Lildon said:
Homework Statement
http://i134.photobucket.com/albums/q100/megajette2/Resistornetwork.png
Homework Equations
parallel resistors share the same nodes or terminals and simplify to (a*b)/(a+b).
in series resistors values are added.
The Attempt at a Solution
http://i134.photobucket.com/albums/q100/megajette2/Resistornetworkwork-1.png
Not sure what I'm doing wrong. I tried doing it other ways but my final answer doesn't come out right.
epenguin has given you a good hint.
I need some way of referring to the nodes so we have a common understanding. The network has 4 nodes above, and 4 nodes below. Starting with the upper row going from left (the open circuit at one end of Req) to right, label them a,b,c,d. With the lower row, going from right to left (the open circuit at the other end of Req), label them e,f,g,h. We have now described the complete network such that the resistance between a and h is Req.
Refer to the resistance between two nodes as R(ab) for example (here R(ab) = 5). When a node is considered electrically equivalent to another node, write d=e (for example), then reduce all relations to the earlier occurring node ('d' in this case). Write parallel as || and series as + (since series resistances are just summed up). All this is useful notation to help you conceptualise and solve future problems.
Electrical equivalence occurs when there's a short (no resistance) between two points. While you've clearly seen that R(de) = 0 and hence d = e, you've missed c = f. Once you see this, you can actually disregard the circuit downstream of nodes c and f (meaning there's no need to have calculated 20||5 = 4 and 10||10 = 5 as you did).
OK, so we have c=f. Everything connected to node f becomes connected to node c. So now we have 2 sets of parallel resistances connected across the 15Ω, i.e.
R(bc) = 20||10 = 20/3
R(gc) = 5||10 = 10/3
So R(bg) = 15||(20/3 + 10/3) = 15||10 = 6
and Req = R(ah) = R(ab) + R(bg) + R(gh) = 5 + 6 + 10 = 21Ω.
Try to do it in this systematic way in the future and you won't go wrong.
