Parallel and Series, Lightbulb Problem

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SUMMARY

The discussion centers on the comparison of brightness between light bulbs connected in series and parallel configurations. In Circuit 1 (series), the total resistance is 4R, while in Circuit 2 (parallel), the resistance is 1/4R. The conclusion is that the individual bulbs in the parallel circuit are not 4 times brighter, as initially thought, but rather the same brightness due to the relationship between power, current, and voltage drop. The key takeaway is to focus on power dissipation rather than just voltage when analyzing such circuits.

PREREQUISITES
  • Understanding of Ohm's Law
  • Knowledge of electrical power equations
  • Familiarity with series and parallel circuit configurations
  • Basic concepts of resistance in electrical circuits
NEXT STEPS
  • Study the relationship between power, current, and voltage in resistive circuits
  • Learn about the implications of resistance in series vs. parallel circuits
  • Explore the concept of equivalent resistance in parallel circuits
  • Investigate how brightness relates to power output in light bulbs
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Homework Statement



4 identical light bulbs are connected either in series (circuit 1) or parallel (circuit 2) to a constant voltage battery with negligible internal resistance as shown.

http://img413.imageshack.us/img413/7250/seriesoz2.th.jpg

http://img413.imageshack.us/img413/327/parallelfe8.th.jpg

(Sorry about the sloppy diagram but that's pretty much the gist of it)

Compared to the individual bulbs in circuit 1, the individual bulbs in circuit 2 are:

A) 1/4 as bright
B) less than 1/4 as bright
C) 4 times brighter
D) the same brightness
E) more than 4 times brighter


The Attempt at a Solution



The resistance in Circuit 1 would be 4R while the resistance in Circuit 2 would 1/4 R. Because the parallel has a lower resistance, it has a greater current therefore there is more power in the parallel configuration. I thought the right answer was C) 4 times brighter but it is wrong and I can't understand why that is. What am I missing here?
 
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Brightness, in that problem, probably corresponds to the power output of the lightbulbs. Remember that power = current * voltage drop
 
Think in terms of power, not in terms of voltage. What is the equation for the power dissipated by a resistive component, in terms of the voltage across it? Hint -- it's not a linear relationship.
 
Rats. EFuzzy was too quick for me!

BTW, remember EFuzzy that on homework help, we should not give out complete answers. We just give out hints and ask the original poster (OP) to do the final work.

Thanks for chiming in, though. More help is always appreciated.
 
I think I got it now. Thank you!
 

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