If you are lucky enough to have A=E and C=D then B=0 ohms, in which case it's super easy ;-)
Dave :)
If you are lucky enough to have A=E and C=D then B=0 ohms, in which case it's super easy ;-)
Dave :)
[snip]
More general but harder to see: e*d=a*c
-- Johannes You can have it: Quick, Accurate, Inexpensive. Pick two.
Here's a closely related problem:
Equality is not a necessary condition for a balanced bridge.
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....
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 :)
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.