Solution :
Given two resistors : [tex]$R_1$[/tex] and [tex]$R_2$[/tex] where [tex]$R_2$[/tex] > [tex]$R_1$[/tex]
When the resistors are connected in series :
Current, [tex]$I_s=\frac{\epsilon}{R_1 + R_2}$[/tex] .............(1)
When the resistors are connected in series :
[tex]$10I_s=\frac{\epsilon}{\left(\frac{R_1R_2}{R_1+R_2}\right)}$[/tex]
[tex]$10I_s=\frac{\epsilon(R_1+R_2)}{R_1R_2}$[/tex] ..................(2)
Therefore, dividing equation (2) by (1), we get
[tex]$10=\frac{(R_1+R_2)^2}{R_1R_2}$[/tex]
Now, since [tex]$r=\frac{R_1}{R_2}$[/tex] , we have [tex]$R_1=rR_2$[/tex]
∴ [tex]$10=\frac{(1+r)^2}{r}$[/tex]
[tex]$\Rightarrow 1+r^2+2r=10r$[/tex]
[tex]$\Rightarrow r^2-8r+1=0$[/tex]
Solving, we get, r = 0.127
