Problem Set 1 - Revision history

Jodi.Hodge at 06:43, 9 February 2010

2010-02-09T06:43:14Z

← Older revision		Revision as of 23:43, 8 February 2010
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	'''~~Problem~~'''		'''PROBLEM'''

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	----		----

	'''~~Solution~~'''		'''SOLUTION'''

	Power is stored in resistors because of their linear behavior.		Power is stored in resistors because of their linear behavior.
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	Both Poc and Voc are giving so solve for R		Both Poc and Voc are giving so solve for R

	<math>R=Voc^{2}/Poc=5V/10W=2.5 Ohms\!</math> '' Answer''		<math>R=Voc^{2}/Poc=5V/10W=2.5 Ohms\!</math> ''''' Answer'''''

	Apparent power relation to voltage and current is		Apparent power relation to voltage and current is
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	<math>Q=Voc^{2}/X\!</math>		<math>Q=Voc^{2}/X\!</math>

	<math>X=Voc^{2}/Q=5V^{2}/224.8VAr=0.11H \!</math> ''Answer''		<math>X=Voc^{2}/Q=5V^{2}/224.8VAr=0.11H \!</math> '''''Answer'''''

Jodi.Hodge at 06:42, 9 February 2010

2010-02-09T06:42:08Z

← Older revision		Revision as of 23:42, 8 February 2010
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	Both Poc and Voc are giving so solve for R		Both Poc and Voc are giving so solve for R

	<math>R=Voc^{2}/Poc=5V/10W=2.5 Ohms\!</math> Answer		<math>R=Voc^{2}/Poc=5V/10W=2.5 Ohms\!</math> '' Answer''

	Apparent power relation to voltage and current is		Apparent power relation to voltage and current is
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	<math>Q=Voc^{2}/X\!</math>		<math>Q=Voc^{2}/X\!</math>

	<math>X=Voc^{2}/Q=5V^{2}/224.8VAr=0.11H \!</math> Answer		<math>X=Voc^{2}/Q=5V^{2}/224.8VAr=0.11H \!</math> ''Answer''

Jodi.Hodge at 06:41, 9 February 2010

2010-02-09T06:41:21Z

← Older revision		Revision as of 23:41, 8 February 2010
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	<math>S^{2}=P^{2}+Q^{2}\!</math>		<math>S^{2}=P^{2}+Q^{2}\!</math>

	~~Use~~ the apparent ~~power~~ and ~~the~~ real power ~~to find the reactive power~~		Solve the above equation for reactive power using the values of apparent and real power.

	<math>Q=\sqrt{S^{2}-P^{2}}= \sqrt{225^{2}-10^{2}}= 224.8VAr\!</math> ~~Answer~~		<math>Q=\sqrt{S^{2}-P^{2}}= \sqrt{225^{2}-10^{2}}= 224.8VAr\!</math>

	Reactive power is stored in non-linear impedance represented as X. In this case X is inductance.		Reactive power is stored in non-linear impedance represented as X. In this case X is inductance.
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	<math>Q=Voc^{2}/X\!</math>		<math>Q=Voc^{2}/X\!</math>

	<math>X=Voc^{2}/Q=5V^{2}/224.8VAr=0.11H \!</math>		<math>X=Voc^{2}/Q=5V^{2}/224.8VAr=0.11H \!</math> Answer

Jodi.Hodge at 06:36, 9 February 2010

2010-02-09T06:36:54Z

← Older revision		Revision as of 23:36, 8 February 2010
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	<math>S=P+iQ\!</math>		<math>S=P+iQ\!</math>

	Apparent power in the imaginary and real coordinate system is the hypotenuse of real and reactive power and ~~are related~~ using the equation of a circle.		Apparent power in the imaginary and real coordinate system is the hypotenuse of real and reactive power and can be expressed using the equation of a circle.

	<math>S^{2}=P^{2}+Q^{2}\!</math>		<math>S^{2}=P^{2}+Q^{2}\!</math>

Jodi.Hodge at 06:34, 9 February 2010

2010-02-09T06:34:38Z

← Older revision		Revision as of 23:34, 8 February 2010
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	Find the core inductance and ~~resitence~~ of a transformer using ~~measurmnets~~ Voc=5V, Ioc=3A, and Poc =10W.		Find the core inductance and resistance of a transformer using measurements Voc=5V, Ioc=3A, and Poc =10W.

