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A 5 H inductor is subjected to an electric current that changes at a rate of 4.5 amps

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    A 5 H inductor is subjected to an electric current that changes at a rate of 4.5 amps

    A 5 H inductor is subjected to an electric current that changes at a rate of 4.5 amps per second. How much voltage will be dropped by the inductor?

    11.25
    22.50
    37.50
    45.00

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    Kalbi_Rob is offline Experienced Member Pro Subscriber
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    Quote Originally Posted by cnowak View Post
    A 5 H inductor is subjected to an electric current that changes at a rate of 4.5 amps per second. How much voltage will be dropped by the inductor?

    11.25
    22.50
    37.50
    45.00
    V = L (di/dt) or L*((I1-I0)/(t1-t0))

    V = 5H ((4.5A-0)/(1s-0s)) = 5H*(4.5A/1s) = 5H*4.5(A/s) = 22.5 V

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    Sep 21' NETA Level II Test Topics

    Topics not covered in/not covered enough the practice tests include:

    Common IEEE Device number names

    Recloser functions and definitions

    MEDIUM VOLTAGE BREAKERS!!!!!

    Operations and construction of amp/volt/ohm/mego meters

    Oil sample analysis elements/ symptoms

    LOTS OF ARC FLASH ratings and boundaries.

    Confined spaces definitions/ rules and regulations/ person in charge

    schematic diagrams/ diagnosing interlocks

    Happy studying

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    Quote Originally Posted by Kalbi_Rob View Post
    V = L (di/dt) or L*((I1-I0)/(t1-t0))

    V = 5H ((4.5A-0)/(1s-0s)) = 5H*(4.5A/1s) = 5H*4.5(A/s) = 22.5 V
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