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 and cII
This is the end of the preview.
access the rest of the document.