How to fix this electric circuit

Click For Summary

Homework Help Overview

The discussion revolves around an electric circuit involving two lamps and resistors. The original poster expresses uncertainty about the circuit's construction and whether both lamps will shine equally brightly. They reference Kirchhoff's and Ohm's laws as relevant equations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to modify the circuit, particularly regarding the placement of resistors and the configuration of the lamps. Some suggest closing the circuit at specific points and question the clarity of the problem's instructions.

Discussion Status

Several participants have offered suggestions for modifying the circuit to achieve equal brightness in the lamps. There is an ongoing exploration of possible configurations, with some participants indicating uncertainty about their proposed solutions. The discussion remains active, with no explicit consensus reached.

Contextual Notes

Participants note constraints such as the requirement not to add components to the circuit and the need to work within the given parameters. There is also mention of the importance of writing out current and voltage equations as part of the problem-solving process.

orpheus
Messages
11
Reaction score
0

Homework Statement



I'm not sure if the construction of this circuit is correct, so that both lamps, which are the same, shine equally brightly. Otherwise only the resistors are given. If it's wrong, what else would it look like? And why?[/B]

1537595868640-940372466-jpg.jpg


Homework Equations


I guess the laws of Kirchoff and Ohm

The Attempt at a Solution



check this link:
[/B]
https://share-your-photo.com/6ef822170a
 

Attachments

  • 1537595868640-940372466-jpg.jpg
    1537595868640-940372466-jpg.jpg
    26 KB · Views: 391
Last edited by a moderator:
Physics news on Phys.org
When I clicked on the photo, google has ads in the way, so that I can not see it completely. Can you please try to find some other way to upload the diagram.
 
Yes, sorry, now it should work. :)
 

Attachments

  • 1537595868640-940372466.jpg
    1537595868640-940372466.jpg
    26 KB · Views: 321
By "fixing" the circuit, you will need to add a resistor at the upper left of some value to be determined =perhaps that value is zero. Suggestion is to try short circuiting that part first and see what you get. Presently, the bulb on the upper right will not light at all. You need to put in some of the work here though. The Physics Forums homework helpers can help you to work to the solution, but we are not allowed to provide the answer. You must do the work to get to the answer.
 
We are not allowed to add something to circuit. We should just switch resistors. I guess. I can't solve it.
 
You are at least going to need to close the circuit at the upper left. The instructions don;t seem to be entirely clear on this problem, but again the first thing I would try is simply close the circuit at the upper left. And you need to write out the current and voltage equations. It takes a little work, but when you get tested on the material, they are not going to write the equations for you. That is what homeworks are for=to learn by trial and error. ## \\ ## One way to get them to shine equally bright is to simply put the bulbs next to each other in series in the circuit=they aren't real clear on what you are allowed to do to the circuit. ## \\ ## Edit: My suggestion of shorting the circuit at the upper left will not work. This one is actually kind of simple now that I have looked it over carefully, and it does not require extensive circuit analysis. If you look at the bulb at the lower left, it is in series circuit with the voltage source with resistances totaling ## R=2.3 \, k \Omega ## . How would you modify the circuit to get the same thing for the other bulb? That will give you the solution you need.
 
Last edited:
See my "Edit" above= (post 6). This one is actually quite simple. And note: The ## R=100 \, \Omega ## at the lower right will be shared by both circuit loops, but that is ok. In the circuit that is the correct solution, that resistor will simply have 2x the current through that all of the other resistors have. The solution involves one simple step.
 
Last edited:
So I am actually not far from solution?
 
I guess I have the solution but I am not sure.
I am not more able draw because I went for a time away from home (now in train), so there is no possibility write it down for me.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
11
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 18 ·
Replies
18
Views
2K