A current of 7.5 A is maintained in a wire for
48 So in this time @ how much charge and
b) how many electrons flow through the wire?
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Answer:
a.q=it=(7.5A)(45s)=337.5C
b. The number of electron N is given by
N=(q)(e)=337.5C1.6×10−19C=2.1×21
where e=1.6×10−19C is the charge of an electron
Explanation:
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