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Another circuit diagram simplifying question (harder)

  1. Mar 25, 2005 #1
    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).
  2. jcsd
  3. Mar 25, 2005 #2
    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

  4. Mar 25, 2005 #3
    i have thought of that : using kirchoff's rules to find the net resistance. but i have no idea how the rules relates to resistance.
  5. Mar 25, 2005 #4
    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

  6. Mar 25, 2005 #5
    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....
  7. Mar 25, 2005 #6
    can you please show me how to do it?
  8. Mar 25, 2005 #7
    i think the diagram can be simplify into this.....but can you please show me what to do next?
  9. Mar 25, 2005 #8
    if you simplify the left side to a 120 ohm resistor, then im pretty sure you can consider the 120 ohm and the 60 ohm on the right as a series resistor
  10. Mar 25, 2005 #9
    Im probably wrong though. I never was good with kirschoff's rules.
  11. Mar 25, 2005 #10
    hm....well can you please tell what you might do FIRST in order to find the net resistance?
  12. Mar 25, 2005 #11
  13. Mar 25, 2005 #12
    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.....
  14. Mar 25, 2005 #13


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    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..
  15. Mar 26, 2005 #14
    oh...i get it. thanks a lot Ouabache for your last implication. thanks very much everyone.
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