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	'''Solution'''		'''Solution'''

	Power is ~~found only through resister~~		Power is stored in resistors because of their linear behavior.

	<math>Poc= Voc^{2}/R\!</math>		<math>Poc= Voc^{2}/R\!</math>
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	Both Poc and Voc are giving so solve for R		Both Poc and Voc are giving so solve for R

	<math>R=Voc^{2}/Poc=5V/10W=2.5 Ohms\!</math>		<math>R=Voc^{2}/Poc=5V/10W=2.5 Ohms\!</math> Answer

	~~To calclate the core inductance we will use the formular for apparent~~ power.		Apparent power relation to voltage and current is

	~~<math>S^{2}=P^{2}+Q^{2}\!</math>~~

	<math>S^{2}=\|VocIoc^{}\|^{2}\!</math>		<math>S^{2}=\|VocIoc^{}\|^{2}\!</math>

	~~Subtituting~~ in values and ~~solving gives~~		The magnitude for current and voltage are given in the problem statement. Subtitute in known values and solve.

	<math>S^{2}=(5V*3A)^{2}=225(VA)^{2}\!</math>		<math>S^{2}=(5V*3A)^{2}=225(VA)^{2}\!</math>

			Apparent power S is complex. It has a real part P, real power, and an imaginary part Q, reactive power.

			<math>S=P+iQ\!</math>

			Apparent power in the imaginary and real coordinate system is the hypotenuse of real and reactive power and are related using the equation of a circle.

			<math>S^{2}=P^{2}+Q^{2}\!</math>

	Use the apparent power and the real power to find the reactive power		Use the apparent power and the real power to find the reactive power

	<math>Q=\sqrt{S^{2}-P^{2}}= \sqrt{225^{2}-10^{2}}= 224.8VAr\!</math>		<math>Q=\sqrt{S^{2}-P^{2}}= \sqrt{225^{2}-10^{2}}= 224.8VAr\!</math> Answer

	Reactive power is stored in non-linear ~~impedence~~ represented as X. In this case X is inductance.		Reactive power is stored in non-linear impedance represented as X. In this case X is inductance.

	<math>Q=Voc^{2}/X\!</math>		<math>Q=Voc^{2}/X\!</math>

	<math>X=Voc^{2}/Q=5V^{2}/224.8VAr=0.11H \!</math>		<math>X=Voc^{2}/Q=5V^{2}/224.8VAr=0.11H \!</math>

Jodi.Hodge at 05:58, 9 February 2010

2010-02-09T05:58:27Z

← Older revision		Revision as of 22:58, 8 February 2010
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	'''Solution'''		'''Solution'''
	\!
	Power is found only through resister		Power is found only through resister

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	<math>Q=\sqrt{S^{2}-P^{2}}= \sqrt{225^{2}-10^{2}}= 224.8VAr\!</math>		<math>Q=\sqrt{S^{2}-P^{2}}= \sqrt{225^{2}-10^{2}}= 224.8VAr\!</math>

	~~Finally~~		Reactive power is stored in non-linear impedence represented as X. In this case X is inductance.

	~~<math>Q=Voc^{2}/~~X~~\!</math>~~

	~~Solve for the~~ inductance

	<math>Q=Voc^{2}/X\!</math>		<math>Q=Voc^{2}/X\!</math>

	<math>X=Voc^{2}/Q=5V^{2}/224.8VAr=0.11H \!</math>		<math>X=Voc^{2}/Q=5V^{2}/224.8VAr=0.11H \!</math>

Jodi.Hodge at 05:50, 9 February 2010

2010-02-09T05:50:49Z

← Older revision		Revision as of 22:50, 8 February 2010
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	'''Solution'''		'''Solution'''
			\!
	Power is found only through resister		Power is found only through resister

	<math>Poc= Voc^{2}/R</math>		<math>Poc= Voc^{2}/R\!</math>

	Both Poc and Voc are giving so solve for R		Both Poc and Voc are giving so solve for R

	<math>R=Voc^{2}/Poc=5V/10W=2.5 Ohms</math>		<math>R=Voc^{2}/Poc=5V/10W=2.5 Ohms\!</math>

	To calclate the core inductance we will use the formular for apparent power.		To calclate the core inductance we will use the formular for apparent power.

