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Can this CT produce enough voltage to trip a relay?

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  1. mmacdonell is offline Junior Member Pro Subscriber
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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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  3. mmacdonell is offline Junior Member Pro Subscriber
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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?

    Not sure how to calculate this, can someone please help?

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    Quote Originally Posted by mmacdonell View Post
    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?

    Not sure how to calculate this, can someone please help?

    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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  7. mmacdonell is offline Junior Member Pro Subscriber
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    Quote Originally Posted by akumaleon View Post
    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?

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    Quote Originally Posted by mmacdonell View Post
    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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  11. mmacdonell is offline Junior Member Pro Subscriber
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    Quote Originally Posted by akumaleon View Post
    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!

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    Quote Originally Posted by mmacdonell View Post
    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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  15. mmacdonell is offline Junior Member Pro Subscriber
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    Quote Originally Posted by akumaleon View Post
    -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?

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    Quote Originally Posted by mmacdonell View Post
    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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    Quote Originally Posted by mmacdonell View Post
    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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