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!

Error in this solution?

  1. Dec 7, 2014 #1

    Zondrina

    User Avatar
    Homework Helper

    1. The problem statement, all variables and given/known data

    Find ##v_o(t)## and plot it.

    Screen Shot 2014-12-07 at 4.07.03 PM.png

    2. Relevant equations


    3. The attempt at a solution

    The solution was provided, and I got all of the same numbers except for one thing. For some reason, in step 3 there is a negative sign in front of ##v_c(t = 0^+)## when calculating ##v_o(t = 0^+)## using a voltage divider.

    Screen Shot 2014-12-07 at 4.10.26 PM.png

    Could someone shed some light to where that came from or is it simply an error?
     
  2. jcsd
  3. Dec 7, 2014 #2

    gneill

    User Avatar

    Staff: Mentor

    Mark the polarity of the capacitor charge for time t < 0. When the switch opens at t = 0, what direction will the current flow through the "output" resistor? So what's the resulting direction of potential drop?
     
  4. Dec 7, 2014 #3

    Zondrina

    User Avatar
    Homework Helper

    So you mean the capacitor discharges like so:

    Screen Shot 2014-12-07 at 6.11.12 PM.png

    So the current will flow head to tail through the 2k.
     
  5. Dec 7, 2014 #4

    gneill

    User Avatar

    Staff: Mentor

    Draw the current through each component.
     
  6. Dec 7, 2014 #5

    Zondrina

    User Avatar
    Homework Helper

    Screen Shot 2014-12-07 at 6.11.12 PM.png

    I hope that looks okay now.
     
  7. Dec 7, 2014 #6

    gneill

    User Avatar

    Staff: Mentor

    You have both currents through the 2k resistors heading downwards into the same node. Where does the current go? (think KCL)
     
  8. Dec 7, 2014 #7

    Zondrina

    User Avatar
    Homework Helper

    Screen Shot 2014-12-07 at 6.11.12 PM.png

    I'm honestly a little confused if that's not the case.
     
  9. Dec 7, 2014 #8

    gneill

    User Avatar

    Staff: Mentor

    You're making things worse! You've changed the polarity of the charge on the capacitor for no reason. Your original figure was correct: the capacitor was charged (at steady state) to the potential difference across the 2k resistor in parallel with it.

    Current flows from higher potential (+) to lower potential (-). Current leaving the + terminal of the capacitor flows around the circuit to make its way back to the - terminal of the capacitor. You will not find current heading away from the - terminal of the capacitor!

    Fig1.gif
     
  10. Dec 7, 2014 #9

    Zondrina

    User Avatar
    Homework Helper

    I realize now that the voltage source was completely messing with my head, even though at ##t = 0^+## I shouldn't even be considering it! Only the current from the capacitor should matter at that point and the way it discharges is given by it's polarity (which can be found by looking at the battery terminals at ##t = 0^-##).

    Thank you for clarifying this for me. Fooled once, never fooled again.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Error in this solution?
  1. Error checks (Replies: 1)

  2. VHDL errors (Replies: 1)

  3. Interpolation error. (Replies: 1)

Loading...