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Use’s Excel Array Formulas to Solve Simultaneous Equations. Simultaneous Linear Equation. Present Example. The example we will consider here is from resistor circuits and goes by the name Kirchhoff’s circuit laws. The variables are currents in different “loops” of the circuit. - PowerPoint PPT Presentation
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PHY 202 (Blum) 1
Use’s Excel Array Formulas to Solve
Simultaneous Equations
Simultaneous Linear Equation
PHY 202 (Blum) 2
Present Example
The example we will consider here is from resistor circuits and goes by the name Kirchhoff’s circuit laws. The variables are currents in different
“loops” of the circuit. The coefficients (numbers multiplying
the variables) are resistances. The constants are voltages.
PHY 202 (Blum) 3
Example with numbers
5 = 3.2 JA – 1.0 JB - 2.2 JC
0 = -1.0 JA + 5.5 JB – 3.0 JC
0 = -2.2 JA – 3.0 JB + 6.9 JC
In the above we have three current variables: JA, JB and JC.
PHY 202 (Blum) 4
5 PHY 202 (Blum)
Loop equations as matrix equation
5 = 3.2 JA – 1 JB - 2.2 JC 0 = -1 JA + 5.5 JB – 3 JC 0 = -2.2JA – 3 JB + 6.9JC
0
0
5
9.632.2
35.51
2.212.3
C
B
A
J
J
J
6 PHY 202 (Blum)
Enter matrix in Excel, highlight a region the same size as the matrix.
7 PHY 202 (Blum)
In the formula bar, enter =MINVERSE(range) where range is the set of cells corresponding to the matrix (e.g. B1:D3). Then hit Crtl+Shift+Enter
8 PHY 202 (Blum)
Result of matrix inversion
9 PHY 202 (Blum)
Prepare the “voltage vector”, then highlight a range the same size as the vector and enter =MMULT(range1,range2) where range1 is the inverse matrix and range2 is the voltage vector. Then Ctrl-Shift-Enter.
Voltage vector
10 PHY 202 (Blum)
Results of Matrix Multiplication
The answer
The current vector is the answer JA=3.152 mA
JB=1.470 mA
JC=1.644 mA
PHY 202 (Blum) 11
Use the matrix approach in Excel to find the solution for 10 = 7 JA – 3 JB - 4 JC
0 = -3 JA + 6 JB – 1 JC
0 = -4 JA – 1 JB + 10JC
PHY 202 (Blum) 12
Atomic Isotope Problem
Atomic MassesIsotopes:The element indium exists naturally as two isotopes. 113In has a mass of 112.9043 amu, and 115In has a mass of 114.9041 amu. The average atomic mass of indium is 114.82 amu. Calculate the percent relative abundance of the two isotopes of indium.
http://eppe.tripod.com/stoictry.htmPHY 202 (Blum) 13
Use the matrix approach in Excel to find the solution for 112.9043 x + 114.9041 y =
114.82x + y = 1where x is the fraction of the first isotope and x is the fraction of the second isotope.
PHY 202 (Blum) 14
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