Resistor Square Problem

blah2222

Joined May 3, 2010
582
No because between points A and B, there is a 3k3 resistor in parallel with 4k7 + 1k5 + 2k7, which has to be accounted for.

Rab = 3k3 || (4k7 + 1k5 + 2k7) = 2k41
 

Thread Starter

yan500

Joined Jul 12, 2011
48
Sorry, I should have been clearer. What I mean is that point A essentially turns into a node that would split a current (if there was a current source) between AB and AC. And that would make it a parallel connection.
 

blah2222

Joined May 3, 2010
582
Sorry, I should have been clearer. What I mean is that point A essentially turns into a node that would split a current (if there was a current source) between AB and AC. And that would make it a parallel connection.
Yes. If a current source were to be connected between nodes A and B. A current divider with two paths, A->B and A->C->D->B, would result.
 
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