The text describes "Internal Weak Pull-Ups and Pull-Downs" in the book Microcontrollers From Assembly Language to C Using the PIC24 Family:
"The term weak is used because the resistance is high enough that an external driver can overpower the pull-up resistor and pull the current to near ground, producing a 0 input. The weak pull-up is implemented as a high-resistance P-transistor and when enabled, the gate of this transistor is 0, turning it on."
I get that a high resistance leads to a weak current output from the resistor. It's the way it is worded that doesn't register based on the previous definition of a pull-up resistor which is simply to ensure that there exists a logic level and not a float, and with the control being dictated by a pushbutton.
How can an external driver [a device that controls high and low output to the input of circuit in question] "overpower" the pull-up resistor? By increasing this current? If that is the case, doesn't V = IR imply that an increase in current increases the voltage? If this is true, how can a 0 input be generated when both scenarios have two different levels of currents flowing, which both take the same path down the pull-up resistor and either to the input, or to ground, based on whether or not the pushbutton was pressed? I am assuming based on my careful analysis of wording that the effect of "pull[ing] the current to near ground, producing a 0 input" can only happen if an external driver "overpowers" the pullup resistor.
Thanks for any help.