Determine the position of the slider on the pot

  • Thread starter Thread starter Harrison01
  • Start date Start date
  • Tags Tags
    Position Slider
Click For Summary
To determine the position of a 10kΩ potentiometer with a 5kΩ load when the voltage across the load is 3V, it's essential to understand that the potentiometer divides voltage based on the ratio of its two resistive parts. The resistance can be expressed as x*10kΩ for one part and (1-x)*10kΩ for the other, where x represents the fraction of the potentiometer used. The output voltage can be calculated using the voltage divider formula, which incorporates the parallel resistance of the load and the potentiometer's resistances. The discussion emphasizes the need to derive equations that relate the known voltage and resistances to solve for the position of the slider on the potentiometer. Understanding these concepts is crucial for accurately determining the potentiometer's setting in the circuit.
  • #31
Harrison01 said:
I just can't get in my thick skull...

Refer to the diagram in post #4.
 
Physics news on Phys.org
  • #32
gneill said:
:confused: What happened to the equation you spent so much time deriving? x should be a unitless number, representing the fraction of the potentiometer that is Ra.

Ok i understand that x and (x-1) are unitles numbers as we yet don't know their value but we do know that their value sums up to 10kohms. We therefore need to work out the ratio. We know we have a Vout of 3v and that Rl is 5kohms. I just can't seem to understand how we use all this data to give us the value of Ra or Rb which will then give us the opposing figure.
 
  • #33
Ra is x*R. Rb is (1-x)R. R is the value of the potentiometer, 10000 Ohms.

Ra + Rb = R

x*R + (1-x)*R = R

x is just a value that "cuts" the potentiometer into two pieces. The two pieces sum to the whole.
 
  • #34
I haven't touched anything like this in years and was following this through ok till i hit post 21 and 22.

Could someone please explain how we specifically get the (Rb*Rl)+(Ra*Rl)+(Rb*Ra) part of Vout= E* Ra*Rl/[(Rb*Rl)+(Ra*Rl)+(Rb*Ra)]

I just can't see why its Rb X Rl + the rest..

Could someone show or explain how these 3 are derived.
I would be very grateful.
 
  • #35
braceman said:
I haven't touched anything like this in years and was following this through ok till i hit post 21 and 22.

Could someone please explain how we specifically get the (Rb*Rl)+(Ra*Rl)+(Rb*Ra) part of Vout= E* Ra*Rl/[(Rb*Rl)+(Ra*Rl)+(Rb*Ra)]

I just can't see why its Rb X Rl + the rest..

Could someone show or explain how these 3 are derived.
I would be very grateful.

Expand the expression for the output of the voltage divider using the given components.
 

Similar threads

  • · Replies 25 ·
Replies
25
Views
7K
Replies
7
Views
3K
  • · Replies 50 ·
2
Replies
50
Views
9K
Replies
1
Views
4K
  • · Replies 7 ·
Replies
7
Views
10K
  • · Replies 43 ·
2
Replies
43
Views
37K
Replies
6
Views
4K
  • · Replies 4 ·
Replies
4
Views
9K
  • · Replies 20 ·
Replies
20
Views
3K
Replies
5
Views
4K