	<math>S^{2}=P^{2}+Q^{2}</math>		<math>S^{2}=P^{2}+Q^{2}\!</math>

	<math>S^{2}=\|VocIoc^{}\|^{2}</math>		<math>S^{2}=\|VocIoc^{}\|^{2}\!</math>

	Subtituting in values and solving gives		Subtituting in values and solving gives

	<math>S^{2}=(5V*3A)^{2}=225(VA)^{2}</math>		<math>S^{2}=(5V*3A)^{2}=225(VA)^{2}\!</math>

	Use the apparent power and the real power to find the reactive power		Use the apparent power and the real power to find the reactive power

	<math>Q=\sqrt{S^{2}-P^{2}}= \sqrt{225^{2}-10^{2}}= 224.8VAr</math>		<math>Q=\sqrt{S^{2}-P^{2}}= \sqrt{225^{2}-10^{2}}= 224.8VAr\!</math>

	Finally		Finally

	<math>Q=Voc^{2}/X</math>		<math>Q=Voc^{2}/X\!</math>

	Solve for the inductance		Solve for the inductance

	<math>Q=Voc^{2}/X</math>		<math>Q=Voc^{2}/X\!</math>

	<math>X=Voc^{2}/Q=5V^{2}/224.8VAr=0.11H </math>		<math>X=Voc^{2}/Q=5V^{2}/224.8VAr=0.11H \!</math>

Jodi.Hodge at 16:57, 5 February 2010

2010-02-05T16:57:13Z

← Older revision		Revision as of 09:57, 5 February 2010
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	Both Poc and Voc are giving so solve for R		Both Poc and Voc are giving so solve for R

	<math>R=Voc^{2}/Poc=5V/10W=2.5 Ohms</!math>		<math>R=Voc^{2}/Poc=5V/10W=2.5 Ohms</math>

	To calclate the core inductance we will use the formular for ~~appearent~~ power.		To calclate the core inductance we will use the formular for apparent power.

	<math>S^{2}=P^{2}+Q^{2}</math>		<math>S^{2}=P^{2}+Q^{2}</math>
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	<math>S^{2}=\|VocIoc^{}\|^{2}</math>		<math>S^{2}=\|VocIoc^{}\|^{2}</math>

	Subtituting in values ~~aand~~ solving gives		Subtituting in values and solving gives

	<math>S^{2}=(5V*3A)^{2}=225(VA)^{2}</math>		<math>S^{2}=(5V*3A)^{2}=225(VA)^{2}</math>

	<math>Q=\sqrt{P^{2}-Q^{2}}</math>		Use the apparent power and the real power to find the reactive power

			<math>Q=\sqrt{S^{2}-P^{2}}= \sqrt{225^{2}-10^{2}}= 224.8VAr</math>

			Finally

			<math>Q=Voc^{2}/X</math>

			Solve for the inductance

			<math>Q=Voc^{2}/X</math>

	<math>~~{Z_{th}}~~=~~\frac{\frac{N_{1}}{N_~~{2}~~}{e_~~{2}~~}}{\frac{N_{2}}{N_{1}}{i_{2}}}~~</math>		<math>X=Voc^{2}/Q=5V^{2}/224.8VAr=0.11H </math>

Jodi.Hodge at 16:48, 5 February 2010

2010-02-05T16:48:14Z

← Older revision		Revision as of 09:48, 5 February 2010
Line 15:		Line 15:
	Both Poc and Voc are giving so solve for R		Both Poc and Voc are giving so solve for R

	<math>R=Voc^{2}/Poc=5V/10W=2.5 Ohms/!</math>		<math>R=Voc^{2}/Poc=5V/10W=2.5 Ohms</!math>

	To calclate the core inductance we will use the formular for appearent power.		To calclate the core inductance we will use the formular for appearent power.

Jodi.Hodge at 16:47, 5 February 2010

2010-02-05T16:47:45Z

← Older revision		Revision as of 09:47, 5 February 2010
Line 15:		Line 15:
	Both Poc and Voc are giving so solve for R		Both Poc and Voc are giving so solve for R

	<math>R=Voc^{2}/Poc=5V/10W=2.5 Ohms</math>		<math>R=Voc^{2}/Poc=5V/10W=2.5 Ohms/!</math>

	To calclate the core inductance we will use the formular for appearent power.		To calclate the core inductance we will use the formular for appearent power.