A battery of emf 2V and internal resistance `0.5(Omega)`is connected across a resistance of `9.5(Omega)`,how many electrons cross through a cross section of the resistance in 1 second?

Explanation

The current in the circuit is
`i=(2V)/(9.5(Omega)+0.5(Omega))=0.2A`.
Thus, a net transfer of 0.2C per second takes place across any cross section in the circuit. The number of elecrrons crossing the section in 1 second is, therefore,
`(2.0C)/(1.6xx10^(-19)C)=0.125xx10^(19)=1.25xx10^(18).`