# What will happen if the middle bulb burns out?

1. Aug 14, 2014

### needingtoknow

1. The problem statement, all variables and given/known data

What will happen if the middle bulb in the picture attached burns out?

I thought the light intensity of the other two bulbs would increase since there is now more current passing through the two bulbs that are not burned out. Originally there is a current of I = V/R going through each bulb so the battery is providing a current of 3I. After the middle bulb burns out isn't the battery still providing a current of 3I and since the bulbs are identical shouldn't the remaining two bulbs have a current of 1.5I passing through each one of them.

As a result of the greater current they are each experiencing, doesn't that mean they also experience greater power P = I^2R and therefore become brighter?

My solutions manual says that after the middle bulb burns out, the total current will decrease so that the current going through each bulb will stay the same, but I don't understand how that works and why my explanation doesn't.

#### Attached Files:

• ###### Brightness of Bulbs.PNG
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2. Aug 14, 2014

### BvU

A battery doesn't "deliver" a current, but a voltage. Ideally a constant voltage, independent of current.
So, in the ideal case, the other two bulbs keep getting the same voltage, draw the same current and remain as bright as they were. (comparable case: if your neigbour turns off his lamp, your lamps won't notice)

In the not-so-ideal case, you can imagine an internal resistance (see other posts...) in the battery and the voltage will go up a little bit if the total current drops. So the other two go a little brighter.