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)