posts: 8211
Loc: SI,New Zealand
If the load is removed from a Current Transformer secondary winding, while there is still current flowing in the Primary winding, the secondary EMF will rise to a high and very dangerous level, this is an attempt to maintain the secondary current in the winding.
Factors that contribute to the magnitude of this voltage would include, the Primary current flowing, the turns ratio between Primary and secondary and the impedance of the secondary winding.
Hope this helps.
Let's face it, these days if you're not young, you're old - Red Green grin
#129363 - 02/02/05 01:05 PM Re: open circuited ct's
Tinkerer Offline
Registered: 12/13/04
Posts: 16
Loc: Birmingham, AL, USA
The CT is designed to cause a current in the secondary in proportion to the current in the primary. If there is a current in the primary the secondary tries to increase the voltage to force the current through the infinite resistance presented by the open circuit. remember v=i*r so when the r is infinite and the current is finite the the voltage will be infinit.
#129364 - 02/02/05 01:05 PM Re: open circuited ct's
pdh Offline
Registered: 01/20/05
Posts: 354
Think of the CT as a step-UP transformer ... because that is what it actually is (stepping current down means stepping voltage up). The difference is that the primary (the conductor going through the window) is "wired" in series with another load (the entire metered load). You have a low number of turns on the primary (usually 1) and a high number of turns on the secondary (for a 600:5 that would be 120 turns). Actual voltage calculation depends on a lot of factors, and under load will drop low. But the turns ratio can determine what you could get in an open circuit condition.

A transformer reflects the impedance of the secondary back upon the primary at the given ratio. That's why a normal transformer primary will draw more or less current depending on the impedance connected to the secondary. As you reflect more impedance back against the current flow going through the CT window, the voltage drop at the window increases. It won't be infinite since it's not perfect coupling, so you can't stop the current in the conductor with an open CT. Potentially (no pun here ... this is lethal) you can get the supply voltage (how hard it will try to make sure the load current flows through the window) times the turns ratio of the CT.

Don't mess with live CT's unless you are specifically trained in the procedure.

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