70CHAPTER 17 7. (b). abcdefIIIIII=>=>=. Charges constituting the current leave the positive terminal of the battery and then split to flow through the two bulbs; thus, aIacIIIe=+. Because the potential difference is the same across the two bulbs and because the power delivered to a device is V∆( )IV=∆PdIe, the 60–W bulb with the higher power rating must carry the greater current, meaning that . Because charge does not accumulate in the bulbs, all the charge flowing into a bulb from the left has to flow out on the right; consequently
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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.