Error in this solution?

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  • #1
STEMucator
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Homework Statement



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

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


Homework Equations




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?
 

Answers and Replies

  • #2
gneill
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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?
 
  • #3
STEMucator
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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?

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.
 
  • #4
gneill
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Draw the current through each component.
 
  • #5
STEMucator
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Draw the current through each component.

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


I hope that looks okay now.
 
  • #6
gneill
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You have both currents through the 2k resistors heading downwards into the same node. Where does the current go? (think KCL)
 
  • #7
STEMucator
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You have both currents through the 2k resistors heading downwards into the same node. Where does the current go? (think KCL)

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


I'm honestly a little confused if that's not the case.
 
  • #8
gneill
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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
 
  • #9
STEMucator
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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!

View attachment 76332

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.
 

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