I have the answer to this question, but I'm trying to understand how to do it.
It's a simple circuit: https://onedrive.live.com/redir?resid=6CDEB7909E9460BA!57382&authkey=!ALscxM1v2OtHZIg&v=3&ithint=photo%2cPNG [Broken]
(note: Image inserted in-thread by moderator)
A) The switch starts open. The switch is closed for a long time and the capacitor charges. What will be the steady state current through the battery?
B) The switch starts open. The switch is closed for a long time and the capacitor charges. What will be the steady state voltage difference across the capacitor?
C) Now the switch is reopened and the capacitor begins to discharge. How long will it take for the voltage on the capacitor to decay to 5% of its fully charged value?
V=IR, Q=CV, T=RC, ΔVc = ΔVo * e^(-t/T)
The Attempt at a Solution
A) Is relatively easy. A fully charged capacitor acts like a resistor with infinite resistance, so essentially it's just R1 - R2 - R3 in series. So Vo=Rtotal * I
So 80=15*I, I = 5.33
B) This is what confuses me. All my examples have been simple RC circuits, one resistor. Per my understanding, the capacitor will charge up to the emf of the battery. So Vc should be 80? The extra resistors just slow down the current and how fast it charges.
Apparently, the correct answer is 42.64. Essentially 80 - R1 * I = 53.35
Then 80 - R1 * I - R3*I = 10.71
Then 53.35 - 10.71 = 42.64.
I understand the math, but don't get why we're doing that. Alternatively, one can do 80 * (8/15). For some reason we're using the resistance in parallel as the driver for charge.
C) ΔVc = ΔVo * e^(-t/T). We want 5% voltage, so e^(-t/T) = 0.05
That part makes sense, but finding T does not.
T=RC. Per my understanding, we use the resistance of the entire circuit, so R=15.
However the correct answer uses 8, i.e. R3 the one in parallel.
Clearly I'm missing something here. All my examples have 1 resistor and 1 capacitor. How do you deal with multiples?
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