Current - Revision history

Jserketich3: Wiki Resource Submission JJS Spring 2022

2022-04-18T03:32:55Z

Wiki Resource Submission JJS Spring 2022

Weerapitch3 at 08:39, 18 April 2016

2016-04-18T08:39:37Z

← Older revision		Revision as of 04:39, 18 April 2016
Line 107:		Line 107:
	<math> I = \|q\|nAv = (1.6⋅10^{-19} C)⋅(2.1 ⋅ 10^{29}m^{-3})⋅(8.04⋅10^{-6}m^2)⋅(2.96⋅10^{-5}m/s)=8.00 A</math>		<math> I = \|q\|nAv = (1.6⋅10^{-19} C)⋅(2.1 ⋅ 10^{29}m^{-3})⋅(8.04⋅10^{-6}m^2)⋅(2.96⋅10^{-5}m/s)=8.00 A</math>

			==Connectedness==
			A current is something which exists is any circuit and affects many other attributes. For example, currents also create magnetic fields. A major application of current is in transformers, since transformers rely on inductance to convert the amount of current and potential difference while maintaining the same amount of power. Fields that deal directly with electricity and circuits, such as electrical and computer engineering may often find themselves dealing with currents.
	==History==		==History==
	A giant leap in our understanding of electric charge and current was advocated by Benjamin Franklin who tried many different experiments with charges. He was the first to coin the terms positive and negative charges. However, evidence and electric currents was not provided until around 1800 when batteries were invented and currents could be investigated further. Initially, scientists thought that charges and currents were different phenomena, but today we understand that currents are simply moving charges.		A giant leap in our understanding of electric charge and current was advocated by Benjamin Franklin who tried many different experiments with charges. He was the first to coin the terms positive and negative charges. However, evidence and electric currents was not provided until around 1800 when batteries were invented and currents could be investigated further. Initially, scientists thought that charges and currents were different phenomena, but today we understand that currents are simply moving charges.

Weerapitch3: /* History */

2016-04-18T08:31:17Z

History

← Older revision		Revision as of 04:31, 18 April 2016
Line 110:		Line 110:
	A giant leap in our understanding of electric charge and current was advocated by Benjamin Franklin who tried many different experiments with charges. He was the first to coin the terms positive and negative charges. However, evidence and electric currents was not provided until around 1800 when batteries were invented and currents could be investigated further. Initially, scientists thought that charges and currents were different phenomena, but today we understand that currents are simply moving charges.		A giant leap in our understanding of electric charge and current was advocated by Benjamin Franklin who tried many different experiments with charges. He was the first to coin the terms positive and negative charges. However, evidence and electric currents was not provided until around 1800 when batteries were invented and currents could be investigated further. Initially, scientists thought that charges and currents were different phenomena, but today we understand that currents are simply moving charges.

	==See also		==See also==

	===Further readings===		===Further readings===
Line 118:		Line 118:

	http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecur.html		http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecur.html

	==References==		==References==
	https://en.wikipedia.org/wiki/Electric_current		https://en.wikipedia.org/wiki/Electric_current

Weerapitch3 at 08:30, 18 April 2016

2016-04-18T08:30:51Z

← Older revision		Revision as of 04:30, 18 April 2016
Line 107:		Line 107:
	<math> I = \|q\|nAv = (1.6⋅10^{-19} C)⋅(2.1 ⋅ 10^{29}m^{-3})⋅(8.04⋅10^{-6}m^2)⋅(2.96⋅10^{-5}m/s)=8.00 A</math>		<math> I = \|q\|nAv = (1.6⋅10^{-19} C)⋅(2.1 ⋅ 10^{29}m^{-3})⋅(8.04⋅10^{-6}m^2)⋅(2.96⋅10^{-5}m/s)=8.00 A</math>

			==History==
			A giant leap in our understanding of electric charge and current was advocated by Benjamin Franklin who tried many different experiments with charges. He was the first to coin the terms positive and negative charges. However, evidence and electric currents was not provided until around 1800 when batteries were invented and currents could be investigated further. Initially, scientists thought that charges and currents were different phenomena, but today we understand that currents are simply moving charges.

