I'm supposed to find the voltage v0(t) for t > 0 for the following circuit:
The first part of the excercise was to find the current io(t). I did this by using the node-voltage method to find the voltage across the 8k resistor and the 10k resistor to find the voltages across the two capacitators. Then I figured since, once the switches open, they'll be in series, I find R*C by finding the total C which was 100 nF.
Then I found the voltage drop across Io which then became 120e^-5000t - 100e^-5000t = 20e^-5000t and divided this by the resistance to find the current, which would be 10e^-5000t mA.
The Attempt at a Solution
That's when things got ugly. I'm thinking that the voltage at v0 will have to be the voltage-drop between Io and the 300nF capacitator. Fair enough. If the current in I0 is 10e^-5000t mA I figure the voltage will have to be 20e^-5000t V. Then I'm thinking the voltage across the 300nF supply will have to be 100e^-5000t V (since the voltage at the 10k resistor was 100v before the switches were opened) which makes me believe the voltage at v0 should be -80e^-5000t v.
However, they gave us the correct answers beforehand, and I know it's supposed to be:
-6.67e^-5000t + 106.67 v.
This has me somewhat stumped. If this is correct, that means even when t -> infinity there will be a voltage of 106.67 v over v0. But then i'm thinking, since these really are capacitators, won't the 150nF capacitator, being the one with the highest voltage, somehow charge the other one? Or what?
I'm thinking some kind of equilibrium here is involved, but I'm having real trouble seeing where those numbers are coming from. A push in the right direction would truly be helpful!
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