Why a 20 Ohm lightbulb won't light up in a closed circuit?

[cbeck]

Homework Statement

1. Why does that happen for circuit A? Use the tools (Voltmeter and Ammeter) to investigate your circuit and explain what happened
2. What is the total resistance the battery sees for circuit A when the switch is open? What is the total resistance the battery sees for circuit A when the switch is closed?

Relevant Equations

V = I*R

When using the voltmeter and ammeter to investigate the circuit, it was found that the batter had a current of 0.67 A, the two 30 Ω bulbs had a current of 0.17 A, and the two 10 Ω bulbs had a current of 0.50 A. In terms of voltage, each bulb had a voltage of 5 V. When the switch was closed, the 20 Ω bulb had a current and voltage of 0.

I was able to make the circuit as instructed, but I just cannot figure out the rationale behind why the bulb won't light up. I thought it had something to do with the fact that the current is not strong enough to divert into the 20 ohm bulb, but I'm just not sure. I know that Kirchhoff's rule in a closed loop is that the change in potential difference is 0, but I don't know if that fits with it. Could someone please walk me through how to figure this out? I'm really hoping to better understand the concepts.

 

[haruspex]

The image is too small. When I blow it up, it is too fuzzy. I cannot see where this switch is.
Anyway, start by removing the 20Ω, leaving that part of the circuit open. This makes it easy to analyse what's left. If you put the bulb back in, what will the voltage across it be immediately?

 

[cbeck]

Does this image work better for you? I'm sorry about that. When I put the bulb back in, the voltage across it is still 0V. I can see that the current doesn't even go through the wires to the 20 ohm bulb. Maybe because the potential is higher here? I know it will flow from high to low potential so perhaps that is an issue?

 

[Vanadium 50]

Maybe because the potential is higher here?

Higher where.

Here's how to solve the problem. (And I bet you won't do it). Draw a picture. A neat and clear picture. Not a screen capture, a picture. Label every point ahead and behind each element (bulb or battery). Now, tell me what the potential (voltage) is ahead and behind the bulb in question.

 

[haruspex]

Does this image work better for you? I'm sorry about that. When I put the bulb back in, the voltage across it is still 0V.

Yes, that's a lot clearer, thanks.
Perhaps I was not clear, though. With that bulb out, so there is no connection across between the middle of the top horizontal and the middle of the middle horizontal, use algebra to solve the circuit. What is the voltage across the gap left by removal of the bulb?

 

[kuruman]

This looks like a balanced Wheatstone bridge to me.

 

[Vanadium 50]

Let's get him to draw a proper labeled picture, and then we can sensibly discuss the potential between A and B. (Or S and Q, or whatever it turns out to be).

 

[cbeck]

I actually was able to figure it out. The potential ahead and behind of the bulb are the same in this case. Which means that there is no potential difference and when there is no potential difference, there can be no current because of change in V = IR. And with no current, it's impossible for there to be light. Thank you for the help!