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Question

A potential difference of 3.2 V is applied across a conductor of resistance 1kΩ. Find the number of electrons flowing through the conductor in 5 minutes.

A
6×1017
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B
6×1018
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C
5×1018
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D
5×1017
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Solution

The correct option is B 6×1018
Current=i=VR=3.21000=3.2×103A

t=5min=5×60=300s

q=It=3.2×103×300=0.96C

q=Ne where N=number of electrons

N=qe=0.961.6×1019

N=6×1018

Answer-(B)

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