In a wire, 8.45x10^20 electrons flow past any point during 3.89s. What is the magnitude of the current
in the wire?
- Anonymous4 months agoFavorite Answer
charge Q = 8.45*10^20 el / (6.24*10^18) el/Co = 135 Co (A*sec)
current I = Q/t = 135 A*sec / 3.89 sec = 34.8 A
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