			==See also

			===Further readings===
			Chabay, Ruth W., and Bruce A. Sherwood. Matter & Interactions. N.p.: n.p., n.d. Print.
			===External Links===
			http://physics.info/electric-current/

			http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecur.html
	==References==		==References==
	https://en.wikipedia.org/wiki/Electric_current		https://en.wikipedia.org/wiki/Electric_current
Line 117:		Line 128:

	http://physics.info/electric-current/		http://physics.info/electric-current/

			http://practicalphysics.org/electric-charge-and-current-short-history.html

	[[Category:Which Category did you place this in?]]		[[Category:Which Category did you place this in?]]

Weerapitch3: /* Examples */

2016-04-18T08:19:57Z

Examples

← Older revision		Revision as of 04:19, 18 April 2016
Line 85:		Line 85:
	Current will not always be what you solve for, though. In this problem, let's use the equations we've discussed to find the drift velocity <math>v</math>.		Current will not always be what you solve for, though. In this problem, let's use the equations we've discussed to find the drift velocity <math>v</math>.
	Given the current is <math>1A</math>, the electron density is <math>8 ⋅ 10^{27}m^{-3}</math>, and the diameter of our wire is <math>1mm</math>.		Given the current is <math>1A</math>, the electron density is <math>8 ⋅ 10^{27}m^{-3}</math>, and the diameter of our wire is <math>1mm</math>.

			'''Solution'''

	First, let's find the area <math>{0.001\over 2}^{2}⋅π=7.85⋅10^{-7}</math>		First, let's find the area <math>{0.001\over 2}^{2}⋅π=7.85⋅10^{-7}</math>
Line 91:		Line 93:

	===Difficult===		===Difficult===
	~~In this problem, we are also solving for something other than the current.~~		The density of mobile electrons in aluminum metal is <math>2.1 ⋅ 10^{29}m^{-3}</math>. You have a wire that is 20 cm long with a radius of 1.6 mm, and it takes 112.6 minutes for a single electron in the electron sea to drift from one end to the other. Given that the electron mobility of aluminum is <math>0.0012 m^2/Vs</math>, find the conventional current (I) and the electric field (E) in the wire.
	The density of mobile electrons in aluminum metal is 2.~~1e-29m~~-3. ~~Suppose that i = 4.2 1018 electrons/s are drifting through~~ a ~~copper~~ wire~~. (This~~ is a ~~typical value for a simple circuit.) The diameter~~ of ~~the wire is 2~~.3 mm. ~~In this case, about how many~~ minutes ~~would it take~~ for a single electron in the electron sea to drift from one end to the other ~~end~~ of a wire ~~22 cm long?~~
			'''Solution'''

			First, find E. Since <math>v=μE, E = v/μ</math>. We can find <math>v</math> by realizing that <math>v=Δx/Δt</math>. <math>Δx = 0.20 m</math> and <math>Δt = 112.6 min = (112.6min)⋅(60 s/min) = 6756 s</math>.

			Then, <math>v=Δx/Δt = (0.20m) / (6756 s) = 2.96⋅10^{-5}m/s</math>.

			Therefore, <math>E = v/μ = ({2.96⋅10^{-5}m/s})/({0.0012 {m^2}/Vs}) = 0.0247 V/m</math>

			Now, we can use <math>I = \|q\|nAv</math> to find the conventional current. Note that <math>A = πr^2 = π(0.0016)^2 = 8.04⋅10^{-6}m^2</math> and <math>q= charge of electron = -1.6⋅10^{-19} C</math>

			<math> I = \|q\|nAv = (1.6⋅10^{-19} C)⋅(2.1 ⋅ 10^{29}m^{-3})⋅(8.04⋅10^{-6}m^2)⋅(2.96⋅10^{-5}m/s)=8.00 A</math>

	==References==		==References==

Weerapitch3: /* Difficult */

2016-04-18T07:44:42Z

Difficult

← Older revision		Revision as of 03:44, 18 April 2016
Line 91:		Line 91:

	===Difficult===		===Difficult===
			In this problem, we are also solving for something other than the current.
			The density of mobile electrons in aluminum metal is 2.1e-29m-3. Suppose that i = 4.2 1018 electrons/s are drifting through a copper wire. (This is a typical value for a simple circuit.) The diameter of the wire is 2.3 mm. In this case, about how many minutes would it take for a single electron in the electron sea to drift from one end to the other end of a wire 22 cm long?

