Anybody who slogged their way through freshman circuit design knows the well-worn formula for "time constant" (tau) being equal to the product of R and C (time in seconds). One tau gets you to 63% of V+, two tau to 86%, three tau to 95%, and so on. (1- (1/e^tau)).
Suppose, though, you had a resistive voltage divider comprised of equal value R. The voltage at the midpoint is, quite naturally, V+/2. Now suppose you put that capacitor across the "bottom" resistor that goes from the midpoint to the minus supply or ground.
Does the time constant equation change? Obviously the capacitor is only going to charge to V+/2, but does it do it in the same percentage as with a single resistor? I've been tossing this one around in my head all afternoon and can argue both sides of the discussion.
Comments appreciated.
Jim