Analyzing the Effect of a Broken Lamp on Current Distribution in a Circuit

  • Thread starter Thread starter ravsterphysics
  • Start date Start date
  • Tags Tags
    Circuit Current
Click For Summary

Homework Help Overview

The discussion revolves around analyzing the effect of a broken lamp on current distribution in a circuit, specifically focusing on how the brightness of the remaining lamps is affected. The subject area includes circuit analysis and electrical power concepts.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the impact of a broken lamp on current distribution, questioning how the brightness of other lamps changes. Some participants attempt to clarify their reasoning regarding the current through different lamps and the resultant resistance in the circuit.

Discussion Status

The discussion is active, with participants sharing their thoughts and corrections. Some have expressed confusion about their answers and are seeking clarification on the effects of the broken lamp, while others have provided insights into the changes in current and resistance.

Contextual Notes

There is mention of specific homework equations and parameters such as the EMF of the battery and the resistance of the lamps, indicating that certain assumptions about the circuit configuration are being examined.

ravsterphysics
Messages
57
Reaction score
1
elec.JPG

Homework Equations


Power = VI

The Attempt at a Solution


[/B]
Since lamp M breaks, lamp L gets more of the current so its brightness increases and lamp N's brightness stays the same because the current at point 2 is the same if the other lamp breaks or not, so N's brightness should not be affected.

So my answer is A but the correct answer is D?? What have I done wrong?
 
Physics news on Phys.org
ravsterphysics said:
View attachment 110973

Homework Equations


Power = VI

The Attempt at a Solution


[/B]
Since lamp M breaks, lamp L gets more of the current so its brightness increases and lamp N's brightness stays the same because the current at point 2 is the same if the other lamp breaks or not, so N's brightness should not be affected.

So my answer is A but the correct answer is D?? What have I done wrong?
The current through N will also change, as the resultant resistance changes when lamp M breaks.
 
ehild said:
The current through N will also change, as the resultant resistance changes when lamp M breaks.

I get it now.

since the parallel combo is no more, resistance has increased across the circuit so less current for both, so N decreases but L increases since it has more current in resistance than in parallel. Thanks.
 
It is correct, but can you derive what are the powers on the lamps in both cases, to prove your statement? The EMF of the battery is E and each lamp has resistance r.
 

Similar threads

  • · Replies 18 ·
Replies
18
Views
2K
Replies
14
Views
880
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 18 ·
Replies
18
Views
2K