A cell of constant emf first connected to a resistance `R_(1)` and then connected to resistance `R_(2)`. If power deliverd in both cases is same, then the internal resistance of 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

Current given by cell
`l= E/(R+r)`
Power delivered in firstcase ltbegt `P_(1)=l^(2)R_(1)=((E )/(R_(1)+r))^(2)R_(1)`
Power delivered in first case
`P_(2)=l^(2)R_(2)=((E )/(R_(2)+r))^(2)R_(2)`
Power delivered is same in the both cases.
`((E )/(R_(1)+r))^(2)R_(1)=((E )/(R_(2)+r))^(2)R_(2)`ltbegt `((E )/(R_(1)+r))^(2)R_(1)=((E )/(R_(2)+r))^(2)R_(2)`
`R_(1)(R_(2)^(2)+r^(2)+2R_(2)r)=R_(2)(R_(1)^(2)+r^(2)+2R_(1)r)`
`R_(1)R_(2)^(2)+R_(1)r^(2)+2R_(1)R_(2)r=R_(2)R_(1)^(2)+R_(2)r^(2)+2R_(1)R_(1)r`
`R_(1)R_(2)^(2)-R_(2)R_(1)^(2)=R_(2)R_(1)^(2)-R_(1)r^(2)`
`R_(1)R_(2)(R_(2)-R_(1))=r^(2)(R^(2)-R_(1))`
`r=sqrtR_(1)R_(2)`