A cell of constant e.m.f. first connected to a resistance `R_(1)` and then connected to a resistance `R_(2)`. If power delivered in both cases is then the internal resistance o f the cell is

Options

A. `sqrt(R_(1)R_(2))`
B. `sqrt(R_(1)//R_(2))`
C. `(R_(1)-R_(2))/(2)`
D. `(R_(1)+R_(2))/(2)`

Answer

A

Explanation

Power dissipated `=I^(2)R`
as we know `I=(E )/(R+r)`
`I=((E )/(R+r))^(2)R`
`:. ((E )/(R_(1)+r))^(2) R_(1)=((E )/(R_(2)+r))^(2)R_(2)`
`rArr R_(2)(R_(2)^(2)+r^(2)+2R_(2)r) =R_(2)(R_(1)^(2)+r^(2)+2R_(1)r)`
`rArr R_(2)^(2)R_(1)+R_(1)r^(2)+2R_(1)R_(2)r=R_(1)^(2)R_(2) +R_(2)r^(2)+2R_(1)R_(2)r`
`rArr (R_(1)-R_(2)) r^(2)=(R_(1)-R_(2))R_(1)R_(2)`
`I=sqrt(R_(1)R_(2))`