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Two light bulbs are in series. If we connect a wire across one of the bulbs in parallel, that bulb will short out because the small resistance associated with this parallel circuit draws a small voltage. Why does the small resistance take a smaller share of the voltage?

Physics
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since they are in paralle they have the same voltage but different cureents kirchoffs laws
so rephrase your question
Connecting a wire across one of the bulbs will short out that bulb because an alternative path for current is created. The alternative path provides less resistance. If voltage is constant, then less resistance will result in greater current. But for a given current, a smaller resistance will result in a smaller voltage. (V = IR)

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Other answers:

Physmath, the "alternative path" frame suggests that the charges aren't flowing through the bulb. I may be wrong, but I think they are. It's just not enough current to power the bulb. The alternative path does provide less resistance, but what I was missing (only truly learned this today) is that in a series circuit, the load with the higher resistance will get a higher share of the voltage. The bulb shorts out because it receives less voltage.
The alternative path indicates that the wire is connected in parallel with the second bulb. If there is a current flowing through the second bulb, then the bulb will light up (P = I2R), but will glow dimmer. There is an equal potential difference across the wire and the second bulb (since they are in parallel); this shows that most (if not all) of the current will flow through the wire (which has less resistance). In a series circuit, current through each load is the same so greater resistance results in greater voltage. This is also why the first light bulb will glow brighter. The wire in parallel has less resistance, which results in less voltage. The second light bulb will not light up if its resistance is significantly greater than the resistance of the wire.

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