#1
Say I have a 1kohm Pot. R1= 1kohm and R2 = 333.333 ohms to establish RL has 3v across it. Since RL (10k) ohms is connected to the 1k pot, the voltage across RL shouldn't change at all, right?
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#4
If you have just a 1k pot as the only resistor (a resistor and DC source) than the voltage drop will be the same (the drop will be the input voltage so that it is 0 at ground) regardless of what value the pot is set at. This help? I'm not sure exactly what you are asking.
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#5
Adding extra resistance in any circuit will change the voltage across the resistors if the voltage isn't raised to compensate. The voltage gets evenly distributed across the resistors that exist (if in series). I dunno if the 1k was already there or not, the wording to your question is a little vague.
#6
Quote by vicd08
I'm not sure exactly what you are asking.


Same here. Try to re-word your question TS so we can actually understand what you are asking us.
#7
Schematic?

Depends on how you connect RL to the pot. Also, I don't think you can make the voltage constant on RL with only 2 resistors.

Edit:

Quote by insurgentsteve
Adding extra resistance in any circuit will change the voltage across the resistors if the voltage isn't raised to compensate. The voltage gets evenly distributed across the resistors that exist (if in series). I dunno if the 1k was already there or not, the wording to your question is a little vague.


Also, this. Ohm's Law.
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Last edited by YourMomToTheMaX at Oct 6, 2009,
#8
Quote by vicd08
If you have just a 1k pot as the only resistor (a resistor and DC source) than the voltage drop will be the same (the drop will be the input voltage so that it is 0 at ground) regardless of what value the pot is set at. This help? I'm not sure exactly what you are asking.


Connected to the 1k pot is a 10k load, I have 9v across the pot without the 10k load. With a 12v Source voltage, I should still keep the 3v across RL, with R1 and R2 staying the same?
BRIGHT LIGHTS PUT ME IN A TRANCE.
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#9
This it? I also don't know what R1 and R2 are. 3 resistors and load in series, parallel?
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#10
Quote by YourMomToTheMaX
Schematic?

Depends on how you connect RL to the pot. Also, I don't think you can make the voltage constant on RL with only 2 resistors.

Edit:


Also, this. Ohm's Law.



The pot has an R1 and and R2.

R1 = 1k because it's a 1k pot
R2 = 333.333 ohms because I wanted to have 9v across the pot and 3v for Vout.

With a source voltage of 12v.



It's basically the circuit on the right, forget everything else. Think of the R1 and R2 as a pot. RL is 10k ohms.

My question is, will I have to change the resistance of my pot to accomodate having 3v across the 10k load still? Or will voltage remain the same through out the circuit because it's in parallel.
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#11
The voltage drop will be the same. 3 volts goes to R2 and the load, so just the current splits. Vout is the same in either case (with or without load) because the voltage drop from after R1 to ground is the same regardless of what is connected. By attaching RL in parallel to R2, you cut the current to the load. Both still see 3 volts.
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Last edited by vicd08 at Oct 6, 2009,
#12
Quote by vicd08
The voltage drop will be the same. 3 volts goes to R2 and the load, so just the current splits. Vout is the same in either case (with or without load) because the voltage drop from after R1 to ground is the same regardless of what is connected.


All I needed, thanks.
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