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Conceptually the power doubles. Look, you just add another source of light, so you need to consume double energy compared to the case of a single bulb, so the power is obviously doubles.
Quantitatively the current through each bulb is I=12/R, where 12 is the voltage, since you have to bulbs in parallel the current you take from the battery doubles I(tot) = 2*12/R
Since the power is given P=IV, the increase of the current by a factor of 2 means that the power increases as well by the same amount. by factor of 2.
From another side you can say that in parallel the net resistance is halved, hence P=V^2/R and once R is halved the power obviously doubles.
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