HI,
I just put together a 12 volt circuit to test adding another pot for throttle/speed ajustment on a motor.
Ohms law applied to the parallel circuit cuts the amperage/current across the visible resistor, in this case the variable resisors/potentionmeters. Have to have both pots all the way open to get 12 volts on the motor.
However, If i exchange the pot for pot with an on off swtich,and turn on the pot to ''off''... does the resistance get masked or no longer applly in the circut?
So that the other pot that is one gets al the current or amperage?
Another way to put it, of a potentionmeter with a pot that has a real on and off switch integratred, in the off potition subtract le resistance from the equation?
Im pretty sure I can use rocker switch on-off-on to isolate both pots, but the client doesnt want another switch in this circut.
I just put together a 12 volt circuit to test adding another pot for throttle/speed ajustment on a motor.
Ohms law applied to the parallel circuit cuts the amperage/current across the visible resistor, in this case the variable resisors/potentionmeters. Have to have both pots all the way open to get 12 volts on the motor.
However, If i exchange the pot for pot with an on off swtich,and turn on the pot to ''off''... does the resistance get masked or no longer applly in the circut?
So that the other pot that is one gets al the current or amperage?
Another way to put it, of a potentionmeter with a pot that has a real on and off switch integratred, in the off potition subtract le resistance from the equation?
Im pretty sure I can use rocker switch on-off-on to isolate both pots, but the client doesnt want another switch in this circut.