Can the Brightness of Bulbs in a Parallel Circuit be Equal?

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SUMMARY

The discussion centers on the behavior of bulbs in a parallel circuit, specifically addressing the question of equal brightness among bulbs when connected. Participants clarify that in a parallel configuration, each bulb receives the same voltage, resulting in equal current through each bulb if they have identical resistance. The misconception regarding equivalent resistance is corrected, emphasizing that the current through each bulb remains consistent regardless of the number of bulbs added, as long as the voltage remains constant. This leads to the conclusion that while bulbs can glow equally bright, the idea of generating infinite light with finite voltage is fundamentally flawed due to practical limitations in real-world circuits.

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ehrenfest
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Can someone help me with 1 (e) at the following site:

http://ocw.mit.edu/NR/rdonlyres/Physics/8-02Electricity-and-MagnetismSpring2002/29B6C858-A92B-431F-A43E-725D58A768EF/0/exam1.pdf

The solutions say that all on-bulbs in 1 (e) glow with equal brightness. But how is that possible because in e-c, the equivalence resistance is

(2/R) so the total current is V * R/2 and the current through each branch is V * R. In e-b, the current is V/R through B, which is different, right?
 
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ehrenfest said:
Can someone help me with 1 (e) at the following site:

http://ocw.mit.edu/NR/rdonlyres/Physics/8-02Electricity-and-MagnetismSpring2002/29B6C858-A92B-431F-A43E-725D58A768EF/0/exam1.pdf

The solutions say that all on-bulbs in 1 (e) glow with equal brightness. But how is that possible because in e-c, the equivalence resistance is

(2/R) so the total current is V * R/2 and the current through each branch is V * R. In e-b, the current is V/R through B, which is different, right?

2/R as the equivalent resistance is wrong. It should be R/2. You don't need to find the equivalent resistance though. The current through a resistor is just the voltage across the resistor divided by resistance. For e-c, the voltage across bulb B is V... the resistance of bulb B is R, so the current through bulb B is V/R... exactly the same way, the current through bulb C is V/R.
 
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I see. But you can just keep adding more and more light bulbs in parallel and they will each glow as bright as a single light bulb? Doesn't that allow you to generate an infinite amount of light (energy) with a finite amount of voltage?

Yes, we are assuming the wires have negligable resistance. But this still seems fishy to me...
 
They are in parallel so it's an infinite amount of light with an infinite amount of current. This isn't a problem as an infinite amount of current at a finitie voltage gives an infinitie amount of energy - of course paying the bill might be tricky!
 

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