×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Are you an
Engineering professional?
Join Eng-Tips Forums!
• Talk With Other Members
• Be Notified Of Responses
• Keyword Search
Favorite Forums
• Automated Signatures
• Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

#### Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

# CT Secondary Voltage2

## CT Secondary Voltage

(OP)
Greetings,

I have some difficulties understanding the CT secondary voltage. I don't quite understand why CT secondary voltage is affected by the burden. The magnetic flux in the core is induced by primary current flowing through the CT and primary current is "constant" (generated by the power system source),say, i=Im*sin(wt). Therefore, d(magnetic flux generated by primary current)/dt should be the same regardless of the CT burden values, so is the induced secondary voltage (Esecondary=N*d(flux)/dt).But by looking at the CT equivalent circuit, it makes sense that the CT secondary circuit voltage is secondary current times burden impedance. I don't know which step I am thinking is wrong. Any help would be really appreciated.

### RE: CT Secondary Voltage

Whatever the secondary current is (within limits) then that current through and times the impedance of the secondary circuit will be the voltage across the secondary impedance.
Ohm's law.

Bill
--------------------
"Why not the best?"
Jimmy Carter

### RE: CT Secondary Voltage

2
The point you are wrong is that the flux isn't constant in the core with respect to the burden or the primary current. The secondary winding provides counter flux, reducing the flux in the core. Core flux increases as the voltage at the secondary increases. A CT with shorted terminals will have very little core flux due to all the couterflux or low voltage developed on the secondary. An open ended CT will have high flux due to having no counter flux.

Here look at slide 27

-----------------------------------
-------------------------------------------------------
If you can't explain it to a six year old, you don't understand it yourself.

#### Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

#### Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Close Box

# Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

• Talk To Other Members
• Notification Of Responses To Questions
• Favorite Forums One Click Access
• Keyword Search Of All Posts, And More...

Register now while it's still free!