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

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The discussion centers on the concept of bulb brightness in a parallel circuit, specifically addressing a problem from an MIT physics exam. Participants clarify that all bulbs can glow with equal brightness because the voltage across each bulb is the same, despite differing currents. The confusion arises from the calculation of equivalent resistance, which some participants argue is incorrectly stated as 2/R instead of R/2. The conversation also touches on the theoretical implications of adding more bulbs in parallel, leading to concerns about generating infinite light and current from a finite voltage. Ultimately, the consensus acknowledges that while the theory holds under ideal conditions, practical limitations exist.
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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