Power dissipated in variable resistor?

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Homework Help Overview

The discussion revolves around the power dissipated in a variable resistor within an electrical circuit, particularly focusing on how this power changes with varying resistance. Participants are exploring concepts related to electromotive force (emf) and internal resistance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the relationship between power dissipation and resistance, particularly why power increases and then decreases. There are inquiries about the constancy of internal resistance and its implications on power calculations. Some suggest deriving expressions for power in terms of resistance and eliminating current from equations.

Discussion Status

The discussion is active, with participants providing guidance on deriving equations and exploring the implications of varying resistance. There is a mix of understanding and confusion, with some participants seeking clarification on specific mathematical relationships and assumptions.

Contextual Notes

Some participants note that internal resistance can be assumed constant for the purposes of the exercise, although it may vary in practical scenarios. There are also references to specific cases of resistance settings affecting power output.

CAH
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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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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?
 
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.
 
I don't understand this still,

Py= VI
V= emf - rI
Py= (emf - Ir)I ...=0
 
What is the current in the circuit in therms of the emf and the resistors?
 
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
 
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
 
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:
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.
 

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