Because of Symmetry, I flowing into the cubical resistor must divide evenly between 3 routes AE, AC,
and AB. So the flow is as follows on the diagram. Then the flows must divide evenly at junctions E, C, and
B. The current is then recombined on junctions D, F, and G. Now using path A, C, G, H we see:
𝑉
=
𝐼𝑅
3
+
𝐼𝑅
6
+
𝐼𝑅
3
𝑉
=
5
∙ 𝐼𝑅
6
Using this information of the applied voltage we use Ohm
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 Fall '06
 Culbertson

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