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Homework Statement
Find V(a) as a function of time in the following circuit when the switch is opened. Originally, the capacitor is charged to Vc = 1/101 * V1.
The Attempt at a Solution
I know that when the switch is opened, the capacitor is going to get charged again, and V(a) is going to be the same as its voltage. I know the general expression for the voltage of a charging capacitor is:
Vc(t) = V0 + V * (1 - e-t/RC)
Where Vc is the voltage of the capacitor, V0 is its original voltage, V is the voltage of the source, R the resistance and C its capacitance.
So, V(a)(t) is going to be the same as that. The problem is that I know that, in its stationary, Vc has to be equal to V, yet the expression I found doesn't do that. In this problem, I found that V0 = 0,29V, V = 30V, C = 2200μF and R... is it 100Ω? 100100Ω?
In an experiment I made (from which this questions comes), after 160 seconds V(a) was around 30V, but if I replace 160s in t in that expression, I get something closer to 17V. I thought that it has to be because I'm not calculating the R of the equivalent RC circuit right. Is that it?
Thanks
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