Varying the voltage of a light bulb

AI Thread Summary
The brightness of a light bulb is determined by the power it receives, which is calculated using the formula P = VI, where P is power, V is voltage, and I is current. In a parallel circuit, removing one bulb does not affect the brightness of the remaining bulbs because the total power supplied remains constant, and the voltage across each bulb stays the same. The removed bulb does not contribute to the overall power, resulting in no change in brightness for the others. This principle highlights that in parallel circuits, individual components operate independently regarding brightness. Thus, the brightness remains unchanged when one part of the circuit is removed.
sanado
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Hey guys, was wondering if you could help me with this question:

What determines the brightness of a lightbulb?

I think it would be power (and therefor voltage and current since P = VI) as power is used in the measurement of energy delivered. Why is it the case then that, from the questions i am doing, it states that if you have a simple circuit such as the one shown in the following link (page 11, question 5), why doesn't the brightness vary when one part of the parrallel circuit is removed?

http://www.vcaa.vic.edu.au/vce/studies/physics/pastexams/physics12004.pdf

Question 5, page 11

Thanks
 
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I don't see the contradiction. There's no power associated with the removed bulb, so it's dark. The power is unchanged for the others*, so their brightness is unchanged.

*P = VI = V2/R, and voltage and resistance are constant.
 
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