Determining the Change in Bulb Brightness by Changing Circuit

[NCanine_1932]

Homework Statement

Suppose bulb E is unscrewed and removed from its socket. (The empty socket remains in the circuit.) Does bulb A get brighter, dimmer, or stay the same brightness? NOTE: A is the left bulb in the first pair of parallel bulbs

Relevant Equations

V = I*R
R_eq (series) = R_1 + R_2 + ...
R_eq (parallel) = [(1/R_1) + (1/R_2) + ....]^-1

I understand that removing bulb E will cause the equivalent resistor to change in the circuit. This change will cause the current across the circuit to change (which will either brighten or dim the bulbs).

I found that R_eq (with E) is 1.167.
R_eq = [1 + (1/1+1)]^-1 + [1 + 1]^-1 = 1.167

I found that R_eq (removing E) is 1.
R_eq = [1 + 1]^-1 + [1+1]^-1 = 1

So, if R_eq is decreasing from when it had E to when it did not have E, then I should increase since V should remain constant for the entire circuit.

Now that I know that current is larger throughout the circuit, I can assume that the current that splits at the first node to go through bulb A should be larger as well. A larger current should mean a larger voltage going through bulb A which means it is brighter.

However, the assignment is saying that bulb A should be dimmer. What am I getting wrong? This thought process worked perfectly fine for a previous question. Thank you!

 

[erobz]

Homework Statement: Suppose bulb E is unscrewed and removed from its socket. (The empty socket remains in the circuit.) Does bulb A get brighter, dimmer, or stay the same brightness? NOTE: A is the left bulb in the first pair of parallel bulbs
Relevant Equations: V = I*R
R_eq (series) = R_1 + R_2 + ...
R_eq (parallel) = [(1/R_1) + (1/R_2) + ....]^-1

I understand that removing bulb E will cause the equivalent resistor to change in the circuit. This change will cause the current across the circuit to change (which will either brighten or dim the bulbs).

I found that R_eq (with E) is 1.167.
R_eq = [1 + (1/1+1)]^-1 + [1 + 1]^-1 = 1.167

I found that R_eq (removing E) is 1.
R_eq = [1 + 1]^-1 + [1+1]^-1 = 1

So, if R_eq is decreasing from when it had E to when it did not have E, then I should increase since V should remain constant for the entire circuit.

Now that I know that current is larger throughout the circuit, I can assume that the current that splits at the first node to go through bulb A should be larger as well. A larger current should mean a larger voltage going through bulb A which means it is brighter.

However, the assignment is saying that bulb A should be dimmer. What am I getting wrong? This thought process worked perfectly fine for a previous question. Thank you!

View attachment 352906

You are thinking that it removes resistance, but think about what is happening in that branch with regards to short and open circuits when the bulb is removed.

 

[NCanine_1932]

You are thinking that it removes resistance, but think about what is happening in that branch with regards to short and open circuits when the bulb is removed.

When the bulb is removed, then the current should split evenly between the left bulb (which is bulb C) and the right bulb (which is bulb D). I understand that this would brighten bulb D and dim bulb C since they are now receiving 0.5I from the current. But, that should have no effect on the above branch. I thought the above branch would only be affected by the equivalent resistance impacting the circuit's total current. Am I correct?

Unless, I am interpreting this wrong. Does removing the bulb cause that part of the circuit to become open (similar to what a switch would do). Because I thought the statement "The empty socket remains in the circuit" tells us that it remains a closed circuit.

 

[erobz]

Unless, I am interpreting this wrong. Does removing the bulb cause that part of the circuit to become open (similar to what a switch would do). Because I thought the statement "The empty socket remains in the circuit" tells us that it remains a closed circuit.

The path goes from what to what? Think about this quite literally...first there is a metal filament (a relatively very good conductor) across the socket, then there is what medium across the socket when that metal filament is removed? Think about older style Christmas light struggles...they are wired such that if one burns out, what happens to the rest of the bulbs.

 

[NCanine_1932]

Okay, I think what you're trying to get at is that since it is an empty socket there is no wire for the current to flow through that part of the circuit. Meaning it becomes open so the current doesn't flow through that way. So it ends up just flowing to the bulb on its own on the left meaning the equivalent resistance is increased (R_eq = 1.5). Then that would make sense as to why bulb A would dim since increased R means decreased I.

I just think this questions could have explicitly said that the empty socket acts as an opened switch. The question before was basically the exact same question then but it just used a switch at that point instead.

 

[erobz]

I just think this questions could have explicitly said that the empty socket acts as an opened switch.

Meh...I know you feel tricked. Playing the devils advocate on this one, they are trying to get you to think. You fell prey to the "just calculate it" alternative answer. It happens! Believe it or not, in the real world its very seldom clear (a main goal becomes just to figure out the problem statement), and you have to either know something about it or figure it out. I think that becomes good practical experience/knowledge when working with electricity for instance.

Heck, (like it or not) you might remember it for a while now. You didn't lose anything by not answering on the first go, you only gained.

The course in swallowing your pride comes after your degree (and it never ends)!

 

[phinds]

I just think this questions could have explicitly said that the empty socket acts as an opened switch. The question before was basically the exact same question then but it just used a switch at that point instead.

I disagree. ALL light bulbs result in an open circuit when removed. I mean, think about it ... if it was a short when the bulb is removed then it would be a short with the bulb in and the bulb would never light.