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