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#1
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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! |
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#2
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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
__________________
"No human being should be allowed to have this much fun." hgmjr Engineering is the discipline of designing in the presence of real world constraints.
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#3
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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. |
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#4
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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
__________________
"No human being should be allowed to have this much fun." hgmjr Engineering is the discipline of designing in the presence of real world constraints.
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#5
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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.
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#6
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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.
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THE BIGGEST ROOM IN THE WORLD IS THE ROOM FOR IMPROVEMENT |
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#7
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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.
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Eddy Current is a little dizzy over Milli Volt. |
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#8
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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 Last edited by rukrazy?; 07-06-2006 at 05:24 PM. |
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#9
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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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