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Value of current through 12V battery

  1. Aug 11, 2013 #1
    1. The problem statement, all variables and given/known data
    a) Calculate the value of the 12V battery
    b) Calculate the power dissipated in R_1,R_2 and R.

    https://www.physicsforums.com/attachment.php?attachmentid=60819&stc=1&d=1376202905

    2. Relevant equations

    I_1+I_2+I_R=0

    3. The attempt at a solution

    For the left-hand circuit,
    13=12-3I_R+1I_1

    (2)
    For the right-hand circuit,
    14=12-3I_R+2I_2

    (3)
    Substituting for I_R in equation (2),

    13=12-3(-I_1-I_2 )+I_1

    13=12+4I_1+3I_2

    1=4I_1+3I_2






    (4)
    Substituting for I_R in equation (3),

    14=12-3(-I_1-I_2 )+2I_2

    14=12+3I_1+5I_2

    2=3I_1+5I_2






    (5)
    Multiplying equation (4) by 3 and equation (5) by 4 and subtracting to find I_2,

    3=12I_1+9I_2

    8=12I_1+20I_2

    5=11I_2

    I_2=5/11 A



    Substituting for I_2 in equation (4) to find I_1,

    1=4I_1+3(5/11)

    I_1=-1/11 A






    (9)
    Substituting for I_1 and I_2 in (1) to find I_R,

    I_R=-(-1/11)-(5/11)

    =-4/11 A

    =-0.3636 A
     

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  3. Aug 11, 2013 #2

    NascentOxygen

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    Staff: Mentor

    I was expecting the current through the 12V to be negative, since the other branches carry higher voltage batteries.

    As a check, you could determine the voltage across each of the three parallel branches, and the voltages should all be equal.

    To determine the power dissipated in each R, use P = I2.R
     
  4. Aug 11, 2013 #3
    For R1
    P=I_1^2 R

    =0.00826 W
    For R2
    P= I_2^2 R

    =0.454*2

    =0.90990 W
    For R
    P=I^2 R

    =0.1322*3

    =0.3967 W
     
  5. Aug 11, 2013 #4
    so to check voltage is just

    V=I/R
     
  6. Aug 11, 2013 #5

    NascentOxygen

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    Staff: Mentor

    The voltage across the left-most branch is 1/11 x 1 + 13 = 131/11 V

    and a similar value should be found for the other branches.
     
  7. Aug 11, 2013 #6
    left hand branch 1/11 x 1 +13 = 13.0909
    right hand branch = 5/11 x 2 + 14 =14.0909

    confused
     
  8. Aug 11, 2013 #7

    NascentOxygen

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    Staff: Mentor

    Their currents are in opposite directions. So figure out where you need a negative sign.
     
  9. Aug 11, 2013 #8
    oh that was easy

    -5/11 x 2 +14 = 13.0909
     
  10. Aug 11, 2013 #9
    The voltage across the left-most branch is

    1/11∙1+13=13 1/11 =13.0909 V

    The voltage across the right-most branch is

    -5/11∙2+14 =13.0909V

    The voltage across the centre is

    4/11∙3+12=13.0909V
     
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