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Re: [Phys-L] circuit analysis : a simpler approach?



"There are cases where series/parallel reduction doesn't work,
such as for mesh structures, but those are relatively rare
in the real world, and inappropriate to the introductory
course. Typically for such structures, Kirchhoff's «laws»
don't work either."

What about the Wheatstone Bridge? (And I'm taking about the general
situation where you have a non-zero current through the bridge.)
The Wheatstone bridge is in introductory texts.

I introduce this circuit to show students circuits do not have to have
parallel nor series connections.


On Thu, Aug 27, 2015 at 12:08 PM, John Denker <jsd@av8n.com> wrote:

On 08/27/2015 08:38 AM, Diego Saravia wrote:

I do not understand, how can yo solve a paralel circuit whithout knowing
that the sum of currents in a node is cero?

For linear circuits ... which are pretty much the only kind
considered in the introductory course ... you write

1
R_eff = ------------------------
1/R_1 + 1/R_2 + 1/R_3

For linear AC circuits, that generalizes nicely to complex
impedances. Replace R with Z.

You don't need to think about the current until afterward,
if at all. This is how any electrical engineer on earth
would do it. It's intuitive and super-easy.

In contrast, in a nontrivial circuit, there are a bazillion
ways of dividing the circuit into loops. The resulting
equations produce neither convenience nor insight.

There are cases where series/parallel reduction doesn't work,
such as for mesh structures, but those are relatively rare
in the real world, and inappropriate to the introductory
course. Typically for such structures, Kirchhoff's «laws»
don't work either.

The whole idea that there are "nodes" is artificial to
begin with. Note the contrast:

++ Conservation of charge is a law of physics. It is
a corollary of the Maxwell equations.
Homework: Prove the corollary.

-- Zero net current into a node is /not/ a law of physics.
It is tantamount to a claim that the self-capacitance
of the node is negligible. It might be true, or it
might not.

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