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: Help Solving Circuit Variables Problem

  1. Mar 3, 2006 #1
    Can Anyone Help Me to solve in this problem the a question
    the one of the time
    thank you,

    Best Regards

    Attached Files:

    • scan.jpg
      File size:
      17.7 KB
  2. jcsd
  3. Mar 3, 2006 #2
    This is not the proper place for homework help.
  4. Mar 3, 2006 #3
    To solve: how does one go about getting an extrema? What does this mean?

    Quick answer: since we know that power may be written as the product of the current and the voltage, you are given the two equations for v(t) and i(t). Write this out and to find the maximum power take the derivative of each side
    [tex] \frac{d}{dt} P= \frac{d}{dt} \left( i(t) \cdot v(t) \right) [\tex]
    set this equal to zero and find the time. Note the initial conditions given in the problem statement. Take this solution for t back into the power equation to find the maximum power. This should get you started.
  5. Mar 4, 2006 #4
    sorry for putting it in the wrong section, am new here but can anyone try to make it more clear for me about this problem , a question?

  6. Mar 5, 2006 #5
    you're given the voltage and the current as functions of time, the product of these two gives an expression for the power.
    P = I V
    Now take the derivative with respect to time, and solve for time, remember from calc, that the critical points are those for which the slope of the tangent line is zero. Once you found a time, remember that for t<0, there's no current -> no power. So this time will answer (a), now take your result for (a) and substitute it into your equation for P, to get the maximum power. For (c) then how do you find the total power delivered to any circuit element. Hint: you have enough information to find this one too. hope this helps, sincerely, x
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook