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# Can this CT produce enough voltage to trip a relay?

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## Can this CT produce enough voltage to trip a relay?

I think yes. 100ft x .0025ohm/ft = 0.25ohm < .9ohm so its ok.

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## 11. A given CT is rated 600:5, C20, 0.3 B 0.9. Can this CT produce enough voltage to

11. A given CT is rated 600:5, C20, 0.3 B 0.9. Can this CT produce enough voltage to trip a relay (with near-zero burden) 50 feet away if connected to the relay with #14 awg (.0025 ohms/ ft) wire?

3. Originally Posted by mmacdonell
11. A given CT is rated 600:5, C20, 0.3 B 0.9. Can this CT produce enough voltage to trip a relay (with near-zero burden) 50 feet away if connected to the relay with #14 awg (.0025 ohms/ ft) wire?

So you have 0.0025 ohms/ft x 50 ft = 0.125 ohms of your total wire circuit.

5A from your CT (600:5) so you can produce: V=I*R= 5*0.125 = 0.625V

I believe that is the voltage you are looking for.

And your total burden in VA would just be 0.625V*5A= 3.125VA

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Originally Posted by akumaleon
So you have 0.0025 ohms/ft x 50 ft = 0.125 ohms of your total wire circuit.

5A from your CT (600:5) so you can produce: V=I*R= 5*0.125 = 0.625V

I believe that is the voltage you are looking for.

And your total burden in VA would just be 0.625V*5A= 3.125VA
So no need to consider the accuracy, accuracy class or max burden?

5. Originally Posted by mmacdonell
So no need to consider the accuracy, accuracy class or max burden?
well I give you what the "circuit" values are. (the Z and voltage drop of your wire)
usually you will have the relay Z or VA to add to your wire... but in this case your relay burden is neglected.

A 0.3 B 0.9 rated metering-class CT means it will operate within 0.3 percent accuracy if the secondary burden does not exceed 0.9 ohms.

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Originally Posted by akumaleon
well I give you what the "circuit" values are. (the Z and voltage drop of your wire)
usually you will have the relay Z or VA to add to your wire... but in this case your relay burden is neglected.

A 0.3 B 0.9 rated metering-class CT means it will operate within 0.3 percent accuracy if the secondary burden does not exceed 0.9 ohms.
Thank you. No need to consider the "C20" or "B" (CT Class)?

Also, the question was "will it produce enough voltage to trip the relay?
- If this is protection class, then the 0.9 means Volts @ 20x CT rating, NOT not to exceed 0.9 ohms
- How do we know how much voltage is required to trip the relay?

Cheers!

7. Originally Posted by mmacdonell
Thank you. No need to consider the "C20" or "B" (CT Class)?

Also, the question was "will it produce enough voltage to trip the relay?
- If this is protection class, then the 0.9 means Volts @ 20x CT rating, NOT not to exceed 0.9 ohms
- How do we know how much voltage is required to trip the relay?

Cheers!

-if this is a relaying CT (protection) ? saying C20 ... means '20' Maximum in Volts @ 20 times CT Amp Rating.

and with a CT ratio of 600:'5'. in that case it's V=I*R where if you need Ohms BURDEN you do R=V/I . Just like you said with 20x CT rating;
20V/ (5A*20) = 0.2 Ohms

for a C100 runing 5A with 20X --> it will be 1 Ohms
C200 --> 2 ohms
C400 --> 4 ohms
C50 --> 0.5 ohms

-To me it looks like your question (trick question) its a Metering CT and its based on how to calculate the total burden of your circuit.
I would have done the following;

0.0025 ohms/ft x 50 ft = 0.125 ohms one way of my wire so 2x = 0.25 ohms
neglect the relays Z etc...
The secondary voltage is my total resistance X CT rated amps = 0.25*5= 1.25 Volts
(on the first post I forgot to double the wire of the circuit sense you need 2 wires for close circuit)

Last edited by akumaleon; January 24, 2020 at 01:04 PM.

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Originally Posted by akumaleon
-if this is a relaying CT (protection) ? saying C20 ... means '20' Maximum in Volts @ 20 times CT Amp Rating.

and with a CT ratio of 600:'5'. in that case it's V=I*R where if you need Ohms BURDEN you do R=V/I . Just like you said with 20x CT rating;
20V/ (5A*20) = 0.2 Ohms

for a C100 runing 5A with 20X --> it will be 1 Ohms
C200 --> 2 ohms
C400 --> 4 ohms
C50 --> 0.5 ohms

-To me it looks like your question (trick question) its a Metering CT and its based on how to calculate the total burden of your circuit.
I would have done the following;

0.0025 ohms/ft x 50 ft = 0.125 ohms one way of my wire so 2x = 0.25 ohms
neglect the relays Z etc...
The secondary voltage is my total resistance X CT rated amps = 0.25*5= 1.25 Volts
(on the first post I forgot to double the wire of the circuit sense you need 2 wires for close circuit)
Thanks, I agree with you there but then ultimately the questions is do you have enough voltage to trip the relay. So how would we know the required voltage?

9. Originally Posted by mmacdonell
Thanks, I agree with you there but then ultimately the questions is do you have enough voltage to trip the relay. So how would we know the required voltage?
V= I * (Rs + Rb)

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Originally Posted by mmacdonell
Thank you. No need to consider the "C20" or "B" (CT Class)?

Also, the question was "will it produce enough voltage to trip the relay?
- If this is protection class, then the 0.9 means Volts @ 20x CT rating, NOT not to exceed 0.9 ohms
- How do we know how much voltage is required to trip the relay?

Cheers!
You do have to consider the C20. That is the maximum voltage that the CT is able to produce. The circuit requires 25V in order to operate. Thus this CT will not work as a relaying CT. However, it will work as a metering CT because the total circuit burden is .25 which is less than the .9 burden rating the CT has.

I attached a link to a great powerpoint that has this exact question in it.

https://nanopdf.com/queue/instrument...E1MC4yNDAuMjE3

Hope this helps

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