Power transfer with movable wiper

AI Thread Summary
The discussion revolves around calculating the power dissipated in a load resistor (RL) connected to a potentiometer in a circuit. Participants clarify that it is not necessary to use Thevenin's theorem, as one can derive the power expression directly by determining the current or voltage across RL. The importance of understanding the potentiometer's wiper mechanism is emphasized, as it affects the resistance values presented in the circuit. The wiper divides the potentiometer into two sections, with its position determining the resistance values of R1 and R2. Overall, the focus is on deriving a formula for power dissipation and understanding the variable nature of resistance in the circuit.
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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
 

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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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