# Basic equivalent resistance problem

Discussion in 'Homework Help' started by fdsa, Jun 28, 2012.

1. ### fdsa Thread Starter New Member

Aug 16, 2011
9
0
Let:
$R1 = 18\Omega\\R_2 = 12\Omega\\R_3 = 9 \Omega\\R_4 = 6\Omega\\$

If I assume $R_1 & R_2$ are in series and $R_{12}$ in parallel with $R_{34}$ the equivalent resistance is correct but I have no solid reasoning behind this conclusion. As I see it the current in the top node connecting $R_1 & R_2$ has two paths to choose from?

2. ### daviddeakin Active Member

Aug 6, 2009
207
27
Perhaps it will be more clear if you redraw the circuit like this:

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3. ### fdsa Thread Starter New Member

Aug 16, 2011
9
0
Thank you, probably need to get more exercise reducing those slightly more complicated resistance problems but that made it completely clear.

4. ### WBahn Moderator

Mar 31, 2012
17,461
4,701
Never be afraid to resketch a circuit in a configuration that makes more sense to you. The human brain is very good at pattern recognition, but that works against us when we see a new pattern that is really just a redrawing of a different pattern -- our brain insists that it is a new pattern with no relationship to previous patterns and we tend to accept it. But as we redraw a circuit, that same tendency biases us toward using patterns with which we are already familiar, enabling us to see them.

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