10-11-2016, 04:32 AM

eNet(M1)

M1 has 6 columns

M1=[[busfrom] [busto] [Resistor] [Independent Voltage] [Independent Current] [Opamp] ]

M1 for linedata: column1=busFrom, column2=busTo, column3=Z, column4=Voltage, column5=Curr.Source, column6=Oamp

BusFrom To BusTo will follow current path

Current flows through the -ve terminal of the battery

"\left(\begin{array}{cccccc}\n1 & 2 & 10 & 0 & 0 & 0 \\\n1 & 1 & 0 & 0 & 5 & 0 \\\n2 & 3 & 40 & 0 & 0 & 0 \\\n1 & 3 & 20 & 0 & 0 & 0 \\\n2 & 2 & 50 & 0 & 0 & 0 \\\n3 & 3 & 0 & 0 & 2 & 0\n\end{array}\right) "

node1 to node2 has 10_Ohms (or node2 to node1 has 10_Ohms )

node1 to node1 has 5_Amps (to node1 to ground, 5 because current flow into Node1)

node2 to node3 has 40_ohms

node1 to node3 has 20_Ohms

node2 to node2 has 50_Ohms (node2 to ground)

node3 to node3 has 2amps (to node3 to ground, 2 because current flow into Node3)

eNet(M1)

results:

"\left(\begin{array}{c}\n404.285714286 \\\n350 \\\n412.857142857\n\end{array}\right) "

interpretative of the the results: type: nkll

CAS programming you can enter variables for elements:

node1 to node2 has r1 (or node2 to node1 has 10_Ohms )

node1 to node1 has Ig1 (to node1 to ground, 5 because current flow into Node1)

node2 to node3 has r2

node1 to node3 has r3

node2 to node2 has r4 (node2 to ground)

node3 to node3 has Ig3 (to node3 to ground, 2 because current flow into Node3)

For Capacitor impedance: 1/s*c1 into resistor column3; (c1, c2.... not c) qualify the variable c

For Inductor impedance: s*l1 into resistor column3

Opamp: [1 2 0 0 0 4] (positive node1 to 1, -ve node2 to 2) 0 in column5 (for ideal Opamp) and node4 for the feedback path;

"\left(\begin{array}{cccccc}

1 & 1 & 0 & 0 & 4 & 0 \\

1 & 1 & 500 & 0 & 0 & 0 \\

1 & 2 & 450 & 0 & 0 & 0 \\

2 & 2 & 1500 & 0 & 0 & 0 \\

2 & 3 & 600 & 0 & 0 & 0 \\

3 & 3 & 550 & 0 & 0 & 0 \\

3 & 3 & 0 & 0 & 5 & 0

\end{array}\right) "

OR

"\left(\begin{array}{cccccc}

1 & 1 & 500 & 0 & 4 & 0 \\

1 & 2 & 450 & 0 & 0 & 0 \\

2 & 2 & 1500 & 0 & 0 & 0 \\

2 & 3 & 600 & 0 & 0 & 0 \\

3 & 3 & 550 & 0 & 0 & 0 \\

3 & 3 & 0 & 0 & 5 & 0

\end{array}\right) "

"\left(\begin{array}{cccccc}

1 & 2 & 4 & 0 & 0 & 0 \\

2 & 3 & 0 & 10 & 0 & 0 \\

1 & 3 & 8 & 0 & 0 & 0 \\

1 & 1 & 0 & 20 & 0 & 0 \\

3 & 3 & 12 & 0 & 0 & 0 \\

2 & 2 & 16 & 0 & 0 & 0

\end{array}\right) "

row2: can be written as

[3 2 0 -10 0 0 ]

"\left(\begin{array}{cccccc}

1 & 1 & 0 & 1 & 0 & 0 \\

1 & 2 & 6000 & 0 & 0 & 0 \\

2 & 2 & 9000 & 0 & 0 & 0 \\

2 & 3 & 18000 & 0 & 0 & 0 \\

3 & 4 & 54000 & 0 & 0 & 0 \\

4 & 4 & 45000 & 0 & 0 & 0 \\

3 & 3 & 0 & 0 & 0 & 4

\end{array}\right) "

last row: Opamp input

M1 has 6 columns

M1=[[busfrom] [busto] [Resistor] [Independent Voltage] [Independent Current] [Opamp] ]

M1 for linedata: column1=busFrom, column2=busTo, column3=Z, column4=Voltage, column5=Curr.Source, column6=Oamp

BusFrom To BusTo will follow current path

Current flows through the -ve terminal of the battery

"\left(\begin{array}{cccccc}\n1 & 2 & 10 & 0 & 0 & 0 \\\n1 & 1 & 0 & 0 & 5 & 0 \\\n2 & 3 & 40 & 0 & 0 & 0 \\\n1 & 3 & 20 & 0 & 0 & 0 \\\n2 & 2 & 50 & 0 & 0 & 0 \\\n3 & 3 & 0 & 0 & 2 & 0\n\end{array}\right) "

node1 to node2 has 10_Ohms (or node2 to node1 has 10_Ohms )

node1 to node1 has 5_Amps (to node1 to ground, 5 because current flow into Node1)

node2 to node3 has 40_ohms

node1 to node3 has 20_Ohms

node2 to node2 has 50_Ohms (node2 to ground)

node3 to node3 has 2amps (to node3 to ground, 2 because current flow into Node3)

eNet(M1)

results:

"\left(\begin{array}{c}\n404.285714286 \\\n350 \\\n412.857142857\n\end{array}\right) "

interpretative of the the results: type: nkll

CAS programming you can enter variables for elements:

node1 to node2 has r1 (or node2 to node1 has 10_Ohms )

node1 to node1 has Ig1 (to node1 to ground, 5 because current flow into Node1)

node2 to node3 has r2

node1 to node3 has r3

node2 to node2 has r4 (node2 to ground)

node3 to node3 has Ig3 (to node3 to ground, 2 because current flow into Node3)

For Capacitor impedance: 1/s*c1 into resistor column3; (c1, c2.... not c) qualify the variable c

For Inductor impedance: s*l1 into resistor column3

Opamp: [1 2 0 0 0 4] (positive node1 to 1, -ve node2 to 2) 0 in column5 (for ideal Opamp) and node4 for the feedback path;

"\left(\begin{array}{cccccc}

1 & 1 & 0 & 0 & 4 & 0 \\

1 & 1 & 500 & 0 & 0 & 0 \\

1 & 2 & 450 & 0 & 0 & 0 \\

2 & 2 & 1500 & 0 & 0 & 0 \\

2 & 3 & 600 & 0 & 0 & 0 \\

3 & 3 & 550 & 0 & 0 & 0 \\

3 & 3 & 0 & 0 & 5 & 0

\end{array}\right) "

OR

"\left(\begin{array}{cccccc}

1 & 1 & 500 & 0 & 4 & 0 \\

1 & 2 & 450 & 0 & 0 & 0 \\

2 & 2 & 1500 & 0 & 0 & 0 \\

2 & 3 & 600 & 0 & 0 & 0 \\

3 & 3 & 550 & 0 & 0 & 0 \\

3 & 3 & 0 & 0 & 5 & 0

\end{array}\right) "

"\left(\begin{array}{cccccc}

1 & 2 & 4 & 0 & 0 & 0 \\

2 & 3 & 0 & 10 & 0 & 0 \\

1 & 3 & 8 & 0 & 0 & 0 \\

1 & 1 & 0 & 20 & 0 & 0 \\

3 & 3 & 12 & 0 & 0 & 0 \\

2 & 2 & 16 & 0 & 0 & 0

\end{array}\right) "

row2: can be written as

[3 2 0 -10 0 0 ]

"\left(\begin{array}{cccccc}

1 & 1 & 0 & 1 & 0 & 0 \\

1 & 2 & 6000 & 0 & 0 & 0 \\

2 & 2 & 9000 & 0 & 0 & 0 \\

2 & 3 & 18000 & 0 & 0 & 0 \\

3 & 4 & 54000 & 0 & 0 & 0 \\

4 & 4 & 45000 & 0 & 0 & 0 \\

3 & 3 & 0 & 0 & 0 & 4

\end{array}\right) "

last row: Opamp input