Question about Power & Lightbulbs (really embarassed dont laugh)

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Homework Help Overview

The discussion revolves around understanding power delivery to a lightbulb, specifically comparing the power when the filament is initially glowing versus when it has reached a steady state. The subject area includes concepts of electricity, power, and resistance in circuits.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to reason through the relationship between current, time, and power, questioning whether more power is delivered when the filament is increasing in glow or when it is steady. Some participants challenge the logic regarding current and time intervals, suggesting a need to consider the filament's resistance as it heats up.

Discussion Status

The discussion is ongoing, with participants providing insights into the relationship between voltage, current, and resistance. There is an exploration of how the filament's temperature affects its resistance and power consumption, but no consensus has been reached yet.

Contextual Notes

Participants note the importance of understanding the behavior of resistance in relation to temperature changes in the filament, which may influence the power delivered to the lightbulb. There is an acknowledgment of basic electrical principles that need to be revisited for clarity.

billiam
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Ok this question is pretty rudimentary but i only have one try so I am not sure.

When is more power delivered to a lightbulb, just after it is turned on and the glow of the filament is increasing or after it has been on for a few seconds and the glow is steady?
a.) When the glow of the filament is increasing.
b.) When the glow of the filament is steady.

Ok the formula for power is P = IV. and I = Q/T

the longer the time interval, the smaller the current, and consequently the lower the power so I am thinking the answer is a.) when the glow of the filament is increasing.

Is this logic totally wrong? thanks.
 
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billiam said:
and I = Q/T

the longer the time interval, the smaller the current,

Is this logic totally wrong? thanks.
Mostly wrong if I=Q/T got you to that smaller current argument.
Instead interpret that to say – To maintain a current I over double a time interval of T will require the moving of twice as many charge units Q.
This has little to do with figuring your power.

On the power issue – deciding what is happening to the resistance of the filament as its temperature changes should do.
 
RandallB said:
On the power issue – deciding what is happening to the resistance of the filament as its temperature changes should do.

so there is resistance to current in the fillament causing heat to be produced known as glowing. I am still not sure how this equates to more or less power being delivered.
 
??
Ok you’re just getting started.
Normal power sources V is constant.
P=IV is right – but what is I?
I=V/R
So P=V (V/R)
Go over that Basic Electricity book or google for some instruction.

To learn how the constant V you’re being supplied, doesn’t always stay constant, read up “Impedance” (internal resistance) of the battery or supply.

Have fun reading.
 
For metals, resitivity generally increases with temperature. As the filament gets hot, its power consumption falls.

- Warren
 

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