PHYSICS questiona 60 watt bulb is connected to a 12 v car batter. When another 60 watt bulb is connected in parallel with the first bulb, the battery's output energy -- a: halves; b:remains the same; c:doubles. Please explain BOTH conceptually and quantitatively. In particular, is there a relationship between power and resistance implicit in this question? If so, how does it play out.
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.
Please let me know if you have more Physics questions, I can help. Just type "For David" before your post
This answer makes sense, especially the Power/Resistance inverse relationship. Just one follow up: I never thought of a battery as changing it's power to accomodate more bulbs in parallel. I guess I just thought that a battery always supplies, say, 12 Volts regardless of resistors. But, from your answer, I should probably think that since voltage is constant, power is an inverse function of resistance in parallel and a direct function of current. So any time you add more resistors to a parallel circuit powered by a battery, the power goes up since the resistance goes down; and, conversely, anytime you add more resistors to a series circuit powered by a battery, the power goes down since the resistance goes up. Is that correct?ThxTom
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PhD in Theoretical Physics