Find and Plot v_o(t) in Solution: Homework Equations & Attempt at a Solution

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The discussion focuses on finding and plotting the output voltage, v_o(t), in a circuit involving a capacitor and resistors. A participant expresses confusion regarding the presence of a negative sign in the voltage divider calculation at t = 0^+. Clarifications are provided about the direction of current flow when the switch opens, emphasizing the importance of the capacitor's polarity and its discharge behavior. The conversation highlights the significance of understanding the steady-state charge on the capacitor and the correct application of Kirchhoff's Current Law (KCL). Ultimately, the participant gains clarity on the role of the voltage source and the capacitor's behavior at the moment the switch is opened.
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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?
 
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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?
 
gneill said:
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.
 
Draw the current through each component.
 
gneill said:
Draw the current through each component.

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


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