	==References==		==References==

Weerapitch3: /* A Mathematical Model */

2016-04-18T07:41:16Z

A Mathematical Model

@@ Line 49: / Line 49: @@
 The current I through the surface is <math>I=∫Nqv⋅dA</math>
 In summary,
@@ Line 55: / Line 65: @@
 <math>I = |q|nAv = |q|nAμE</math> (conventional current)
 ==Examples==

Weerapitch3: /* Examples */

2016-04-18T07:35:28Z

Examples

← Older revision		Revision as of 03:35, 18 April 2016
Line 75:		Line 75:

	Then, plug in the rest: <math>1=\|-1.6⋅10^{-19}\|⋅5⋅10^{28}⋅7.85⋅10^{-7}⋅v, v = 9.9477.85⋅10^{-4}{m\over s}</math>		Then, plug in the rest: <math>1=\|-1.6⋅10^{-19}\|⋅5⋅10^{28}⋅7.85⋅10^{-7}⋅v, v = 9.9477.85⋅10^{-4}{m\over s}</math>

			===Difficult===

	==References==		==References==

Weerapitch3: /* References */

2016-04-18T07:31:26Z

References

← Older revision		Revision as of 03:31, 18 April 2016
Line 82:		Line 82:

	http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecur.html		http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecur.html

			Chabay, Ruth W., and Bruce A. Sherwood. Matter & Interactions. N.p.: n.p., n.d. Print.

			http://physics.info/electric-current/

	[[Category:Which Category did you place this in?]]		[[Category:Which Category did you place this in?]]

Weerapitch3: /* The Main Idea */

2016-04-18T07:28:36Z

The Main Idea

← Older revision		Revision as of 03:28, 18 April 2016
Line 16:		Line 16:
	Before scientists knew that electrons were the common charge carriers, they discovered current. And, with only two choices, they chose to treat the moving charges as positive. This may seem annoying at first, but there are some benefits to using the conventional current over the directly opposed "electron current". For one, it flows from the positive end of a battery toward the negative end, and from high to low energy.		Before scientists knew that electrons were the common charge carriers, they discovered current. And, with only two choices, they chose to treat the moving charges as positive. This may seem annoying at first, but there are some benefits to using the conventional current over the directly opposed "electron current". For one, it flows from the positive end of a battery toward the negative end, and from high to low energy.

			===A Mathematical Model===

			Assume uniform density <math>n</math> of mobile electrons (uniform current) moving with an average drift speed <math>v</math> in a wire of cross-sectional area <math>A</math>

			<math>i=density⋅area⋅length/time = nAv</math>

			with units of

	===~~A Mathematical Model~~===		<math>electrons/m^3⋅m^2⋅m/s=electrons/s</math>

			Since I (conventional current) is defined as the amount of charge moving pass a point per second, we multiply the number of electrons by positive charge <math>\|q\|</math> to get:

			<math>I = \|q\|⋅i = \|q\|nAv </math>

			with units of

			<math>coulombs⋅electrons/s = C/s = Ampere</math>

			Another way to express the formulas for current is in terms of the electron mobility <math>μ</math>of the wire and the electric field <math>E</math> inside the metal. This can be done by recognizing that the average drift speed <math>v = μE</math>. Substituting this into the equations for electron and conventional current, we get:

	~~Assume uniform density~~ <math>n</math> ~~of electrons~~ (~~uniform~~ current) ~~in a wire of area <math>A</math>~~		<math>i = nAv = nAμE</math> (electron current)

	~~<math>i=density⋅volume/seconds =nA v</math>~~		and

			<math>I = \|q\|nAv = \|q\|nAμE</math> (conventional current)

	If a charge distribution of density <math>ρ</math> moves with the velocity <math>v</math>, the charge per unit time through <math>ΔA</math> is <math>ρv⋅nΔA</math>		If a charge distribution of density <math>ρ</math> moves with the velocity <math>v</math>, the charge per unit time through <math>ΔA</math> is <math>ρv⋅nΔA</math>
Line 32:		Line 50:
	The current I through the surface is <math>I=∫Nqv⋅dA</math>		The current I through the surface is <math>I=∫Nqv⋅dA</math>

	<math>I = ~~\|q\|~~nAv</math>		In summary,

			<math>i = nAv = nAμE</math> (electron current)

	<math>i = nAv</math>		<math>I = \|q\|nAv = \|q\|nAμE</math> (conventional current)

	==Examples==		==Examples==