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

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    Brightness Circuit
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Homework Help Overview

The discussion revolves around the behavior of light bulbs in a parallel circuit, specifically addressing the conditions under which multiple bulbs can glow with equal brightness. Participants are examining the implications of equivalent resistance and current distribution in the circuit.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants are questioning the validity of the claim that all bulbs glow with equal brightness despite differing current calculations. There is also a discussion about the implications of adding more bulbs in parallel and the resulting current and energy considerations.

Discussion Status

Some participants are exploring the reasoning behind the equal brightness of bulbs, while others are raising concerns about the physical implications of infinite current and energy generation in the circuit. There is no explicit consensus, but the discussion is actively engaging with the underlying principles.

Contextual Notes

Participants are operating under the assumption that the wires have negligible resistance, which is influencing their interpretations of the circuit behavior. There is also a reference to specific problems from an exam, which may impose certain constraints on the discussion.

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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