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If you want to lower a resistance by pressing a switch, then you can add a parallel resistor with a series switch to a resistor.
When the switch is open (not pressed), then current will only flow through the resistor that is not in series with the switch.
Since there is 9V powering the circuit, and assuming that we are lighting an LED that drops 2V, then 7V will be across the current limiting resistor. Since the main current setting resistor is 1,000 ohms (1k), then there will be about 7mA (0.007A) of current flowing through the resistor and LED while the switch is open.
Closing the switch (by pressing it) puts that same 7V across the 1k resistor in series with the resistor in series with the switch (both of the parallel to the main current setting resistor). The second resistor will also pass about 7mA of current through it and the LED while the switch is closed.
The 2 separate currents flowing through the separate resistor combine before flowing through the LED. Therefore, there will be about 14mA of current flowing through the LED while the switch is closed.
7V/0.014A = 500Ω of equivalent resistance while the switch is closed.
I am planning on using the resistor with parallel resistor plus switch set up to replace the trimpot of a 555 timer controlled servo. That will give me 2 servo positions instead of a wide range of servo positions.
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