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So keep in mind that I’m a beginner, this is a secondary exam in my university and i just started with exercises so be kind with me :)

Here is the circuit

Here is the “obscure” step

I can’t figure out what happened in that step. It seems like they used some parallel resistor transformation but really don’t know how.

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  • \$\begingroup\$ It is math: they make a common denominator (the denominator is the number under the 'division bar') \$\endgroup\$
    – Oldfart
    Jul 1, 2018 at 16:06
  • \$\begingroup\$ Yeah now i understood, but the inverse of the Rvet is very similar to the formula in the parallel resistor transformation so i messed up. Thanks for saying what a denominator is, Oldfart. Thanks. \$\endgroup\$
    – neilpare2
    Jul 1, 2018 at 16:10

3 Answers 3

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Make use of the Norton theorem!

BTW: The question seems to have an fault, since J_2 is given, but not shown in the circuit diagram.

schematic

simulate this circuit – Schematic created using CircuitLab

It's much easier!

Regarding your step

The matrix is

$$ \mathbf{Y} \cdot \mathbf{V} = \mathbf{J}\, .$$

To get V, you have to invert the Y matrix:

$$ \mathbf{Y}^{-1} \mathbf{Y} \cdot \mathbf{V} = \mathbf{Y}^{-1} \mathbf{J} \\ \mathbf{V} = \mathbf{Y}^{-1} \mathbf{J} \, . $$

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Looks like it inverses the matrix here. You can try to inverse the matrix based on the equation

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You have Rmat.Vvec = Jvec.

What they've done is pre-multiply both sides by the inverse of Rmat, to get Rmat_inv.Rmat.Vvec = Rmat_inv.Jvec, which simplifies (as a matrix times its inverse is unity) to Vvec = Rmat_inv.Jvec.

The exciting things happening to the R terms are the result of taking the matrix inverse.

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