Marechiare wrote:
> First, we are talking only about rising voltage on a capacitor, we ARE
> NOT talking about how the capacitor will be DISCHARGED through the
> resistors, please, re-read my previous post.

It doesn't matter since the two are symmetric.  We are assuming a perfect
voltage source, therefore 0 impedence, driving the input to the resistor
divider.

> How, the hell on the earth, lower R2 value, that is increase in
> current leakage to ground from the capacitor would help the DC source
> to charge the capacitor faster through the same constant resistor R1?

Because "faster" is in the context of the final steady state voltage, and is
not the same absolute value as you change the resistance.


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