2_Ch 17 College Physics ProblemCH17 Current and Resistance

2_Ch 17 College Physics ProblemCH17 Current and Resistance...

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70 CHAPTER 17 7. (b). abcde f IIIIII =>=>= . Charges constituting the current leave the positive terminal of the battery and then split to flow through the two bulbs; thus, a I ac III e = + . Because the potential difference is the same across the two bulbs and because the power delivered to a device is V ( ) IV =∆ P d I e , 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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