1.

Show that one ampere is equivalent to a flow of 6.25xx10^(8) elementary charges per second.

Answer»

SOLUTION :`I=1A,t=1S,e=1.6xx10^(-19)C`
`"q=NE"`
`I=q/t=("ne")/(t)`
Number of electrons, `n=(If)/(e)=(1xx1)/(1.6xx10^(-19))=6.25xx10^(18)`


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