Problem of voltage & light bulbs

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Connecting a 9V source to a light bulb rated for 6V can lead to excess current flowing through the bulb, causing it to heat up and potentially burn out. The light bulb is designed with a specific resistance to allow the correct current at its rated voltage. When subjected to a higher voltage, the bulb will glow brighter but is at risk of failure due to overheating. The extra voltage does not flow back to the battery; instead, it increases the current through the bulb. Ultimately, using a higher voltage than specified can damage the bulb quickly.
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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
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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