Determine the position of the slider on the pot

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Discussion Overview

The discussion revolves around determining the position of a slider on a 10kΩ potentiometer in a circuit with a 5kΩ load, specifically when the voltage across the load is 3V. Participants explore the theoretical and practical aspects of voltage dividers, potentiometers, and the relationships between resistances in the circuit.

Discussion Character

  • Homework-related
  • Technical explanation
  • Exploratory
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the problem and seeks simpler explanations regarding the division of the potentiometer's resistance.
  • Another participant explains that the potentiometer can be represented as two resistors, x10kΩ and (1-x)10kΩ, where x represents the position of the slider.
  • A participant questions the absence of a supply voltage and circuit diagram, which are crucial for solving the problem.
  • There is a discussion about the concept of a rheostat and whether the configuration can be treated as such, with some participants suggesting that the potentiometer functions similarly.
  • One participant proposes a method of calculating the voltage drop across the potentiometer based on a supply voltage, but another cautions that this approach may not be valid due to the presence of the load resistor.
  • Participants discuss the need to derive an equation for the output voltage in terms of the resistances involved, acknowledging the complexity introduced by the load resistor.
  • There is an ongoing exploration of how to express the combined resistance of the potentiometer and load in order to apply the voltage divider formula correctly.
  • Some participants express frustration with understanding the ratio of resistances and how to incorporate the load resistor into their calculations.
  • One participant attempts to clarify the relationship between the resistances and voltage output, but acknowledges the need for further refinement of their approach.

Areas of Agreement / Disagreement

Participants generally agree on the need to understand the relationships between the resistances in the circuit and how to apply the voltage divider principle. However, there is no consensus on the specific approach to take or the correct method for calculating the output voltage, indicating multiple competing views and ongoing uncertainty.

Contextual Notes

Participants mention the importance of knowing the supply voltage and the configuration of the circuit, which are not fully provided. There are unresolved mathematical steps regarding the calculation of combined resistances and the application of the voltage divider formula.

  • #31
Harrison01 said:
I just can't get in my thick skull...

Refer to the diagram in post #4.
 
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  • #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.
 

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