Problem of voltage & light bulbs

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SUMMARY

When a 9V voltage is applied to a light bulb designed for 6V, the excess 3V does not flow back to the battery; instead, it causes an increase in current through the bulb. The bulb's resistance is calibrated for 6V, and applying a higher voltage results in excessive current that heats the filament, leading to a brighter glow and a significant risk of burnout. The circuit will not break due to the extra voltage; however, the bulb will likely fail quickly due to overheating.

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if the applied voltage is 9V, and it is connected in series with a light bulb with designed voltage 6V only, what will happen to the extra 3V??
will the 3V be used up by the light bulb also or by the wire (heat up)? or the 3V just flow back to the battery?

and will the circuit break because of the extra V?
 
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What do you mean when you say that a light bulb is designed for 6V ?
 
There isn't an extra 3V - all 9v goes the light bulb.
The bulb is designed with a certain resistance so that with 6V the correct amount of current will flow to heat up the filament and make it glow.
If you put more voltage across the bulb then more current will flow and the filament will get hotter and the bulb will glow much brighter - until it burns out.
Depending on the bulb - it might burn out very quickly
 

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