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: Maximizing voltage across a load resistor?

  1. Feb 22, 2010 #1
    Hey, so I have this loop circuit that has some input voltage V in series with a resistor with a resistance of R and another in series resistor that is my load resistance, lets call it G. V and R are not variables, only G is.

    I know that because of Jacobi's law, to maximize the power dissipated by the load resistance, you have to have G = R.

    But I'm wondering how i should maximize the voltage drop across the load resistor? I just can't figure out how to write an equation and go from there.

    Also, how can i maximize the current going into the load resistor?

    Please help,
    Thanks
     
  2. jcsd
  3. Feb 22, 2010 #2

    berkeman

    User Avatar

    Staff: Mentor

    What is the voltage divider equation? That is what you use to do what you are asking about.

    Assuming that V and Rs are fixed, you maximize the output voltage with an ______ circuit, and maximize the output current (a different situation) with a _______ circuit.
     
  4. Feb 23, 2010 #3
    You mean kirchoffs voltage eqn?

    Its V - IR - IG = 0

    im starting to think that i can maximize the voltage drop on the load resistor G by maximizing the resistance of G though...still don't know if thats right.
     
  5. Feb 23, 2010 #4
    The voltage drop across the load resistor is given by VG=VG/(R+G), to find the maximum value of this function find when the derivative equals zero, so when R/(R+G)2= 0, which is when G equals infinity...basically the bigger the resistance the bigger the voltage drop across it.. if I am understanding your question..now to maximize the current going into the load resistor, well I=V/(R+G)...what would give a maximum value for I if G is the only variable?
     
  6. Feb 23, 2010 #5
    OH! so minimizing R+G for current..i see.

    Thanks a lot guys.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook