# Brightness of Light Bulbs in a Circuit

• cvc121

## Homework Statement

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Given the circuit diagram (attached below), predict how unscrewing each bulb separately affects the brightness of the other two bulbs?

N/A

## The Attempt at a Solution

I predict that by unscrewing one of the bulbs, the other two will get brighter. This is because there is less resistance in the circuit overall, and more current which is distributed to the other two bulbs, making them brighter. Am I on the right track here? I am only at a basic high school level so this is the best explanation I could think of given my knowledge. Any help is very much appreciated!

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It depends on which bulb you unscrew! Go ahead and try to figure out what would happen if you were to unscrew each of the three bulbs.

I predict that by unscrewing one of the bulbs, the other two will get brighter. This is because there is less resistance in the circuit overall
What does the circuit look like if you unscrew, say, the bulb on the horizontal wire, B2?

Would B1 and B3 be in parallel with one another?

Would B1 and B3 be in parallel with one another?
Have you heard of the expression "Christmas tree lights failure"?

No, I have not.
Let's just say that I were to unscrew B1. Wouldn't B2 and B3 both get brighter because there is less resistance in the circuit overall? However, I am not sure if that's true because of the fact that B1 is in parallel with B2 and B3 and B2 is in series with B3.

Let's just say that I were to unscrew B1. Wouldn't B2 and B3 both get brighter because there is less resistance in the circuit overall? However, I am not sure if that's true because of the fact that B1 is in parallel with B2 and B3 and B2 is in series with B3.
Consider the voltage across B1. How is that affected by removing B1?
Have you tried drawing the circuit with B1 removed?

Decrease

Decrease
Why?

Pardon me. I believe that voltage would not change because isn't the only thing that would change the voltage be to replace the battery with a more or less powerful one. I am becoming very confused now. I have just been exposed to this topic and it has taken me a while to understand it. Sorry if I sound really dumb.

voltage would not change
Right, but I still suspect you have not understood what the circuit will look like. What do you see where the bulb used to be before removing it?

The Battery Voltage should change very little when connecting and removing bulbs.

Mentor note: the remainder of this post has been removed because members are not permitted to provide near-complete answers to homework. [emoji998]

Thanks so much Jon B. I would just like some clarification as to why removing B1 has a negligible effect on B2 and B3 but yet removing either B2 or B3 has a negligible effect on B1 and the other bulb is no longer lit. Does it have anything to do with how the bulbs are arranged in the circuit ? (B1 is in parallel with B2 and B3, and B2 is in series with B3?)

Thanks so much Jon B. I would just like some clarification as to why removing B1 has a negligible effect on B2 and B3 but yet removing either B2 or B3 has a negligible effect on B1 and the other bulb is no longer lit. Does it have anything to do with how the bulbs are arranged in the circuit ? (B1 is in parallel with B2 and B3, and B2 is in series with B3?)
Try to answer my question in post #11

Thanks so much Jon B. I would just like some clarification as to why removing B1 has a negligible effect on B2 and B3 but yet removing either B2 or B3 has a negligible effect on B1 and the other bulb is no longer lit. Does it have anything to do with how the bulbs are arranged in the circuit ? (B1 is in parallel with B2 and B3, and B2 is in series with B3?)
It has everything to do with the Battery not being effected by changes in load. Its Voltage will remain relatively constant.