- #1
deftonix
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Homework Statement
The bit I'm struggling with is to determine the time constant [tex]\tau[/tex] for the referenced circuit after the switch is closed. Constants are given in the image, which can be found at https://www.physicsforums.com/attachment.php?attachmentid=12737&d=1203627574.
Homework Equations
[tex]\tau=RC[/tex]
The Attempt at a Solution
I know that the answer is [tex]100k\Omega \cdot 10.0\mu F = 1.00s[/tex] and that it is because current is going to flow in the loop that goes from the cap, through the switch, through the [tex]100k\Omega[/tex] resistor, and back to the cap; since the only resistor that affects it is [tex]100k\Omega[/tex] one, it's just that times C. I also understand that there should be a similar loop on the battery side of the circuit. What I don't understand is why the loop that doesn't go through the switch(i.e., the one containing everything) doesn't affect the time constant. Wouldn't the battery be pushing current into that loop as the capacitor's voltage dropped?