A current of 4.8A is flowing in a conductor. The number of electrons passing per second through the conductor will be :
3×1020
76.8×1020
7.68×10−19
3×1019
A
3×1020
B
7.68×10−19
C
76.8×1020
D
3×1019
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Solution
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Given, e=1.6×10−19C, I=4.8A Let number of electrons per second be n. The current is defined as rate of flow of charge, I=qt Hence q=n×e Therefore, I=n×et Substituting values, 4.8 =n×1.6×10−19 Hence, n=4.81.6×10−19=3×1019 electrons per second.
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