Question about Power & Lightbulbs (really embarassed dont laugh)

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More power is delivered to a lightbulb when the filament is increasing in glow, as the resistance changes with temperature. The relationship between power (P), current (I), and voltage (V) is defined by the formula P = IV, where I can also be expressed as I = V/R. As the filament heats up, its resistance increases, leading to a decrease in current and thus lower power consumption. Understanding the concept of impedance and how it affects voltage supply is crucial for grasping these dynamics. Overall, the initial increase in glow corresponds to higher power delivery before the filament stabilizes.
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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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