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NETA 4 exam questions I am unsure about

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  1. mmacdonell is offline Junior Member Pro Subscriber
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    NETA 4 exam questions I am unsure about

    Here are a couple I am unsure about. Can someone answer and explain? Thank you!

    A CT circuit with a CT resistance of 15mohm goes to a SEL-751 (.27VA @ 5A) through 125ft of #12 SIS (.988 ohms per 100ft). What is the appropriate CT class?

    C10
    C50
    C100
    C200

    Equipment rated above 1200 amperes or more and over 6 ft wide containing overcurrent devices requires an entrance of at least _____ wide at each end of the working space.

    22 inches
    24 inches
    18 inches
    26 inches


    Which test method measures Hydrolyzable Flouride content in SF6 gas?

    ASTM D2029
    ASTM D2284
    ASTM D2685
    ASTM D2556

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    Equipment rated above 1200 amperes or more and over 6 ft wide containing overcurrent devices requires an entrance of at least _____ wide at each end of the working space.

    22 inches
    24 inches
    18 inches
    26 inches

    ANSWER= 24"
    2017 NEC 110.26(C)(2)

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  5. mcmahanrf is offline Junior Member Pro Subscriber
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    Answers

    I think this is how you do this:

    A CT circuit with a CT resistance of 15mohm goes to a SEL-751 (.27VA @ 5A) through 125ft of #12 SIS (.988 ohms per 100ft). What is the appropriate CT class?

    C10
    C50
    C100
    C200

    The number after the C in the ratings of these CT's is the max burden in volts at 20 times the CT amp rating. So we need to find the burden of this circuit and figure out which ones have a big enough rating and then pick the smallest of that list.

    To find the burden we need to add the burdens of the CT, relay, and interconnecting wiring. We are told the CT burden is 0.015 ohms. The relay and wires need some math, for the relay:

    VA = 0.27 at 5 AMP so
    VA = I*V = (I^2)*R therefor
    R = (VA/(I^2)) = (0.27/(5^2)) = 0.0108 ohms

    now the wires:
    R = (resistivity per foot)*(length in feet) = (0.988 ohms/100ft)*125ft = 1.235 ohms

    I think this part of the question is poorly written because a lot of times they give you the distance between the relay and the CT only and you have to double that distance to get the round trip of the wire to and from the CT. I don't think its clear that that is not what was meant except for that if you double this number none of the CT's are big enough. Just food for thought.

    Now that we have all of those burdens we can combine them for the entire burden of the circuit:

    Total burden = CT burden + Wire Burden + Relay Burden = 0.015 ohms + 1.235 ohms + 0.0108 ohms = 1.2608 ohms

    So remembering what the rating numbers mean we need to see at 1.2608 ohms and 100 amps (5 amp rating that the don't tell you to assume but it intended to be assumed * 20) what the voltage would be:

    V = I*R = 100 Amps*1.2608 Ohms = 126.08 Volts

    So we now know the voltage rating required for a CT in this circuit needs to be equal to or grater than 126.08. And since only D at C200 meets the requirement then that must be the answer.


    Equipment rated above 1200 amperes or more and over 6 ft wide containing overcurrent devices requires an entrance of at least _____ wide at each end of the working space.

    22 inches
    24 inches
    18 inches
    26 inches

    The answer is 24 inches, this answer is found in article 110.26(C)(2) of the NEC NFPA 70


    Which test method measures Hydrolyzable Flouride content in SF6 gas?

    ASTM D2029
    ASTM D2284
    ASTM D2685
    ASTM D2556

    The answer is ASTM D2284, you can confirm this with google but it is also listed in the ATS Table 100.13

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