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Thread: Test Questions

  1. #11
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    Quote Originally Posted by Unbridled View Post
    #10 THWN Solid= 10380 CM
    Using Vd= 1.73*12.9*27*420 / 10380
    You get 2.72 V
    Way less than 3% of 208V which is 6.24V

    12.9 = K
    27= Load Current
    420= Length of conductors (3X140)
    1.73=Sq. Root of 3

    Ampacity per 430.22 must be 125% of the FLC = 33.75 A

    #10 THW Ampacity @ 75 Degrees = 35 A
    You only use the line-to-load length of one conductor. Not the length of all three conductors summed.

    sqrt(3) × 12.9 × 27A × 140FT ÷ (208V × 3%) = 13,535cmil
    I will have achieved my life's goal if I die with a smile on my face.

  2. #12
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    Quote Originally Posted by jimdavis View Post
    FYI- The Utah journeyman exam is actually 3 separate exams- code examination (80 questions), theory examination (50 questions), and a practical examination (7 practical exercises). There are 10 voltage drop calculation questions on the theory exam.
    Thanks for the information everyone it's very helpful! The 3% voltage job factor is just a factor be used on this particular question. I don't think it applies to any real world situation

  3. #13
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    Test question

    The following loads are connected in series, 5 ohms, 20 ohms, 10 ohms, and 15 ohms. The voltage total is 50v. Find the voltage drop on the 10 ohm resistor.

    Thanks in advance!

  4. #14
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    Quote Originally Posted by sclement View Post
    The following loads are connected in series, 5 ohms, 20 ohms, 10 ohms, and 15 ohms. The voltage total is 50v. Find the voltage drop on the 10 ohm resistor.
    So how do you think it is done?
    I will have achieved my life's goal if I die with a smile on my face.

  5. #15
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    Quote Originally Posted by Smart $ View Post
    So how do you think it is done?
    Well To be honest I don't know? That's kinda why I'm throwing out the problems that I'm struggling with in hopes I'll get some help with them.

  6. #16
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    Quote Originally Posted by sclement View Post
    Well To be honest I don't know? That's kinda why I'm throwing out the problems that I'm struggling with in hopes I'll get some help with them.
    Resistors connected in series sum arithmetically. Apply Ohm's law to the result. Current is the same through all the resistors.

    That's the most I can give you without providing the actual answer.
    I will have achieved my life's goal if I die with a smile on my face.

  7. #17
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    Quote Originally Posted by sclement View Post
    The following loads are connected in series, 5 ohms, 20 ohms, 10 ohms, and 15 ohms. The voltage total is 50v. Find the voltage drop on the 10 ohm resistor.

    Thanks in advance!
    Is the part in red correct or did they/you mean 50 ohms?
    If you aim at nothing, you will hit it every time!

  8. #18
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    Quote Originally Posted by Little Bill View Post
    Is the part in red correct or did they/you mean 50 ohms?
    Since they're in series the current would be the same at each resistor but the voltage drop across each resistor would be different. The sum of all of the voltage drops would be 50 volts, at least that's how I read it.
    Last edited by infinity; 08-13-17 at 08:58 PM. Reason: missing word
    Rob

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    All responses based on the 2014 NEC unless otherwise noted

  9. #19
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    Quote Originally Posted by infinity View Post
    Since they're in series the current would be the same at each resistor but the voltage across each resistor would be different. The sum of all of the voltage drops would be 50 volts, at least that's how I read it.
    I got to thinking after I posted that there were no other voltages mentioned so that must be correct.
    If you aim at nothing, you will hit it every time!

  10. #20
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    Quote Originally Posted by Little Bill View Post
    Is the part in red correct or did they/you mean 50 ohms?
    They probably meant 50V so it's 50V across 50 ohms to make the math easy.

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