Use of linear algebra in electrical circuit engineering

Use of linear algebra in electrical circuit engineering

To begin with we shall present Kirchhoff 's Laws

Kirchhoff 's Current Law (First Law

The algebraic sum of all currents at any branch point is zero

Kirchhoff 's Voltage Law (Second Law

The algebraic sum of all voltage changes around a loop is zero

Electrical engineers use Kirchoff`s current law and Kirchoff`s voltage law to write a system of linear equations . This procedure is going to be described below 1 ) It is necessary to draw the circuit and assign labels to the

unknown quantities , including currents in each branch . Also it is necessary to assign directions to currents . The answer resulting from the analysis will come out positive if the direction of a particular unknown current is guessed correctly and come out negative otherwise . In both cases the magnitude of the current will be right

2 ) Then the Kirchhoff 's Current Law is applied to as many junctions in the circuit as possible to obtain all independent relations

3 ) As the next step the Kirchhoff 's Voltage Law is applied to as many loops in the circuit as necessary in to solve for the unknowns . If one has n unknowns in a circuit one will need n independent equations In general there will be more loops present in a circuit than one needs to solve for all the unknowns

4 ) And finally it is necessary to solve the resulting set of simultaneous equations for the unknown quantities . The simplest solution for this set of equations is executed within the linear algebra methods .Problem 2

Finding Unknowns

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To write down Kirchhoff 's Laws we have to choose the directions of summation of circuits and the directions for currents

We choose following directions for currents

I1 (segment g-a-b ) direction bottom-up

I2 (segment b-c-d-e ) direction from left to right

I3 (segment f-e ) direction from left to right

I4 (segment g-f ) direction bottom-up

Now we choose the directions of summation of circuits

circuit a-b-f-g-a in a clockwise direction

circuit g-f-e-h-g against a clockwise direction

circuit b-c-d-e-f in a clockwise direction

Now we write down the equations for currents retrieved on the base of the Kirchhoff 's Current Law (First Law

branch b : I1 - 19 - I2 0

branch f : 19 - I3 I4 0

branch e : I2 I3 50 0

Now we write down the equations retrieved on the base of the Kirchhoff 's Voltage Law (Second Law

circuit a-b-f-g-a : E1 10I3 -5I1 0

circuit g-f-e-h-g : E2 8I3 10I4 0

circuit b-c-d-e-f : -2I2 8I3 0

Now we select equations which contain only currents and write down the equation system

I1 - 19 - I2 0

19 - I3 I4 0

I2 I3 50 0

-2I2 8I3 0

Let 's rewrite them into more convenient form

I1 - I2 19

I2 I3 -50

-2I2 8I3 0

- I3 I4 - 19

These equations in a matrix format are the following

Now we convert A to an upper-triangular

Thus...