RC and RL Exponential Responses - Revision history

Hwang: /* Charging */

2010-01-27T03:15:40Z

Charging

← Older revision		Revision as of 22:15, 26 January 2010
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	<math>v_C(t)=V_0(1-e^{-\frac{t}{RC}})</math>		<math>v_C(t)=V_0(1-e^{-\frac{t}{RC}})</math>

	where <math>V_0</math> is the final voltage across the capacitor.		where <math>V_0=V_S</math>, the final voltage across the capacitor.

	===Discharging===		===Discharging===

Hwang at 20:12, 6 July 2006

2006-07-06T20:12:27Z

← Older revision		Revision as of 15:12, 6 July 2006
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	[[Image:RC_charge_schematic.gif]]		[[Image:RC_charge_schematic.gif]]

	In this example, <math>R=100\Omega</math> and <math>C=10 \mu F</math>.

	Current flows into the capacitor and accumulates a charge there. As the charge increases, the voltage rises, and eventually the voltage of the capacitor equals the voltage of the source, and current stops flowing. The voltage across the capacitor is given by:		Current flows into the capacitor and accumulates a charge there. As the charge increases, the voltage rises, and eventually the voltage of the capacitor equals the voltage of the source, and current stops flowing. The voltage across the capacitor is given by:
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	[[Image:RL_charge_schematic.gif]]		[[Image:RL_charge_schematic.gif]]

	In this example, <math>R=100\Omega</math> and <math>L=200mH</math>.

	The inductor initially has a very high resistance, as energy is going into building up a magnetic field. Once the magnetic field is up and no longer changing, the inductor acts like a short circuit. The current at steady state is equal to <math>I_0=V_s/R</math>. Since the inductor is acting like a short circuit at steady state, the voltage across the inductor then is 0. The current through the inductor is given by:		The inductor initially has a very high resistance, as energy is going into building up a magnetic field. Once the magnetic field is up and no longer changing, the inductor acts like a short circuit. The current at steady state is equal to <math>I_0=V_s/R</math>. Since the inductor is acting like a short circuit at steady state, the voltage across the inductor then is 0. The current through the inductor is given by:

Hwang: /* Charging */

2006-07-06T19:58:22Z

Charging

← Older revision		Revision as of 14:58, 6 July 2006
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	[[Image:RL_charge_schematic.gif]]		[[Image:RL_charge_schematic.gif]]

			In this example, <math>R=100\Omega</math> and <math>L=200mH</math>.

	The inductor initially has a very high resistance, as energy is going into building up a magnetic field. Once the magnetic field is up and no longer changing, the inductor acts like a short circuit. The current at steady state is equal to <math>I_0=V_s/R</math>. Since the inductor is acting like a short circuit at steady state, the voltage across the inductor then is 0. The current through the inductor is given by:		The inductor initially has a very high resistance, as energy is going into building up a magnetic field. Once the magnetic field is up and no longer changing, the inductor acts like a short circuit. The current at steady state is equal to <math>I_0=V_s/R</math>. Since the inductor is acting like a short circuit at steady state, the voltage across the inductor then is 0. The current through the inductor is given by:

	<math>i_L(t)=I_0(1-e^{-\frac{R}{L}t})</math>		<math>i_L(t)=I_0(1-e^{-\frac{R}{L}t})</math>


	===Discharging===		===Discharging===

Hwang: /* Charging */

2006-07-06T19:57:27Z

Charging

← Older revision		Revision as of 14:57, 6 July 2006
Line 17:		Line 17:

	[[Image:RC_charge_schematic.gif]]		[[Image:RC_charge_schematic.gif]]

			In this example, <math>R=100\Omega</math> and <math>C=10 \mu F</math>.

	Current flows into the capacitor and accumulates a charge there. As the charge increases, the voltage rises, and eventually the voltage of the capacitor equals the voltage of the source, and current stops flowing. The voltage across the capacitor is given by:		Current flows into the capacitor and accumulates a charge there. As the charge increases, the voltage rises, and eventually the voltage of the capacitor equals the voltage of the source, and current stops flowing. The voltage across the capacitor is given by:

Hwang at 19:53, 6 July 2006

2006-07-06T19:53:26Z

Hwang at 16:29, 5 July 2006

2006-07-05T16:29:22Z

← Older revision		Revision as of 11:29, 5 July 2006
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	Remember that for an inductor, <math>v(t)=L*di/dt</math>. Note that the voltage across the inductor can change instantly at ''t=0'', but the current changes slowly.		Remember that for an inductor, <math>v(t)=L*di/dt</math>. Note that the voltage across the inductor can change instantly at ''t=0'', but the current changes slowly.

