1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Batteries in Parallel and series

  1. May 6, 2012 #1
    1. The problem statement, all variables and given/known data
    You are given two circuits with two batteries of emf "E" and internal resistance R1 each. Circuit A has the batteries connected in series with a resistor of resistance R2, and circuit B has the batteries connected in parallel to an equivalent resistor.
    Note that the symbol E should be entered in your answers as EMF.

    What is the power dissipated by the resistor of resistance R2 for circuit A, given that
    EMF=10 R1=300 ohms R2=5000 Ohms

    Calculate the power to two significant figures.


    2. Relevant equations
    P=I^2.R
    P=V^2/R


    3. The attempt at a solution

    well i used the second equation, and from previous question worked out my current in algebraic equation,
    I=2E/2R1+R2 so I= 2x10/2x300+5000 =1/280 R=5000+2x300= 5600

    P=I^2.R

    P=(1/280)^2.5600= 0.07142857143

    how ever the correct answer to 2.s.f is 0.064
     
  2. jcsd
  3. May 6, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    They're only looking for the power dissipated by R2. You've included all the resistance in your R.
     
  4. May 6, 2012 #3
    oh ok thanks, thread can be closed now.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook