Question

There is a current of 3.2 A in a conductor. The number of electrons that cross any section normal to the direction of flow per second is:

Answer 1

CURRENT = no. of electrons * charge on 1 e
3.2 = n * 1.6 * 10^-19
n = 3.2 / 1.6*10^-19
= 2 * 10^19

Answer 2

HEY.......
HERE IS U R ANSWER
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CURRENT = n * 1.6 * 10^-19
3.2 = n * 1.6*10^-10
n = 2*10^19


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