			==References==
			Hayt, William H. Jr., Jack E. Kemmerly, and Steven M. Durbin. <u>Engineering Circuit Analysis</u>. 6th ed. New York:McGraw-Hill, 2002.

Hwang: /* RL Circuits */

2006-06-20T22:02:42Z

RL Circuits

← Older revision		Revision as of 17:02, 20 June 2006
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	If the inductor is initially uncharged and we want to charge it by inserting a voltage source <math>V_s</math> in the RL circuit:		If the inductor is initially uncharged and we want to charge it by inserting a voltage source <math>V_s</math> in the RL circuit:

	[[Image:RL_charge_schematic.~~jpg~~]]		[[Image:RL_charge_schematic.gif]]

	The inductor initially has a very high resistance, as energy is going into building up a magnetic field. Once the magnetic field is up and no longer changing, the inductor will act like a short circuit. The current at steady state will be equal to <math>I_0=V_s/R</math>. Since the inductor is acting like a short circuit at steady state, the voltage across the inductor then will be 0. The current through the inductor is given by:		The inductor initially has a very high resistance, as energy is going into building up a magnetic field. Once the magnetic field is up and no longer changing, the inductor will act like a short circuit. The current at steady state will be equal to <math>I_0=V_s/R</math>. Since the inductor is acting like a short circuit at steady state, the voltage across the inductor then will be 0. The current through the inductor is given by:
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	In the following circuit, the inductor initially has current <math>I_0=V_s/R</math> flowing through it; we replace the voltage source with a short circuit at <math>t=0</math>.		In the following circuit, the inductor initially has current <math>I_0=V_s/R</math> flowing through it; we replace the voltage source with a short circuit at <math>t=0</math>.

	[[Image:RL_discharge_schematic.~~jpg~~]]		[[Image:RL_discharge_schematic.gif]]

	After we cut out the voltage source, the voltage across the inductor will be <math>I_0*R</math>, but the higher voltage will be at the negative terminal of the inductor. Thus, <math>I_0=-V/R</math>. The the current flowing through the inductor at time ''t'' is given by:		After we cut out the voltage source, the voltage across the inductor will be <math>I_0*R</math>, but the higher voltage will be at the negative terminal of the inductor. Thus, <math>I_0=-V/R</math>. The the current flowing through the inductor at time ''t'' is given by:
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	! !!Voltage !! Current		! !!Voltage !! Current
	\|-		\|-
	!Charge !! [[Image:RL_charge_voltage.~~jpg~~]] \|\| [[Image:RL_charge_current.~~jpg~~]]		!Charge !! [[Image:RL_charge_voltage.gif]] \|\| [[Image:RL_charge_current.gif]]
	\|-		\|-
	!Discharge !! [[Image:RL_discharge_voltage.~~jpg~~]] \|\| [[Image:RL_discharge_current.~~jpg~~]]		!Discharge !! [[Image:RL_discharge_voltage.gif]] \|\| [[Image:RL_discharge_current.gif]]
	\|-		\|-
	\|}		\|}

Hwang: /* Discharging */

2006-06-20T21:58:49Z

Discharging

← Older revision		Revision as of 16:58, 20 June 2006
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	Consider the following circuit:		Consider the following circuit:

	[[Image:RC_discharge_schematic.~~jpg~~]]		[[Image:RC_discharge_schematic.gif]]

