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When a battery is connected to the resistance of 10Omega the current in the circuit is 0.12 A. The same battery gives 0.07A current with20Omega Calculate e.m.f. and internal resistance of the battery.

Answer»

Solution :We KNOW that `E=Ir+IR`
`I_(1)R+I_(1)R_(1)=I_(2)r+I_(2)R_(2), r=(I_(2)R_(2)-I_(1)R_(1))/(I_(1)-I_(2))`
`r=(0.07xx20-0.12xx10)/(0.12-0.07)=(1.4-1.2)/(0.05)=(0.2)/(0.05)=4Omega`
`"INTERNAL resistance r"=4Omega,"e.m.fE"=Ir+IR`
`0.12xx4+0.12xx10=0.48+1.2, E="1.68 volt."`


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