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Homework Help: 2 Problems

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

    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 by a moderator: May 1, 2017
  2. jcsd
  3. Mar 23, 2004 #2

    Doc Al

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    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

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    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

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    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!
     
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