Varying the voltage of a light bulb

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SUMMARY

The brightness of a light bulb is determined by the power it receives, which is calculated using the formula P = VI (Power = Voltage x Current). In a parallel circuit, when one bulb is removed, the power supplied to the remaining bulbs remains unchanged, as voltage and resistance are constant. Therefore, the brightness of the other bulbs does not vary despite the removal of one bulb. This understanding clarifies the relationship between voltage, current, and brightness in electrical circuits.

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  • Familiarity with Ohm's Law (V = IR)
  • Knowledge of power calculations in electrical systems
  • Concept of parallel circuits and their characteristics
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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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