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: Emf and Terminal Voltage

  1. Feb 2, 2004 #1
    Okay here is the question:

    a) what is the internal resistance of a voltage source if its terminal voltage drops by 2 V when the current supplied increases by 5 A.

    b) Can the emf of the voltage source be found with the information supplied?

    I know that E = V - Ir, and I am pretty sure that a simple algebraic equation can be put together. So what formulat am I missing??

    Thanks
    Nautica
     
  2. jcsd
  3. Feb 2, 2004 #2

    HallsofIvy

    User Avatar
    Science Advisor

    You know that E = V - Ir so ΔE= ΔV- ΔI r. You are told that ΔE= 2volts when &Dekta;I= -5 amps. Of course, ΔV= 0 since the voltage source is fixed.
    Thatis: 2= -5r. Can you solve for r now? :smile:.

    The answer to b is NO. In effect V is the "constant of integration" when you "integrate" dE= -rdI. There is no way to find that from the information given.
     
    Last edited by a moderator: Feb 2, 2004
  4. Feb 2, 2004 #3
    I am in College Physics, we can not use calculus for an answer. So is there a reason algebraically.

    thanks
    nautica
     
  5. Feb 3, 2004 #4
    If u know the potential diff b/w two terminals, then u can find the emf.

    With this information u cannot find
     
  6. Feb 3, 2004 #5
    So ther is no way to find emf without Potential difference???

    thanks
    nautica
     
  7. Feb 4, 2004 #6
    Okay For a) here is what I got

    V=E-Ir

    or dV=dE - dIr

    2V = -5r

    so r = -.4Ohms

    BUT, my instructor said this was wrong b/c there I did not no the Rload and there are too many unknowns for this to be solved.

    I can see how part B could not be solved but this looked so simple for part a)

    He said this was the other equation I should have considered, but I do not see why it was needed.

    I = E / (Rload + r)

    Thanks
    nautica
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook