Another circuit diagram simplifying question (harder)

[F|234K]

here's the problem

i want to find the net resistance. the only problem i think i have is that i don't know how to change the diagram so that the battery isn't in the middle of the circuit (i only know how to do questions where the battery is connected like a single complete path, which this particular problem sort of has 2 paths).

 

[marlon]

It's easy and starightforexard to calculate the substitution resitance for the triangle on the right hand side from the battery right? Once you have done that you will need to apply Kirchhoff's laws expressing the conservation of charge and energy.

The battery can be in the middle, that should not be a problem

regards
marlon

 

[F|234K]

i have thought of that : using kirchhoffs rules to find the net resistance. but i have no idea how the rules relates to resistance.

 

[marlon]

i have thought of that : using kirchhoffs rules to find the net resistance. but i have no idea how the rules relates to resistance.

The conservation of energy requires that the sum of all potentials are zero in each chain. Then use Ohm's law to bring the resistance. it will work like that. Don't forget to apply conservation of charge also...

good luck

marlon

 

[F|234K]

i think i did it the way you told me to, but then i got the wrong answer. my answer is 43.3333, but the right answer is 50...

 

[F|234K]

can you please show me how to do it?

 

[F|234K]

i think the diagram can be simplify into this...but can you please show me what to do next?

 

[whozum]

if you simplify the left side to a 120 ohm resistor, then I am pretty sure you can consider the 120 ohm and the 60 ohm on the right as a series resistor

 

[whozum]

Im probably wrong though. I never was good with kirschoff's rules.

 

[F|234K]

hm...well can you please tell what you might do FIRST in order to find the net resistance?

 

[whozum]

http://www.public.asu.edu/~hyousif/circuitreduction2.JPG

Im sure of all that. What happens after that I don't know.

 

[F|234K]

i got to the point where you got. but then i don't know how to find the net resistance when the battery is in the middle...

 

[Ouabache]

if you simplify the left side to a 120 ohm resistor, then I am pretty sure you can consider the 120 ohm and the 60 ohm on the right as a series resistor

Whosum is partly right in taking the 3 resistors on the left and noticing there are in series. Once you have that combined, notice that the right side (60ohms) is now in parallel with the left side (because they share a common node at both their ends)..

Once you combine the left side in parallel with the right, the rest should become clear..

 

[F|234K]

oh...i get it. thanks a lot Ouabache for your last implication. thanks very much everyone.