Example: Ideal Transformer Exercise: Difference between revisions
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<math>e_2=\frac{N_2}{ |
<math>e_2=\frac{N_2}{N_1}e_1=\frac{2000}{500}(120)=480\text{ V}</math> |
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==Reviewed By== |
==Reviewed By== |
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Tyler Anderson |
*Tyler Anderson |
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Jimmy Apablaza-Lorca |
*Jimmy Apablaza-Lorca |
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==Read By== |
==Read By== |
Latest revision as of 23:13, 31 January 2010
Author
John Hawkins
Problem Statement
An ideal transformer has a primary winding with 500 turns and a secondary winding with 2000 turns. Given that and , find the load impedance, and the Thevenin equivalent, .
Solution
We could find the Thevenin impedance directly, but we will save that until the end as a checking mechanism. First, we will find the actual load impedance by finding the current and voltage in the secondary winding and finding their ratio. The equations used are those derived in class by Professor Frohne.
As mentioned at the beginning, this should be the impedance found using the ratio of the primary voltage and current. Using this method, we find that
This is the same answer as above, which verifies the solutions.
Reviewed By
- Tyler Anderson
- Jimmy Apablaza-Lorca
Read By
Comments
Tyler Anderson: it may be helpful to the readers if you referenced what equations you are using. For example: Otherwise it looks sound to me.
John Hawkins: I didn't use the textbook, so such a reference is not required. I agree that it would be useful for those in the class, but I don't have the same textbook as everyone else, and I doubt anyone would care to know my book's equation numbers. Thanks for reminding me about references, though. I mentioned the class derivation above in the text.
Tyler Anderson: haha fair enough then. props for that. perhaps I could barrow your book sometime? cause ours is absolute crap.
John Hawkins: I hate the book I'm using as well, but if you want to use it sometime that would be fine.
J. Apablaza: Everything looks sound to me. Perhaps, you should include an image so that you can earn some extra points.