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
When lamp M breaks in the circuit, lamp L receives more current, increasing its brightness, while lamp N's brightness initially appears unaffected. However, the overall resistance of the circuit changes due to the break, leading to a decrease in current for lamp N. The correct understanding reveals that both lamps' brightness is affected, with N decreasing and L increasing. The discussion emphasizes the need to analyze the power distribution and resistance changes in the circuit to fully understand the effects of the broken lamp. The conversation concludes with a request to derive the powers of the lamps to validate the findings.
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
14
Views
673
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · 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
2K
  • · Replies 18 ·
Replies
18
Views
2K