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Current and power in restisor combo

  1. Oct 22, 2009 #1
    1. The problem statement, all variables and given/known data
    The battery in the figure below has negligible internal resistance.
    26-50.gif
    (a) Find the current in each resistor.
    3 ohm resistor-
    4 ohm resistor-
    2 ohm vertical resistor-
    2 ohm diagonal resistor-
    (b) Find the power delivered by the battery.

    2. Relevant equations
    Loop Rule, Junction Rule, Resistor Rule. ?
    P=VI=RI2=(V2/R)
    R=V/I
    I=V/R
    Req=(V1+V2/I)


    3. The attempt at a solution
    6/3=2A 6/4= 1.5A, 6/2=3A ect. all wrong
    For I1: -3omhs (since it's along the current) I1-2omhs(along current)I1+6V=0 solving for I1=1 wrong
    Am I leaving stuff out?
     
  2. jcsd
  3. Oct 22, 2009 #2

    cepheid

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    Staff Emeritus
    Science Advisor
    Gold Member

    You tried to use the loop rule for the loop involving the 3 ohm and the vertical 2 ohm resistor only. It didn't work because the loop is not a closed, isolated loop. After the current has passed from the positive end of the battery through the 3 ohm resistor, it has the option of going into three different branches. Therefore, the current in the 3 ohm resistor is not the same as the current in the vertical 2 ohm resistor. That's why the loop rule didn't work.

    But I did just give you a hint about the arrangement of the vertical 2 ohm, diagonal 2 ohm, and 4 ohm resistors that should make this problem easy.
     
  4. Oct 22, 2009 #3
    Isub1(-3ohms-2ohms-2ohms-4ohms)+6V=0 Solve for I=.55A Current only changes at intersections so all the current has to pass threw the 3 ohms then branch into 3 parts it reconnects with the 2 diagonal and the 4 ohms then they reconnect with the 2 vertical. Which must equal the incoming current? right. I think I get the idea it's just the setup that is confusing me a bit.
     
  5. Oct 23, 2009 #4
    Hey, first of all, I'd start with finding the Main current I.

    The 2 ohm(R2), 2 ohm(R3) and 4 ohm(R4) resistors are all in parallell with each other, and in serial with the 3 ohm (R1), so:

    R1 + (R2||R3||R4) = Rtot
    3 ohm + ((1/2 ohm)+(1/2 ohm)+(1/4 ohm)) = 3.8 ohm

    You'll find the current via Itot=E/Rtot => 6 volt / 3.8 ohm = 1.58A

    From there, you can use current divider to find the current going through each resistor(if you don't know what current dividing is, there's a good article on wikipedia)

    Ir1 = 1.58A (the whole main current is going through it)
    Ir2 = 1.58A * ((2 ohm||4 ohm)/((4 ohm||2 ohm) + 2 ohm) = 0.632A
    Ir3 = 1.58A * ((2 ohm||4 ohm)/((4 ohm||2 ohm) + 2 ohm) = 0.632A

    Imain - Ir2 - Ir3 - Ir4= 0
    Imain - Ir2 - Ir3 = Ir4
    1.58A - 0.632A - 0.632A = 0.316A

    Power delivered by the battery => P=UI => 6v*1.58A = 9.48 W

    Hope I could help, cheers.
     
  6. Oct 24, 2009 #5
    Thanks, I asked my professor and he gave me that same method.
     
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