An Ideal Transformer Example - Revision history

Tyler.anderson: /* Reviwed By */

2010-01-25T00:34:54Z

Reviwed By

← Older revision		Revision as of 17:34, 24 January 2010
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	===Reviwed By===		===Reviwed By===
	Andrew Sell - Chris, everything looks fine, though I would do some extra formatting if possible to help make the problem flow a little smoother as you read it, and locate the picture a little higher to help bring the solution together.		Andrew Sell - Chris, everything looks fine, though I would do some extra formatting if possible to help make the problem flow a little smoother as you read it, and locate the picture a little higher to help bring the solution together.

			Tyler Anderson - Looks good.

	===Read By===		===Read By===

	John Hawkins		John Hawkins

2010-01-21T17:58:13Z

Solution

← Older revision		Revision as of 10:58, 21 January 2010
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	<math>\ =(45+j27)\Omega</math>		<math>\ =(45+j27)\Omega</math>

	Since <math>{i_{1}}=\frac{e_{1}}{Z_{th}}</math>,		Since this is an ideal transformer, it can be modeled by this simple circuit:
			[[Image: Ideal_Circuit.jpg]]

			Therefore, <math>{i_{1}}=\frac{e_{1}}{Z_{th}}</math>,

	<math>{i_{1}}=\frac{120\sqrt{2}}{45+j27} A</math>		<math>{i_{1}}=\frac{120\sqrt{2}}{45+j27} A</math>

	~~Since this is an ideal transformer, it can be modeled by this simple circuit:~~
	~~[[Image: Ideal_Circuit.jpg]]~~

	===Contributors===		===Contributors===

2010-01-21T17:56:05Z

← Older revision		Revision as of 10:56, 21 January 2010
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	<math>=(\frac{N_{1}}{N_{2}})^2{Z_{L}}</math>		<math>=(\frac{N_{1}}{N_{2}})^2{Z_{L}}</math>

	Now, ~~substituting~~:		Now, plugging in the given values:

	<math>\ {Z_{th}} = 3^2(5+j3)</math>		<math>\ {Z_{th}} = 3^2(5+j3)</math>

2010-01-21T17:55:23Z

Solution

2010-01-21T17:55:02Z

Solution

← Older revision		Revision as of 10:55, 21 January 2010
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	So we can substitute, <math>{Z_{th}}=\frac{\frac{N_{1}}{N_{2}}{e_{2}}}{\frac{N_{2}}{N_{1}}{i_{2}}}</math>		So we can substitute, <math>{Z_{th}}=\frac{\frac{N_{1}}{N_{2}}{e_{2}}}{\frac{N_{2}}{N_{1}}{i_{2}}}</math>

	<math>=(\frac{N_{1}}{N_{2}})^2{R_{L}}</math>		<math>=(\frac{N_{1}}{N_{2}})^2{Z_{L}}</math>

	Now, substituting:		Now, substituting:

2010-01-21T17:54:44Z

Solution

← Older revision		Revision as of 10:54, 21 January 2010
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	<math>{Z_{th}} = \frac{e_{1}}{i_{1}}</math>		<math>{Z_{th}} = \frac{e_{1}}{i_{1}}</math>

	Since this is an ideal transformer <math>{e_{1}}=\frac{N_{1}}{N_{2}}{e_{2}}</math> and <math>\{i_{1}}=\frac{N_{2}}{N_{1}}{i_{2}}</math>		Since this is an ideal transformer <math>{e_{1}}=\frac{N_{1}}{N_{2}}{e_{2}}</math> and <math>{i_{1}}=\frac{N_{2}}{N_{1}}{i_{2}}</math>

	So we can substitute, <math>=\frac{\frac{N_{1}}{N_{2}}{e_{2}}}{\frac{N_{2}}{N_{1}}{i_{2}}}</math>		So we can substitute, <math>{Z_{th}}=\frac{\frac{N_{1}}{N_{2}}{e_{2}}}{\frac{N_{2}}{N_{1}}{i_{2}}}</math>

	<math>=(\frac{N_{1}}{N_{2}})^2{R_{L}}</math>		<math>=(\frac{N_{1}}{N_{2}})^2{R_{L}}</math>

