Power transfer with movable wiper

In summary: The wiper determines how much of the 5K is R1 and how much is R2. In summary, the conversation discusses a circuit with a potentiometer connected across a load resistor and the need to find an expression for the power dissipated in the load resistor for any value of R1. The use of Thevenin is not necessary and it is sufficient to pick an expression for power dissipated and apply it to RL. The wiper on the potentiometer divides the total resistance of 5 kΩ between R1 and R2, depending on its position.
  • #1
gfd43tg
Gold Member
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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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  • #2
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?
 
  • #3
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
 
  • #4
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".
 
  • #5
Simply if R1 = 5K then R2 = 0R; If R1 = 2.5K then R2 = 2.5K; if R1 = 0.1K then R2 = 4.9K.
 

FAQ: Power transfer with movable wiper

1. What is power transfer with movable wiper?

Power transfer with movable wiper is a type of electrical circuit where power is transferred from one source to another through the use of a movable wiper arm. This wiper arm is used to make contact with different points along the circuit, allowing the transfer of power.

2. How does power transfer with movable wiper work?

In a power transfer with movable wiper circuit, the power source is connected to one end of the circuit and the load is connected to the other end. The movable wiper arm is then used to make contact with different points along the circuit, allowing the flow of electricity from the source to the load.

3. What are the benefits of using power transfer with movable wiper?

One of the main benefits of using power transfer with movable wiper is its versatility. The wiper arm can be moved to different positions, allowing for the control of power transfer and the ability to switch between different power sources. It also allows for efficient power management and can reduce the need for additional components in a circuit.

4. What are some common applications of power transfer with movable wiper?

Power transfer with movable wiper is commonly used in electronic devices such as audio equipment, power supplies, and variable resistors. It is also used in more complex circuits, such as motor controllers and industrial automation systems.

5. Are there any limitations to using power transfer with movable wiper?

One limitation of power transfer with movable wiper is that it can introduce resistance into the circuit, which can affect the efficiency of power transfer. It also requires regular maintenance to ensure the wiper arm is making proper contact with the circuit. Additionally, it may not be suitable for high power applications as the wiper arm may not be able to handle heavy currents.

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