Example Problems with Transformers: Difference between revisions
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==Transformer Example== |
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'''Kevin Starkey, Nick Christman, Aric Vyhmeister''' |
'''[[Kevin Starkey EMEC]], [[Nick Christman]], [[Aric Vyhmeister]]''' |
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<u>'''Problem:'''</u> |
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⚫ | A step down transformer has a winding of <math> N_1 = 10 \text{ turns and } N_2 = 2 </math> turns. (a) If the input voltage is 1200V, what is the resulting output voltage? (b) If the input and output currents are <math>5A \mbox{ and } 20A</math>, respectively, what is the current loss due to the leakage inductance, <math>i_{m}(t)</math>. |
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<math> e_2 = \frac{N_2}{N_1}e_1 </math>. |
<math> e_2 = \left( \frac{N_2}{N_1} \right) e_1 </math>. |
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With the information |
With the information provided we can now determine the output voltage: |
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<math> e_2 = \frac{2}{10} |
<math> e_2 = \left( \frac{N_2}{N_1} \right) e_1 = \left( \frac{2}{10} \right) 1200V = 240V </math> |
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b) From the figure above, we need to find the true value of current (i.e. the current that is not lost by leakage inductance). To accomplish this we will use equation 5-40 (Mohan 5-23) to obtain the value of <math>i_{1}^{'}</math>. That is, |
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<math>i_{1}^{'} = \left( \frac{N_2}{N_1} \right) i_{2} = \left( \frac{2}{10} \right) 20A = 4A</math>. |
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Now from equation 5-42 (Mohan 5-23) we can obtain the current loss due to the leakage inductance, which is |
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<math>i_{m} = i_{1} - i_{1}^{'} = 5A - 4A = 1A</math>. |
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Therefore, the current loss in our transformer is <math>1A</math>, which means this specific transformer is very "leaky." |
Latest revision as of 17:03, 26 January 2010
Transformer Example
Kevin Starkey EMEC, Nick Christman, Aric Vyhmeister
Problem:
A step down transformer has a winding of turns. (a) If the input voltage is 1200V, what is the resulting output voltage? (b) If the input and output currents are , respectively, what is the current loss due to the leakage inductance, .
Solution:
a) By modifying equation 5-39 (Mohan 5-22) we can obtain an equation for the output voltage. That is,
.
With the information provided we can now determine the output voltage:
b) From the figure above, we need to find the true value of current (i.e. the current that is not lost by leakage inductance). To accomplish this we will use equation 5-40 (Mohan 5-23) to obtain the value of . That is,
.
Now from equation 5-42 (Mohan 5-23) we can obtain the current loss due to the leakage inductance, which is
.
Therefore, the current loss in our transformer is , which means this specific transformer is very "leaky."