Power transfer with movable wiper

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SUMMARY

The discussion focuses on calculating the power dissipated in a load resistor (RL) connected to a potentiometer with a total resistance of 5 kΩ. Participants clarify that it is not necessary to use Thevenin's theorem for this problem; instead, one can derive the power expression by determining the current (I) or voltage (V) across RL. The relationship between the wiper position and the resistance values R1 and R2 is emphasized, illustrating how the wiper alters the effective resistance in the circuit.

PREREQUISITES
  • Understanding of basic electrical circuits
  • Knowledge of potentiometers and their function
  • Familiarity with power equations for resistors
  • Ability to analyze circuit components and their relationships
NEXT STEPS
  • Learn how to derive power equations for resistors in series and parallel configurations
  • Study the principles of Thevenin's theorem and its applications
  • Explore the characteristics and applications of potentiometers in circuits
  • Practice plotting power versus resistance graphs for various circuit configurations
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Electrical engineering students, circuit designers, and anyone involved in analyzing or designing circuits with variable resistors such as potentiometers.

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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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