Quick Question about secondary current over a 100 W load resistor

Discussion in 'Homework Help' started by C_Love, Jun 28, 2006.

  1. C_Love

    Thread Starter New Member

    Jun 28, 2006
    3
    0
    Evening folks,
    Just found your website on google.com and screamed in thanks for finding someplace that might be able to help me with this. I'm going for my Bachelor's in Computer Engineering and the class I'm taking "Fundamental Properties of AC Circuits" has been killing me. A question that I've run into tonight is running me ragged. Your help with this would be greatly appreciated.

    Question: What is the secondary current in an AC circuit with 120Vrms, 100W load resistor, and a 3:1 turns ratio?

    I've tried every equation I can think of and have been able to find in my textbook and just haven't been able to come up with the answer.

    Thanks guys!
     
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Can you post your attempt at a solution so that our members can assist you in finding where you might have gotten of the track?

    hgmjr
     
  3. C_Love

    Thread Starter New Member

    Jun 28, 2006
    3
    0
    I've been trying to concentrate on the following formula:

    Vsec=n(Vpri), where n=.33 and Vpri=120.

    Where I'm losing it is where the 100W Load Resistor falls into it. Basically the question as a whole confuses me because it's throwing in a concept that I've never dealt with before.
     
  4. hgmjr

    Moderator

    Jan 28, 2005
    9,030
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    Is it possible that the resistor is a 100 ohm resistor instead of 100 watt resistor?

    Is this resistor connected across the secondary of the transformer?

    hgmjr
     
  5. C_Love

    Thread Starter New Member

    Jun 28, 2006
    3
    0
    By looking at the question it doesn't give me enough information to even begin to speculate where in the circuit the resistor is located, but the question also states 100W for the resistor. It may be a typo by the publisher, but at this point, even it if was a 100 Ohm resistor I'd still be confused.
     
  6. mozikluv

    AAC Fanatic!

    Jan 22, 2004
    1,437
    1
    hi,

    use Ohms Law to get your answer. you already have the turns ratio and the load spec. (100ohms or 100w) to derive your Sec. current value.

    what's your answer on the secondary voltage?

    as to the question of "hgmjr" to you, presume it as place across the secondary since the resistor in question is acting as a load in your circuit.
     
  7. windoze killa

    AAC Fanatic!

    Feb 23, 2006
    605
    24
    If the 120Vrms is the secondary voltage the you don't care about the turns ratio. If it is the primary voltage then you need to calculate the secondary voltage.

    Second point. You have stated there is a 100W load resistor. I think what it should read is a resistor that is producing 100W. You can have a 100W load resistor that is only dissapating 1W.

    Assuming it is a resistor that is dissapating 100W you can use the formula P=IE. You have calculated E using the transformer ratio, you know P=100W. Should be easy from here.
     
  8. rukrazy?

    Member

    Mar 5, 2005
    21
    0
    Very easy, but I warn you, I have only a two year stint in electronics at college without a degree.
    120vrms is on the primary
    the ratio is 3 to 1
    so for every volt in there is .3 volts out.
    So for 120v *.3 = 36V on secondary
    If the resistor is dissipating 100 watts then using OHMS law
    I = W/V or 100w/36v = 2.7 amps
    Don't make it complicated, hope you can use this and good luck.
    If it were an inductive or capacitive load, you would have a lot more calculations involved.

    My motto KISS
     
  9. nomurphy

    AAC Fanatic!

    Aug 8, 2005
    567
    12
    If one presumes the 120V RMS to be on the primary (which brings the turns ratio into the problem), and the load to be on the secondary (naturally), then the secondary voltage would be 120/3 = 40V RMS.

    100W RMS / 40V RMS = 2.5A RMS

    Don't leave off the RMS units as stated, since it is an AC circuit.

    Obviously, this question leaves a lot to be assumed.
     
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