Three resistances `R_(1),R_(2)and R_(3)` are coonected in parallel . This combination is then connected to a celll of negligble internal resistance .Applying Kirchhoff's law prove that the equivalent resistance of the whole combination is give by ,
`R=(R_(1)R_(2)R_(3))/(R_(1)R_(2)+R_(2)R_(3)+R_(1)R_(3)`

Explanation

Applying Kirchhoff's second law in the closed loop `AR_(1)BEA` ,
`i_(1)R_(1)-E=0`
[ where emf of the cell is E and internal resistance is zero ]
`therefore i_(1)=(E)/(R_(1))" " ....(1) `

Again applying Kirchhoff's second law in `AR_(2)BEA` ,
`i_(2)R_(2)-E=0`
`therefore i_(2)=(E)/(R_(2))" " ....(2)`
and similarly , `i_(3)=(E)/(R_(3)) " "... (3) ` ltbr gt From equatios (1) ,(2) and (3) we get ,
`i_(1)+i_(2)+i_(3)=E[(1)/(R_(1))+(1)/(R_(2))+(1)/(R_(3))]=i[i="total current"] `
If R be the equivalent reistance then
`R=(E)/(i)=(1)/((1)/R_(1)+(1)/R_(2)+(1)/R_(3))=(R_(1)R_(2)R_(3))/(R_(1)R_(2)+R_(2)R_(3)+R_(1)R_(3))`