1. Not finding help here? Sign up for a free 30min 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!

2 Problems

  1. Mar 23, 2004 #1
    https://hw.utexas.edu/tmp/Muddam1/1080079116Xuj.pdf

    for #5:
    I know you have to use this equation:
    [tex]Ve^\frac{-t}{RC}[/tex]
    R=3e6+2e6=5e6
    So I did [tex]11e^\frac{-1.5}{5e6*1e-6}= 8.149000428[/tex]
    but it's wrong and I don't know why.


    for #8:
    q= 3.376e-6
    I= 1.79839e-6
    (both values are right)
    C=1.1e-6
    q=CV
    V=q/C
    3.06909=V
    P=IV
    5.5194224e-6=1.79839e-6*3.06909
    because the answer should be in to the 1e-6 I didvided the answer by 1e-6, so my final answer is 5.5194224, which is also wrong.

    Thanks in advance.
     
    Last edited: Mar 23, 2004
  2. jcsd
  3. Mar 23, 2004 #2

    Doc Al

    User Avatar

    Staff: Mentor

    First Problem

    It looks like you found the voltage across the capacitor at t=1.5 sec, but the problem asks for the voltage across the 3M Ohm resistor. Hint: find the current, then use Ohm's law.
     
  4. Mar 23, 2004 #3
    So you want me to find current using this equation:
    [tex]\frac{V}{R}e^\frac{-t}{RC}[/tex]
    I should just use 3e6 for R? Then use V=IR?
     
  5. Mar 23, 2004 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Second Problem

    This problem asks for the power delivered by the battery, which is always P=IV, where V is the voltage of the battery.
     
  6. Mar 23, 2004 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Yes, where R is the total resistance.
    To find the voltage across a resistor, use V=IR for that resistor.
     
  7. Mar 23, 2004 #6
    ...

    Thanks a LOT!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: 2 Problems
  1. 2 problems (Replies: 1)

  2. 2 problems (Replies: 4)

  3. Problem (2) (Replies: 1)

Loading...