Example: Step Down Transformer: Difference between revisions
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From the given data of the step down transformer we can see that |
From the given data of the step down transformer we can see that |
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<math>~~~a=2~and~s=48000~\angle \cos^{-1}(0.7)</math> |
<math>~~~a=2~and~s=48000~\angle \cos^{-1}(0.7)</math> |
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<math>I_2~=~\frac{s^*}{v_2^*}~=~200\angle -46^\circ </math> |
<math>I_2~=~\frac{s^*}{v_2^*}~=~200\angle -46^\circ </math> |
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<math> R_H~+~a^2~R_l~=~1.6</math> |
<math> R_H~+~a^2~R_l~=~1.6</math> |
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<math>j~(X_H~+A^2~X_L)~=~J~4~=~\frac{1}{a}~\vec I_2~\left((R_H~+~a^2~R_L)~+~j(X_H~+~a^2X_L))+a\vecV_2\right)</math> |
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<math> |
<math>j~(X_H~+A^2~X_L)~=~J~4~=~\frac{1}{a}~\vec I_2~\left((R_H~+~a^2~R_L)~+~j(X_H~+~a^2X_L))+a\vec V_2\right)</math> |
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<math>P_{core}~=~ P_c~=~250 w = \frac{v_1^2}{R_{cH}}=\frac{893^2}{3200}</math> |
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<math>VR~=~\frac{V_1-a~V_2}{a~V_2}</math> |
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Conclusion: |
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<math>n~=~67~% </math> |
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<math> VR~=~86~%</math> |
Latest revision as of 02:55, 29 January 2010
Problem Statement
Given a 48 kVA, 480/240 V/V step down transformer operating at rated load with a power factor of 0.7 (lag), determine the efficiency and voltage regulation.
Given:
- this problem is from EMEC EXAM 1, Winter 08 by legendary Dr. Cross
Solution
From the given data of the step down transformer we can see that
Conclusion: