Network Analysis for Electric Circuits

Determine the currents I₁,I₂, and I₃ for the electrical network.

There are two junctions shown, but both of them are equal, so there is no need to name them as J1 and J2.

The answer is I₂ = I₁+I₃, when we use Network Analysis.

After that, we will use a matrix to solve for the Amperes (I). We will combine the resistances (R_n) shown on the Amperes (I_n) and make them equal to the amount of their voltage (v) for determining the entries for the matrix. The answers are:

I₁ – I₂ + I₃ = 0 Volts

4I₁ + 3I₂ = 3 Volts 

1I₃ + 3I₂ = 4 Volts

Then we use them as entries for the matrix.

[   1  -1   1   0   ]

[   4   3   0   3   ]     -4(R₁) + R₂ ---> R₂

[   0   3   1   4   ]

[   1  -1   1   0   ]

[   0   7  -4   3   ]     -2(R₃) + R₂ ---> R₂     

[   0   3   1   4   ]

[   1  -1   1   0   ]

[   0   1  -6  -5   ]     -3(R₂) + R₃ ---> R₃

[   0   3   1   4   ]

[   1  -1   1   0   ]

[   0   1  -6  -5   ]      1/19 (R₃)

[   0   0   19 19  ]

[   1  -1   1   0   ]

[   0   1  -6  -5   ]     R₁ + R₂ ---> R₁

[   0   0   1   1    ] 

[   1   0  -5  -5   ]

[   0   1  -6  -5   ]     5(R₃) + R₁ ---> R₁

[   0   0   1   1    ]

[   1   0   0   0   ]

[   0   1  -6  -5  ]     6(R₃) + R₂ ---> R₂

[   0   0   1   1   ]

[   1   0   0   0   ]

[   0   1   0   1   ]     

[   0   0   1   1   ]

Now we know that:

I₁ = 0 Amperes

I₂ = 1 Ampere

and

I₃ = 1 Ampere

Network Analysis

Below is an example of a network analysis.

In this example, we assume that the circles in the upper part of the picture are 1 and 2 from left to right, while the ones below are 3 and 4 from left to right. The answers are as follows:

J1 300 = x₁+x₂

J2 x₁+x₃ = 150+x₄

J3 200+x₂ = x₃+x₅

J4 x₅+x₄ = 350

J is known as the junction. The ones within the junctions are the numbers on the arrows going toward the junction, while the ones not within the junction in the equation are the numbers on the arrows going away from the junction. For example, in J1, 300 is the number on the arrow going toward the junction, while x₁ and x₂ are the numbers on the arrows going away from the said junction. The equation can also be J1 x₁+x₂ = 300, since it is also produces the same result.

Translating Large x-Values Before Curve Fitting

Find a polynomial that fits the points

                       (2005, 1)    (2006, 2)   (2007,3)   (2009,4)   (2010,10)

Because the given x-values are large, use the translation z = x - 2007 to obtain
   
                                 (-2, 1)   (-1, 2)   (0, 3)   (2, 4)   (3, 10)

This makes the polynomial functions as

p(-2) = a₀ + a₁(-2) + a₂(-2)² + a₃(-2)³ + a₄(-2)⁴ = a₀ - 2a₁ + 4a₂ – 8a₃ + 16a₄ = 1

p(-1) = a₀ + a₁(-1) + a₂(-1)² + a₃(-1)³ + a₄(-1)⁴ = a₀ - a₁ + a₂ –  a₃ + a₄ = 2

p(0) = a₀ + a₁(0) + a₂(0)² + a₃(0)³ + a₄(0)⁴ = a₀ = 3
p(2) = a₀ + a₁(2) + a₂(2)² + a₃(2)³ + a₄(2)⁴ = a₀ + 2a₁ + 4a₂ + 8a₃ + 16a₄ = 4
p(3) = a₀ + a₁(3) + a₂(3)² + a₃(3)³ + a₄(3)⁴ = a₀ + 3a₁ + 9a₂ + 27a₃ + 81a₄ = 10

Then use the Gauss-Jordan Elimination. At the end, the solution is a₀=3, a₁=29/60, a₂=-67/120, a₃=1/15. and a₄=13/120. This means that p(x) = 3 + 29/60x - 67/120x² + 1/15x³ + 13/120x⁴. But because we used z to translate the x-value, the p(x) is now written as 3 + 29/60(x-2007) - 67/120(x-2007)² + 1/15(x-2007)³ + 13/120(x-2007)⁴

  
                                                                              
                                                           

1	-2	4	-8	16	1
1	-1	1	-1	1	2
1	0	0	0	0	3
1	2	4	8	16	4
1	3	9	27	81	10

    
                       

-1(R1) + R2 ---> R2

1	-2	4	-8	16	1
0	1	-3	7	-15	1
1	0	0	0	0	3
1	2	4	8	16	4
1	3	9	27	81	10

-1(R1) + R3 ---> R3

1	-2	4	-8	16	1
0	1	-3	7	-15	1
0	2	-4	8	-16	2
1	2	4	8	16	4
1	3	9	27	81	10

-1(R1) + R4 ---> R4

1	-2	4	-8	16	1
0	1	-3	7	-15	1
0	2	-4	8	-16	2
0	4	0	16	0	3
1	3	9	27	81	10

-1(R1) + R5 ---> R5

1	-2	4	-8	16	1
0	1	-3	7	-15	1
0	2	-4	8	-16	2
0	4	0	16	0	3
0	5	5	35	65	9

-2(R2) + R3 ---> R3

1	-2	4	-8	16	1
0	1	-3	7	-15	1
0	0	2	-6	14	0
0	4	0	16	0	3
0	5	5	35	65	9

-4(R2) + R4 ---> R4

1	-2	4	-8	16	1
0	1	-3	7	-15	1
0	0	2	-6	14	0
0	0	12	-12	60	-1
0	5	5	35	65	9

-5(R2) + R5 ---> R5

1	-2	4	-8	16	1
0	1	-3	7	-15	1
0	0	2	-6	14	0
0	0	12	-12	60	-1
0	0	20	0	140	4

1/2(R3) ---> R3

1	-2	4	-8	16	1
0	1	-3	7	-15	1
0	0	1	-3	7	0
0	0	12	-12	60	-1
0	0	20	0	140	4

-12(R3) + R4 ---> R4

1	-2	4	-8	16	1
0	1	-3	7	-15	1
0	0	1	-3	7	0
0	0	0	24	-24	-1
0	0	20	0	140	4

-20(R3) + R5 ---> R5

1	-2	4	-8	16	1
0	1	-3	7	-15	1
0	0	1	-3	7	0
0	0	0	24	-24	-1
0	0	0	60	0	4

1/24(R4)

1	-2	4	-8	16	1
0	1	-3	7	-15	1
0	0	1	-3	7	0
0	0	0	1	-1	-1/24
0	0	0	60	0	4

-60(R4) + R5 ---> R5

1	-2	4	-8	16	1
0	1	-3	7	-15	1
0	0	1	-3	7	0
0	0	0	1	-1	-1/24
0	0	0	0	60	13/2

1/60(R5) ---> R5

1	-2	4	-8	16	1
0	1	-3	7	-15	1
0	0	1	-3	7	0
0	0	0	1	-1	-1/24
0	0	0	0	1	13/120

1(R5) + R4 ---> R4

1	-2	4	-8	16	1
0	1	-3	7	-15	1
0	0	1	-3	7	0
0	0	0	1	0	1/15
0	0	0	0	1	13/120

-7(R5) + R3 ---> R3

1	-2	4	-8	16	1
0	1	-3	7	-15	1
0	0	1	-3	0	-91/120
0	0	0	1	0	1/15
0	0	0	0	1	13/120

15(R5) + R2 ---> R2

1	-2	4	-8	16	1
0	1	-3	7	0	21/8
0	0	1	-3	0	-91/120
0	0	0	1	0	1/15
0	0	0	0	1	13/120

-16(R5) + R1 ---> R1

1	-2	4	-8	0	-11/15
0	1	-3	7	0	21/8
0	0	1	-3	0	-91/120
0	0	0	1	0	1/15
0	0	0	0	1	13/120

3(R4) + R3 ---> R3

1	-2	4	-8	0	-11/15
0	1	-3	7	0	21/8
0	0	1	0	0	-67/120
0	0	0	1	0	1/15
0	0	0	0	1	13/120

-7(R4) + R2 ---> R2

1	-2	4	-8	0	-11/15
0	1	-3	0	0	259/120
0	0	1	0	0	-67/120
0	0	0	1	0	1/15
0	0	0	0	1	13/120

8(R4) + R1 ---> R1

1	-2	4	0	0	-1/5
0	1	-3	0	0	259/120
0	0	1	0	0	-67/120
0	0	0	1	0	1/15
0	0	0	0	1	13/120

3(R3) + R2 ---> R2

1	-2	4	0	0	-1/5
0	1	0	0	0	29/60
0	0	1	0	0	-67/120
0	0	0	1	0	1/15
0	0	0	0	1	13/120

-4(R3) + R1 ---> R1

1	-2	0	0	0	61/30
0	1	0	0	0	29/60
0	0	1	0	0	-67/120
0	0	0	1	0	1/15
0	0	0	0	1	13/120

-2(R2) + R1 ---> R1

1	0	0	0	0	3
0	1	0	0	0	29/60
0	0	1	0	0	-67/120
0	0	0	1	0	1/15
0	0	0	0	1	13/120