TF EIM Chapt1 - Revision history

Foretony: /* Find currents when Resistances are known */

2015-01-20T19:14:56Z

Find currents when Resistances are known

2015-01-20T19:09:52Z

Find currents when Resistances are known

2015-01-20T19:08:22Z

Find currents when Resistances are known

2015-01-20T19:07:25Z

Find currents when Resistances are known

2015-01-20T19:06:44Z

Find currents when Resistances are known

2015-01-20T18:40:58Z

Charge

2015-01-20T18:40:35Z

Charge

2015-01-20T18:39:58Z

Charge

2015-01-20T18:39:00Z

Charge

2015-01-20T17:59:30Z

Electric Potential Energy

@@ Line 574: / Line 574: @@
 ;Matrix Inversion/Gauss-Jordan elimination
-If <math>\tilde{\beta}</math>  and <math>\tilde{a}</math> are the vectors for the problem and <math>\tilde{A}</math> is a matrix such that
+If <math>\vec V</math>  and <math>\vec I</math> are vectors and <math>\tilde{R}</math> is a matrix such that
-: <math>\tilde{\beta} = \tilde{a} \tilde{A}</math>
+: <math>\vec V = \tilde{R} \vec I</math>
-then one can find <math>\tilde{a}</math>  if  <math>\tilde{\beta}</math> and <math>\tilde{A}</math> are known by using the following matrix math
+then one can find <math>\vec I</math>  if  <math>\vec V</math> and <math>\tilde{R}</math> are known by using the following matrix math
 : <math>\vec V = \tilde A \vec I</math>

@@ Line 577: / Line 577: @@
 : <math>\tilde{\beta} = \tilde{a} \tilde{A}</math>
-then one can find <math>\tilde{a}</math>  if  <math>\tilde{\beta}</math> and <math>\tilde{a}</math> are known by using the following matrix math
+then one can find <math>\tilde{a}</math>  if  <math>\tilde{\beta}</math> and <math>\tilde{A}</math> are known by using the following matrix math
 : <math>\left ( \tilde{\beta} = \tilde{a} \tilde{A} \right) \tilde{A}^{-1}</math>

← Older revision		Revision as of 19:08, 20 January 2015
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−		+	: <math>\vec V = \tilde A \vec I</math>
−	:<math>\left( \begin{array}{c} 0 \\ V \\0 \end{array} \right) = \left( \begin{array}{ccc} -1 & 1 & 1\\ R_1 & R_2 & 0 \\0 & -R_2 & R_3 \end{array} \right)\left( \begin{array}{c} I_1 \\ I_2 \\ I_3\end{array} \right) ~~= \tilde{A} \vec{I}~~</math>	+	:<math>\left( \begin{array}{c} 0 \\ V \\0 \end{array} \right) = \left( \begin{array}{ccc} -1 & 1 & 1\\ R_1 & R_2 & 0 \\0 & -R_2 & R_3 \end{array} \right)\left( \begin{array}{c} I_1 \\ I_2 \\ I_3\end{array} \right) </math>

← Older revision		Revision as of 19:07, 20 January 2015
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−	:<math>\tilde{A} \left( \begin{array}{c} 0 \\ V \\0 \end{array} \right) = \left( \begin{array}{c} I_1 \\ I_2 \\ I_3\end{array} \right) = \vec I</math>	+	:<math>\tilde{A}^{-1} \left( \begin{array}{c} 0 \\ V \\0 \end{array} \right) = \left( \begin{array}{c} I_1 \\ I_2 \\ I_3\end{array} \right) = \vec I</math>

@@ Line 585: / Line 585: @@
 so find the inverse of the matrix  <math>\tilde{A}^{-1} </math> and the solution will be
 : <math> \tilde{a}=\tilde{\beta}\tilde{A}^{-1} </math>
 :<math>\tilde{A}=\left( \begin{array}{ccc} -1 & 1 & 1\\ R_1 & R_2 & 0 \\0 & -R_2 & R_3 \end{array} \right)</math>
@@ Line 605: / Line 614: @@
 <math>\left( \begin{array}{c} I_1 \\ I_2 \\ I_3\end{array} \right)= \left( \begin{array}{ccc}  -R_2R_3/(R_1R_2+R_1R_3+R_2R_3)&(R_2+R_3)/(R_1R_2+R_1R_3+R_2R_3) &R_2/(R_1R_2+R_1R_3+R_2R_3) \\ R_1R_3/(R_1R_2+R_1R_3+R_2R_3)&R_3/(R_1R_2+R_1R_3+R_2R_3)&-R_1/(R_1R_2+R_1R_3+R_2R_3)\\R_1R_2/(R_1R_2+R_1R_3+R_2R_3)&R_2/(R_1R_2+R_1R_3+R_2R_3) &R_1+R_2/(R_1R_2+R_1R_3+R_2R_3)  \end{array} \right)\left( \begin{array}{c} 0 \\ V \\0 \end{array} \right) </math>
 :<math>=\left( \begin{array}{c} V(R_2+R_3)/(R_1R_2+R_1R_3+R_2R_3)  \\ VR_3/(R_1R_2+R_1R_3+R_2R_3)  \\VR_2/(R_1R_2+R_1R_3+R_2R_3)  \end{array} \right)</math>
 ==Convention summary==

@@ Line 24: / Line 24: @@
 <math>F = 9.64846 \times 10^{4}</math>Coul/mole
-<math>N_A = 6.02257 \times 10^{23}</math>
+<math>N_A = 6.02257 \times 10^{23}</math> Particles(Atoms)/mole
 Experiment 2: Oil drop experiment

@@ Line 22: / Line 22: @@
 Q = charge required to accumulate mass <math>\Delta m</math>
-<math>F = 9.64846 \times 10^{6}</math>
+<math>F = 9.64846 \times 10^{4}</math>Coul/mole
 <math>N_A = 6.02257 \times 10^{23}</math>

← Older revision		Revision as of 18:39, 20 January 2015
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	:<math>e = \frac{F}{N_A}</math>		:<math>e = \frac{F}{N_A}</math>
		+
		+	The experiment is Electrolysis.
		+
		+	You build up material on a copper cathode that is immersed in an ion solution along with an anode using a battery.
		+
		+	Overtime an ion reaches the cathode a charge passes through the circuit. If we know how much charge passed through the circuit and we measure how much mass accumulated on the cathode we can find Faradays constant.
		+
		+	:<math>F= \frac{Q}{Z} \frac{M}{\Delta m}</math>
		+
		+	M= molar mass (g/mole)
		+	Z = charge of ion used
		+	Q = charge required to accumulate mass <math>\Delta m</math>

	Experiment 2: Oil drop experiment		Experiment 2: Oil drop experiment

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 From the work energy theorem, Work is equal to the change in energy of a system
-;<math>W = \Delta</math> Energy
+;<math>W = \Delta</math> Kinetic  Energy
 The work done moving a change along an electric field line is