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Calculating current from diagram.

  1. Jul 19, 2011 #1
    1. The problem statement, all variables and given/known data

    from the diagram i drew I want to calculate the current in the battery for the circuit, can someone help me with this. Want to use symbols too. R1 and R2 are in series, they are in parallel with R3, and R4 is in series with R1, R2, and R3. The diagram is attached.

    2. Relevant equations



    3. The attempt at a solution

    Im not sure where to start with this problem, can ya get me jump started. The problem is that I realize that the diagram I have drawn has 3 resistors that are in parallel, then one that they will always be in series with. So i'm confused as to how to calculate the current when I have both parallel and series connections? I can't just use 1/r1 + 1/r2 + 1/r3/ ... etc, or series as well just series = r1+r2+r3.. etc. Please give me some help and stick with me with this today.
     

    Attached Files:

  2. jcsd
  3. Jul 19, 2011 #2

    gneill

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    Staff: Mentor

    You want to work step by step, replacing sets of resistors that you can combine (series or parallel) with single resistors, until you are left with just a single resistor representing the whole resistor network.

    To begin with, you've already identified R1 and R2 as being in series, so what single resistance value can replace them?
     
  4. Jul 20, 2011 #3
    Okay could we call R1 and R2 10 Ohms?
     
  5. Jul 20, 2011 #4

    gneill

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    Staff: Mentor

    Yes. So replace R1 and R2 with a single resistor of 10 Ohms. What's next? Do you see another series or parallel opportunity to pursue?
     
  6. Jul 21, 2011 #5
    Okay well 10 Ohms is in parellel with R3 could we add those together?
     
  7. Jul 21, 2011 #6
    yes we can ... and it'll give ___ .
     
  8. Jul 21, 2011 #7
    25 Ohms, so now I have 25 Ohms of parallel circuitry with an 8 Ohm series.

    Whats next.
     
  9. Jul 21, 2011 #8
    how will it give 25Ω?

    what formula do we use for parallel resistor combination?
     
  10. Jul 21, 2011 #9
    Oh I see for the parallel connects, we must use this: (1/10 + 1/15) = 1/66_7 = 5.9999 Ohms? Now what about the last series, just add that together?
     
  11. Jul 21, 2011 #10

    gneill

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    Staff: Mentor

    Call it 6 Ohms: (1/10 + 1/15)-1 = 10*15/(10 + 15) = 6

    If resistors are in series you add them...
     
  12. Jul 21, 2011 #11
    Oh okay, now its just two resistors in series, so 6 Ohms + 8 Ohms

    So the total Ohms of the circuit: 14 Ohms
     
  13. Jul 21, 2011 #12

    gneill

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    Staff: Mentor

    Right. So there's a battery (of some unspecified voltage V) connected across 14 Ohms. What's the current?
     
  14. Jul 22, 2011 #13
    V=i*r

    So current of this circuit would be (Battery Voltage/ 14 Ohms)= current (Amps)
     
  15. Jul 22, 2011 #14
    Yes that's correct.
     
  16. Jul 22, 2011 #15
    Thankyou!
     
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