	In the circuit, the capacitor is initially charged and has voltage <math>V_0</math> aross it, and the switch is initially open. At time <math>t=0</math>, we close the circuit and the capacitor will discharge through the resistor. The voltage across a capacitor discharging through a resistor as a function of time is given as:		In the circuit, the capacitor is initially charged and has voltage <math>V_0</math> aross it, and the switch is initially open. At time <math>t=0</math>, we close the circuit and the capacitor will discharge through the resistor. The voltage across a capacitor discharging through a resistor as a function of time is given as:
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	! !!Voltage!! Current		! !!Voltage!! Current
	\|-		\|-
	!Charge !! [[Image:RC_charge_voltage.~~jpg~~]] \|\| [[Image:RC_charge_current.~~jpg~~]]		!Charge !! [[Image:RC_charge_voltage.gif]] \|\| [[Image:RC_charge_current.gif]]
	\|-		\|-
	!Discharge !! [[Image:RC_discharge_voltage.~~jpg~~]] \|\| [[Image:RC_discharge_current.~~jpg~~]]		!Discharge !! [[Image:RC_discharge_voltage.gif]] \|\| [[Image:RC_discharge_current.gif]]
	\|-		\|-
	\|}		\|}

Hwang: /* Charging */

2006-06-20T21:57:25Z

Charging

← Older revision		Revision as of 16:57, 20 June 2006
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	If the capacitor is initially uncharged and we want to charge it with a voltage source <math>V_s</math> in the RC circuit:		If the capacitor is initially uncharged and we want to charge it with a voltage source <math>V_s</math> in the RC circuit:

	[[Image:RC_charge_schematic.~~jpg~~]]		[[Image:RC_charge_schematic.gif]]

	Current will flow into the capacitor and accumulate a charge there. As the charge increases, the voltage rises, and eventually the voltage of the capacitor will equal the voltage of the source, and current will stop flowing. The voltage across the capacitor is given by:		Current will flow into the capacitor and accumulate a charge there. As the charge increases, the voltage rises, and eventually the voltage of the capacitor will equal the voltage of the source, and current will stop flowing. The voltage across the capacitor is given by:

Hwang: /* RL Circuits */

2006-06-20T15:53:30Z

RL Circuits

← Older revision		Revision as of 10:53, 20 June 2006
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	[[Image:RL_charge_schematic.jpg]]		[[Image:RL_charge_schematic.jpg]]

	The inductor initially has a very high resistance, as energy is going into building up a magnetic field. Once the magnetic field is up and no longer changing, the inductor will act like a short circuit. The current at steady state will be equal to <math>~~V_0~~/R~~</math> and in this case, <math>V_0=V_s~~</math>. Since the inductor is acting like a short circuit at steady state, the voltage across the inductor will be 0. The current through the inductor is given by:		The inductor initially has a very high resistance, as energy is going into building up a magnetic field. Once the magnetic field is up and no longer changing, the inductor will act like a short circuit. The current at steady state will be equal to <math>I_0=V_s/R</math>. Since the inductor is acting like a short circuit at steady state, the voltage across the inductor then will be 0. The current through the inductor is given by:

		⚫	<math>i_L(t)=I_0(1-e^{-\frac{R}{L}t})</math>

⚫	<math>i_L(t)=I_0~~(1-e^{-\frac{R}{L}t})=\frac{V_0}{R}~~(1-e^{-\frac{R}{L}t})</math>

	===Discharging===		===Discharging===

← Older revision		Revision as of 14:53, 6 July 2006
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	[[Image:RC_charge_schematic.gif]]		[[Image:RC_charge_schematic.gif]]

	Current ~~will flow~~ into the capacitor and ~~accumulate~~ a charge there. As the charge increases, the voltage rises, and eventually the voltage of the capacitor ~~will equal~~ the voltage of the source, and current ~~will stop~~ flowing. The voltage across the capacitor is given by:		Current flows into the capacitor and accumulates a charge there. As the charge increases, the voltage rises, and eventually the voltage of the capacitor equals the voltage of the source, and current stops flowing. The voltage across the capacitor is given by:

	<math>v_C(t)=V_0(1-e^{-\frac{t}{RC}})</math>		<math>v_C(t)=V_0(1-e^{-\frac{t}{RC}})</math>
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	[[Image:RC_discharge_schematic.gif]]		[[Image:RC_discharge_schematic.gif]]

	In the circuit, the capacitor is initially charged and has voltage <math>V_0</math> ~~aross~~ it, and the switch is initially open. At time <math>t=0</math>, we close the circuit and the capacitor ~~will~~ discharge through the resistor. The voltage across a capacitor discharging through a resistor as a function of time is given as:		In the circuit, the capacitor is initially charged and has voltage <math>V_0</math> across it, and the switch is initially open. At time <math>t=0</math>, we close the circuit and allow the capacitor to discharge through the resistor. The voltage across a capacitor discharging through a resistor as a function of time is given as:

	<math>v_C(t)=V_0e^{-\frac{t}{RC}}</math>		<math>v_C(t)=V_0e^{-\frac{t}{RC}}</math>
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	where <math>V_0</math> is the initial voltage across the capacitor.		where <math>V_0</math> is the initial voltage across the capacitor.

