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: Electric potential

  1. Mar 5, 2008 #1
    1. The problem statement, all variables and given/known data

    how much energy is wasted when 35 C of charge flow through a 120 V toaster in 5.0 S?
    how much power?

    2. Relevant equations
    work=qv i think...

    power= work /time

    3. The attempt at a solution

    i thought to find energy wasted i would find the work used. (?????yes?)

    so i guess that would be 35(120) whihc is 4200 J

    so then do i power = 4200/5=840

    is this correct?
  2. jcsd
  3. Mar 5, 2008 #2
    well P = IV

    and they give you the time...so you know the units of Power = (Joules/second), so given the time you can find the joules wasted...

    both problems solved with one equation!

  4. Mar 5, 2008 #3
    wait i didnt really understand what u just said.
    did i do this right or is it wrong?
  5. Mar 5, 2008 #4
    no recheck your work and look at the formula I gave you and the units of power, energy is in JOULES, and power is in Watts or Joules/second...it should be straightforward from there
  6. Mar 5, 2008 #5
    well a watt = J/sec
    so i got energy in joules correct right??

    then those Joules per 5 sec gives u 840 J/ a sec
  7. Mar 6, 2008 #6
    ok another formula that will help is to know that I (current) is equal to (change in Q- the charge)/ (change in time - seconds), both of these informations are given in the problem...

    once you find I...you can solve for P = IV, using the information given in the problem...
  8. Mar 6, 2008 #7
    oh and I forget to mention that the change in potential difference (V) is equal to the potential-energy change per unit charge in moving across the toaster...so U = Vq...you know volts and know q so you can find the energy lost.
  9. Mar 6, 2008 #8
    so isnt i=7
    and then 7(120)=840.......
  10. Mar 6, 2008 #9
    ddint i do that in the proble, above??????
  11. Mar 6, 2008 #10
    ohh..lol my bad, yes your work is correct, sorry for the explanation again above...wasn't thinking so simply...
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook