Power transfer with movable wiper

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Discussion Overview

The discussion revolves around a homework problem involving a potentiometer connected across a load resistor (RL). Participants are tasked with deriving an expression for the power dissipated in RL as a function of the resistance R1, and plotting this power against R1 as the wiper of the potentiometer is moved.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant expresses uncertainty about whether to use Thevenin's theorem to solve the problem and seeks guidance on starting the calculation for power in RL.
  • Another participant suggests that Thevenin's theorem is not necessary and recommends finding an expression for power dissipated by RL using current (I) or voltage (V) across it.
  • A participant points out an assumption made in an earlier post regarding the resistance presented by the potentiometer, indicating that it varies with the position of the wiper.
  • Some participants describe the function of the wiper in a potentiometer, explaining that it divides the resistor into two sections and that the position of the contact affects the resistance values.
  • One participant provides specific examples of resistance values for R1 and R2 based on different positions of the wiper, illustrating how these values change.

Areas of Agreement / Disagreement

Participants generally agree on the mechanics of the potentiometer and its wiper, but there is uncertainty regarding the approach to solving the problem, particularly about the use of Thevenin's theorem and the assumptions made about resistance values.

Contextual Notes

There are limitations in understanding how the wiper affects the total resistance and the assumptions regarding the fixed resistance of the potentiometer in earlier posts. The discussion does not resolve these uncertainties.

Who May Find This Useful

Students working on circuit analysis involving potentiometers, those interested in power dissipation in resistive circuits, and individuals seeking clarification on the application of Thevenin's theorem in practical scenarios.

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Homework Statement



In the circuit shown below, a potentiometer is connected across the load resistor RL. The total resistance of the potentiometer is R = R1+R2 = 5 kΩ.
(a) Obtain an expression for the power PL dissipated in RL for any value of R1.
(b) Plot PL versus R1 over the full range made possible by the potentiometer’s wiper.

Homework Equations





The Attempt at a Solution


I don't understand if I need to find the thévenin resistance for this, also I don't know how to really start the problem and need a hint for finding the power for any value of R1
 

Attachments

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Maylis said:

Homework Statement



In the circuit shown below, a potentiometer is connected across the load resistor RL. The total resistance of the potentiometer is R = R1+R2 = 5 kΩ.
(a) Obtain an expression for the power PL dissipated in RL for any value of R1.
(b) Plot PL versus R1 over the full range made possible by the potentiometer’s wiper.

Homework Equations





The Attempt at a Solution


I don't understand if I need to find the thévenin resistance for this, also I don't know how to really start the problem and need a hint for finding the power for any value of R1

You don't necessarily need to use Thevenin here. It is sufficient to pick an expression for the power dissipated by a resistor and apply it to RL, after finding expressions for I, or V, or both for RL (depends upon what equation you pick for the power dissipated).

EDIT: In the attempt that you attached you've made the assumption that the resistance to the right of RL is always 5 kΩ. This is not correct since moving the wiper on the potentiometer will clearly vary the amount of resistance "presented" between the rails. What part of the potentiometer makes up the net resistance that it presents?
 
I don't understand the wiper, we have never seen it before, so I don't know really how to go about doing this problem
 
Maylis said:
I don't understand the wiper, we have never seen it before, so I don't know really how to go about doing this problem

It's essentially a resistor where a sliding contact divides it into two sections. Externally it looks like two resistors end-to-end with the central junction available via the sliding contact. The position of the contact determines how much of the total resistance lies in each section. Look up "potentiometer".
 
Simply if R1 = 5K then R2 = 0R; If R1 = 2.5K then R2 = 2.5K; if R1 = 0.1K then R2 = 4.9K.
 

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