	The term ''RC'' is the resistance of the resistor multiplied by the capacitance of the capacitor, and known as the ''time constant'', which is a unit of time. The ~~value of the~~ function ~~will be~~ 63% of the final ~~value~~ at <math>t=1RC</math>, and over 99.99% of the ~~final value~~ at <math>t=5RC</math>.		The term ''RC'' is the resistance of the resistor multiplied by the capacitance of the capacitor, and known as the ''time constant'', which is a unit of time. The function completes 63% of the transition between the initial and final states at <math>t=1RC</math>, and completes over 99.99% of the transition at <math>t=5RC</math>.

	The voltage and current of the capacitor in the circuits above are shown in the graphs below, from ''t=0'' to ''t=5RC''. Note the polaritiy—the voltage is the voltage measured at the "+" terminal of the capacitor relative to the ground (0V). A positive current flows into the capacitor from this terminal; a negative current flows out of this terminal.:		The voltage and current of the capacitor in the circuits above are shown in the graphs below, from ''t=0'' to ''t=5RC''. Note the polaritiy—the voltage is the voltage measured at the "+" terminal of the capacitor relative to the ground (0V). A positive current flows into the capacitor from this terminal; a negative current flows out of this terminal.:
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	[[Image:RL_charge_schematic.gif]]		[[Image:RL_charge_schematic.gif]]

	The inductor initially has a very high resistance, as energy is going into building up a magnetic field. Once the magnetic field is up and no longer changing, the inductor ~~will act~~ like a short circuit. The current at steady state ~~will be~~ equal to <math>I_0=V_s/R</math>. Since the inductor is acting like a short circuit at steady state, the voltage across the inductor then ~~will be~~ 0. The current through the inductor is given by:		The inductor initially has a very high resistance, as energy is going into building up a magnetic field. Once the magnetic field is up and no longer changing, the inductor acts like a short circuit. The current at steady state is equal to <math>I_0=V_s/R</math>. Since the inductor is acting like a short circuit at steady state, the voltage across the inductor then is 0. The current through the inductor is given by:

	<math>i_L(t)=I_0(1-e^{-\frac{R}{L}t})</math>		<math>i_L(t)=I_0(1-e^{-\frac{R}{L}t})</math>
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	[[Image:RL_discharge_schematic.gif]]		[[Image:RL_discharge_schematic.gif]]

	After we cut out the voltage source, the voltage across the inductor ~~will be~~ <math>I_0*R</math>, but the higher voltage ~~will~~ be at the negative terminal of the inductor. Thus, <math>I_0=-V/R</math>. The ~~the~~ current flowing through the inductor at time ''t'' is given by:		After we cut out the voltage source, the voltage across the inductor is <math>I_0*R</math>, but the higher voltage is now at the negative terminal of the inductor. Thus, <math>I_0=-V/R</math>. The current flowing through the inductor at time ''t'' is given by:

	<math>i_L(t)=I_0e^{-\frac{R}{L}t}</math>		<math>i_L(t)=I_0e^{-\frac{R}{L}t}</math>
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	The ''time constant'' for the RL circuit is equal to <math>L/R</math>.		The ''time constant'' for the RL circuit is equal to <math>L/R</math>.

	The voltage and current of the inductor for the circuits above are given by the graphs below, from ''t=0'' to ''t=5L/R''. The voltage is measured at the "+" terminal of the inductor, relative to the ground. A positive current flows into the inductor from this terminal; a negative current flows out of this terminal.:		The voltage and current of the inductor for the circuits above are given by the graphs below, from ''t=0'' to ''t=5L/R''. The voltage is measured at the "+" terminal of the inductor, relative to the ground. A positive current flows into the inductor from this terminal; a negative current flows out of this terminal:

	{\| border="1" cellspacing="0" cellpadding="5" align="center"		{\| border="1" cellspacing="0" cellpadding="5" align="center"