Ideal Transformer Example: Difference between revisions
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New page: An idea transformer has a 150-turn primary and 750-turn secondary. The primary is connected to a 240-V, 50-Hz source. The secondary supplies a load of 4 A at a lagging power factor of 0.8.... |
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<math>\ =\frac{150}{750}</math> | <math>\ =\frac{150}{750}</math> | ||
<math>\ =0.2</math> | <math>\ =0.2</math> | ||
(B) Because <math>\ {I_{2}}=4 A</math>, the current in the primary is... | (B) Because <math>\ {I_{2}}=4 A</math>, the current in the primary is... | ||
<math>\ {I_{1}}=\frac{I_{2}}{turns-ratio}</math> | <math>\ {I_{1}}=\frac{I_{2}}{turns-ratio}</math> | ||
<math>\ =\frac{4}{0.2}</math> | <math>\ =\frac{4}{0.2}</math> | ||
<math>\ =20 A</math> | <math>\ =20 A</math> | ||
(C) <math>\ {V_{2}}=\frac{V_{1}}{turns-ratio}</math> | (C) <math>\ {V_{2}}=\frac{V_{1}}{turns-ratio}</math> | ||
<math>\ =\frac{240}{0.2}</math> | <math>\ =\frac{240}{0.2}</math> | ||
<math>\ =1200 V</math> | <math>\ =1200 V</math> | ||
Therefore, the power supplied to the load is... | |||
<math>\ {P_{L}}=V_{2} I_{2}\cos(\theta)</math> | |||
<math>\ =1200 * 4 * 0.8</math> | |||
<math>\ =3840 W</math> | |||
(D) <math>\ {\phi_{m}}=\frac{E_{1}}{4.44 f N_{1}}</math> | |||
<math>\ =\frac{V_{1}}{4.44 f N_{1}}</math> | |||
<math>\ =\frac{240}{4.44 * 50 * 150}</math> | |||
<math>\ =7.21 mWb</math> | |||
Revision as of 01:27, 20 January 2010
An idea transformer has a 150-turn primary and 750-turn secondary. The primary is connected to a 240-V, 50-Hz source. The secondary supplies a load of 4 A at a lagging power factor of 0.8. Find the turns-ratio, the current in the primary, the power supplied to the load, and the flux in the core.
Solution
(A)
(B) Because , the current in the primary is...
(C)
Therefore, the power supplied to the load is...
(D)