ETA practice exam couple questions

Discussion in 'Homework Help' started by woody136, Jan 13, 2015.

  1. woody136

    Thread Starter New Member

    Jan 13, 2015
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    Hello,
    I've been studying for quite awhile now for the ETA associates exam, now it's tomorrow and my coworkers and I are stumped on a couple questions on the practice exam. Please help!

    They are questions 62 and 63.
    Link to the exam: http://www.eta-i.org/practice_exams/aststudy.htm

    The answer to 62 is 4.7V and 63 is 5A.

    Thanks!
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    You need to show your work so that we can use that as a starting point, particularly so we can see how you go about working a problem so that we can make suggestions using the concepts and techniques you are familiar with.

    For Q63, I don't see where they define what Iz is.
     
  3. woody136

    Thread Starter New Member

    Jan 13, 2015
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    Sorry I've tried so many different things to get 4.7 V not even sure what's right anymore, I solved for Xl and Xc and found Z using those two and the 270 ohm resistance.
    Xl = 414.48
    Xc = 24.13
    Z = 474.63 (Total impedance I believe)

    If the numbers are right I can't figure out what to do from there everything I've tried doesn't end at 4.7V.

    Iz is the current for the circuit, they put the Z on because it involves reactance numbers I believe.
     
  4. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    I did it using complex numbers:
    Zl=2*pi*440*0.150j
    Zc=1/(2*pi*440*0.000015j)
    I found equivalent impedance of the Zl and Zc in parallel.
    Then I used voltage divider to find voltage across the equivalent impedance. It came out to be 0.44548-4.69850j, this is complex number, which means that it contains both magnitude and angle. The magnitude is equal to 4.719 volts. It seems to match the provided answer.
     
  5. woody136

    Thread Starter New Member

    Jan 13, 2015
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    It's relieving to know the answer they gave wasn't a typo. Thank you, I do have a question though. I'm unfamiliar with the voltage divider, looking quick online I'm not sure how to apply it to this and my test is tomorrow. Could you break it down a bit?

     
  6. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    The usual way to use voltage divider is to have two resistors in series.

    Lets say I have 50 volt source. Across it I have two resistors in series, 2 Ohm and 3 Ohm.

    What is the voltage across 2 Ohm resistor?
    V 2 ohm=[2/(2+3)]*50=20 volts

    What is voltage across 3 Ohm resistor?
    V 3 ohm=[3/(2+3)]*50=30 volts

    We can double check these answers. How? We know that resistors in series add up. So 2 Ohm in series with 3 Ohm add up to 2+3=5 Ohm. The same is true for the voltage across the resistors. So 20 volts across 2 Ohm resistor add to 30 volt across 3 Ohm resistor and give us 50 volts. What was the input into voltage divider? 50 volts. So we put 50 volts in and we received 50 volts out.

    The point of voltage divider is to enable user to receive a specific portion of the input. For example my input is 50 volts. But I want to power a LED. I don't need all of 50 volts, I only need a small portion of it. So I would throw together a couple of resistors and get the portion of 50 volts that I need.

    Here is article about it from this website: http://www.allaboutcircuits.com/vol_1/chpt_6/1.html
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    The reactance of a capacitor is negative, so you should get -24.13Ω for the capacitive reactance. Now, if your material has both as positive and then has things like (Xl-Xc) scattered around in formulas, then that will work but it is asking for trouble.

    I don't know how you got Z of 474.63Ω because you don't show your work. But we can ask if this answer makes sense by considering that you have a resistance of 270Ω in series with the effective reactance of the inductor and capacitor. If it was a second resistor instead of reactive elements, then to get 475Ω the second resistor would have to be 205Ω. Since it IS reactive, the magnitude of the reactance would have to be

    <br />
|Z| \; = \; \sqrt{|R|^2+|X|^2}<br />
\,<br />
|X| \; = \; \sqrt{|Z|^2-|R|^2} \; = \; \sqrt{|475 \Omega |^2-|270 \Omega|^2} \; = \; 391 \Omega<br />

    While it's a bit difficult to see, there is no way that a 24Ω capacitive reactance in parallel with a 414Ω inductive reactance is going to result in a nearly 400Ω reactance. When the two reactances differ by a significant amount, the smaller one will dominate (just like with resistors in parallel). So you should expect to see the parallel combination look fairly close to a 25Ω capacitive reactance, making the total impedance somewhere in the ballpark of 270Ω (slightly more than 270Ω and significantly less than 300Ω).

    Once you have the total impedance, you can calculate the total current. Once you have the total current you can multiply that by the effective reactance of the parallel combination of the inductor and capacitor to get the magnitude of the voltage across them.
     
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