2010-01-21T17:53:48Z

Solution

← Older revision		Revision as of 10:53, 21 January 2010
Line 15:		Line 15:
	<math>{Z_{th}} = \frac{e_{1}}{i_{1}}</math>		<math>{Z_{th}} = \frac{e_{1}}{i_{1}}</math>

	Since this is an ideal transformer <math>{e_{1}}=\frac{N_{1}}{N_{2}}{e_{2}}</math> and <math>{i_{1}}=/frac{N_{2}}{N_{1}}{i_{2}}</math>		Since this is an ideal transformer <math>{e_{1}}=\frac{N_{1}}{N_{2}}{e_{2}}</math> and <math>\{i_{1}}=\frac{N_{2}}{N_{1}}{i_{2}}</math>

	So we can substitute, <math>=\frac{\frac{N_{1}}{N_{2}}{e_{2}}}{\frac{N_{2}}{N_{1}}{i_{2}}}</math>		So we can substitute, <math>=\frac{\frac{N_{1}}{N_{2}}{e_{2}}}{\frac{N_{2}}{N_{1}}{i_{2}}}</math>

2010-01-21T17:53:22Z

Solution

← Older revision		Revision as of 10:53, 21 January 2010
Line 15:		Line 15:
	<math>{Z_{th}} = \frac{e_{1}}{i_{1}}</math>		<math>{Z_{th}} = \frac{e_{1}}{i_{1}}</math>

	Since this is an ideal transformer <math>{e_{1}}=\frac{N_{1}}{N_{2}}{e_{2}}</math>		Since this is an ideal transformer <math>{e_{1}}=\frac{N_{1}}{N_{2}}{e_{2}}</math> and <math>{i_{1}}=/frac{N_{2}}{N_{1}}{i_{2}}</math>

	So we can substitute, <math>=\frac{\frac{N_{1}}{N_{2}}{e_{2}}}{\frac{N_{2}}{N_{1}}{i_{2}}}</math>		So we can substitute, <math>=\frac{\frac{N_{1}}{N_{2}}{e_{2}}}{\frac{N_{2}}{N_{1}}{i_{2}}}</math>

2010-01-21T17:51:57Z

Solution

← Older revision		Revision as of 10:51, 21 January 2010
Line 15:		Line 15:
	<math>{Z_{th}} = \frac{e_{1}}{i_{1}}</math>		<math>{Z_{th}} = \frac{e_{1}}{i_{1}}</math>

	Since this is an ideal transformer <math>{e_{1}}=\frac{N_{2}}{N_{1}}{e_{2}}</math>		Since this is an ideal transformer <math>{e_{1}}=\frac{N_{1}}{N_{2}}{e_{2}}</math>

	So we can substitute, <math>=\frac{\frac{N_{2}}{N_{1}}{e_{2}}}{\frac{N_{2}}{N_{1}}{i_{2}}}</math>		So we can substitute, <math>=\frac{\frac{N_{1}}{N_{2}}{e_{2}}}{\frac{N_{2}}{N_{1}}{i_{2}}}</math>

	<math>=(\frac{N_{1}}{N_{2}})^2{R_{L}}</math>		<math>=(\frac{N_{1}}{N_{2}})^2{R_{L}}</math>

2010-01-21T17:51:34Z

Solution

← Older revision		Revision as of 10:51, 21 January 2010
Line 15:		Line 15:
	<math>{Z_{th}} = \frac{e_{1}}{i_{1}}</math>		<math>{Z_{th}} = \frac{e_{1}}{i_{1}}</math>

	Since this is an ideal transformer <math>{e_{1}}=\frac{N_{2}}{N_{1}}{e_{2}}		Since this is an ideal transformer <math>{e_{1}}=\frac{N_{2}}{N_{1}}{e_{2}}</math>

	So we can substitute, <math>=\frac{\frac{N_{1}}{N_{2}}{e_{2}}}{\frac{N_{2}}{N_{1}}{i_{2}}}</math>		So we can substitute, <math>=\frac{\frac{N_{2}}{N_{1}}{e_{2}}}{\frac{N_{2}}{N_{1}}{i_{2}}}</math>

	<math>=(\frac{N_{1}}{N_{2}})^2{R_{L}}</math>		<math>=(\frac{N_{1}}{N_{2}})^2{R_{L}}</math>