What's the total resistance of this network?

If you are lucky enough to have A=E and C=D then B=0 ohms, in which case it's super easy ;-)

Dave :)

Reply to
David L. Jones
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[snip]

More general but harder to see: e*d=a*c

--
Johannes
You can have it:
Quick, Accurate, Inexpensive.
Pick two.
Reply to
John F

Here's a closely related problem:

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Reply to
kell

Equality is not a necessary condition for a balanced bridge.

Reply to
Robert Baer

And how some moron can conclude that B must be zero is beyond me. The David L. Jones sounds a lot like that fag Grise in many ways with the trivial non-informational content of his posts and the inclusion of the smiley, which is quite sickening when you think of the marijuana stained brown teeth....

Reply to
Fred Bloggs

You are all class "Fred"

In case you can't figure it out on your own and you really want to know... a common (and often taught) way to solve these resistor problems is by finding equipotential nodes, and once found, the technique is to short those nodes together, thus reducing the complexity of the circuit and making it easier to analyse.

So in this case, if it happened that A=E and C=D, then the two inner nodes are obviously equipotential so you could short them out, thus giving B=0 using this technique.

I was not implying that B *must* be 0, in fact, if A=E and C=D then B can be infinite as well.

Dave :)

Reply to
David L. Jones

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