Power dissipated in variable resistor?

In summary, the conversation discusses the relationship between power dissipation and variable resistors, as well as the role of internal resistance. The participants also mention the importance of impedance matching and deriving mathematical expressions for power. They also consider the behavior of power at different values of the variable resistor.
  • #1
CAH
48
0
See photo!

Why does the power dissipated increase and then decrease (see graph)? Is it something to do with emf?
Also can internal resistance change as the variable resistor increases/decreases or is it always constant?

- this isn't one of the questions they ask I'm just wondering
 

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  • #2
CAH said:
See photo!

Why does the power dissipated increase and then decrease (see graph)? Is it something to do with emf?
Also can internal resistance change as the variable resistor increases/decreases or is it always constant?

- this isn't one of the questions they ask I'm just wondering

You can derive the power on the variable resistor in terms its resistance. Keep the emf and internal resistance constant.
It is a very important result that the power is maximum at a certain loading resistance - how is it related to the internal resistance of the source?
Have you heard about impedance matching?
 
  • #3
Internal resistance can be assumed as being fixed for this exercise, though in practice it can be expected to change slightly.

Can you derive an expression for the power in Y, perhaps assume a particular value for r. Then sketch your graph, power versus value of Y.
 
  • #4
I don't understand this still,

Py= VI
V= emf - rI
Py= (emf - Ir)I ...=0
 
  • #5
What is the current in the circuit in therms of the emf and the resistors?
 
  • #6
CAH said:
See photo!

Why does the power dissipated increase and then decrease (see graph)? Is it something to do with emf?
Also can internal resistance change as the variable resistor increases/decreases or is it always constant?

- this isn't one of the questions they ask I'm just wondering
to answer your question, the internal resisitance won't change with respect to the variable resistor
 
  • #7
CAH said:
I don't understand this still,

Py= VI
V= emf - rI
Py= (emf - Ir)I ...=0
this is a good start. Your problem is you need to get rid I.

In your second equation, try to get rid of I using Y and r
 
  • #8
CAH said:
See photo!

Why does the power dissipated increase and then decrease (see graph)? Is it something to do with emf?
Also can internal resistance change as the variable resistor increases/decreases or is it always constant?

-
 
Last edited:
  • #9
Eliminate current from your equations and see what function you get relating the power with the load Y..Also see whether this power is total power or power consumed by load Y..
 
  • #10
Why does the power dissipated increase and then decrease (see graph)?

Power = voltage * current.

Consider two cases...

1) The variable resistor is set to a very high value (eg open circuit) => The current is very low or zero => Power is very low.
2) The variable resistor is set to a very low value (eg short circuit) => Voltage is very low or zero => Power is very low.

So at both ends of the range the power is very low. So no surprise it's higher in the middle.
 

What is a variable resistor?

A variable resistor, also known as a potentiometer, is an electronic component that can change its resistance value, allowing for the control of electrical current or voltage in a circuit.

How is power dissipated in a variable resistor?

Power is dissipated in a variable resistor through the conversion of electrical energy into heat. As the resistance of the variable resistor increases, more energy is converted into heat, resulting in a higher power dissipation.

What factors affect the power dissipated in a variable resistor?

The power dissipated in a variable resistor is affected by its resistance value, the current flowing through it, and the voltage drop across it. Other factors such as temperature and material composition can also play a role.

What is the formula for calculating power dissipation in a variable resistor?

The formula for calculating power dissipation in a variable resistor is P = I^2 * R, where P is power in watts, I is current in amperes, and R is resistance in ohms.

Why is it important to consider power dissipation in a variable resistor?

It is important to consider power dissipation in a variable resistor because excessive heat can damage the component and surrounding circuitry. It is also important for determining the appropriate size and rating of the resistor for a given application.

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