Two incadescent bulbs are connected in series to a 3V battery. Bulb B lights up,

Question

Two incadescent bulbs are connected in series to a 3V battery. Bulb B lights up, while bulb A does not. Assume that both bulbs are equally bright when they output the same power. What can you conclude? A. Electrons flow through bulb B but not through A B.Bulb A has a higher resistance than bulb B, but it is not infinite. C. Bulb A has a broken filament ( Acts like an open circuit). D. If you switched the placse of two bulbs in the circuit, bulb A would now light up while bulb B would not. E. Bulb A has a lower resistance than bulb B. Why is the answer E? if they are equally bright and output the same power. and since they are in series, they have the same current. and P = I^2 R. If P and I are identical, wouldn't the resistance of both A and B have to be the same?

Explanation / Answer

E. Bulb A has a lower resistance than bulb B


They will equal current in through them as they are in series.

Power = I2R

so bulb with high resistance will use more power and bulb with low resistance will use less power.
As bulb B is brighter it should have more resistance.


# power is not same in 2 bulbs. It says if power is equal they will be equally bright.

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