Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Power supplied by battery in a multiple loop Circuit

  1. Feb 13, 2008 #1
    1. The problem statement, all variables and given/known data
    I am given the circuit, which is shown in the attachment, and I need to calculate the power supplied from each of the 12V batteries.


    2. Relevant equations
    I have already solved for the current in each resistor and the potential difference between a and b.

    The only equation I know to use is P= IV, where I am calculating I by dividing V/R(equivalence)



    3. The attempt at a solution

    I have tried solving this multiple times, all without success. My most current attempt at the Power supplied from the right side was finding R(equivalence) = 9 the using I=V/R, then using the new R (4/3) in P=IV P=(4/3 A)*12V, where P=8, which I thought was a rather low number.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Feb 13, 2008 #2
    ur attachment is pending approval.. why dont u upload it somewhere.. like: http://bayimg.com [Broken] and then post it here so that it becomes available immediately.
     
    Last edited by a moderator: May 3, 2017
  4. Feb 13, 2008 #3
    Sorry, I just didn't think about it.

    F26-59.jpg

    http://i2.photobucket.com/albums/y30/snoweangel27/F26-59.jpg" [Broken]
     
    Last edited by a moderator: May 3, 2017
  5. Feb 13, 2008 #4
    Use Kirchoff's law to find the amount of current flowing in each branch i.e. through each resistor. Once you've done that, use the formula:

    [tex]
    P = I^2R
    [/tex]

    to find the power dissipated.
     
  6. Feb 13, 2008 #5
    Should I calculate the Power dissipated from each resistor in the loop then sum them to get the total power?
     
    Last edited: Feb 13, 2008
  7. Feb 14, 2008 #6
    If you sum [tex] I^2R[/tex] for all the resistances in the circuit that will give you the total power delivered by both voltage sources. But you need the power of each of them separately.

    A voltage source with potential difference [tex]V[/tex] that delivers a current [tex]I[/tex] delivers a power [tex]VI[/tex]
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook