
#1
Jan1912, 10:54 PM

P: 12

1. The problem statement, all variables and given/known data
http://i134.photobucket.com/albums/q...tornetwork.png 2. Relevant equations parallel resistors share the same nodes or terminals and simplify to (a*b)/(a+b). in series resistors values are added. 3. The attempt at a solution http://i134.photobucket.com/albums/q...workwork1.png Not sure what I'm doing wrong. I tried doing it other ways but my final answer doesn't come out right. 



#2
Jan2012, 03:17 AM

HW Helper
P: 1,932

I would think you would do well to first of all eliminate all the resistanceless branches and fuse the nodes connected by them.
I think you then may find there is a part of the circuit with no voltages and no currents. The fact I think so is far from a guarantee so I am within the rules saying so. Anyway look to simplify the circuit in this way. 



#3
Jan2012, 08:00 PM

HW Helper
P: 2,875

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 205 = 4 and 1010 = 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) = 2010 = 20/3 R(gc) = 510 = 10/3 So R(bg) = 15(20/3 + 10/3) = 1510 = 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. 



#4
Jan2112, 05:00 PM

P: 12

Equivalent resistors in resistor networks
Thanks so much for your explanation. I'm pretty new to this stuff and kind of having a hard time. I'll definitely use this on the rest of my homework. Thanks again.



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