Answer
The Value of r = 0.127
Explanation:
The mathematical representation of the two resistors connected in series is
[tex]R_T = R_1 +R_2[/tex]
And from Ohm law
[tex]I_s =\frac{ V}{R_T}[/tex]
[tex]I_s = \frac{V_0}{R_1 +R_2} ---(1)[/tex]
The mathematical representation of the two resistors connected in parallel is
[tex]R_T = \frac{1}{R_1} +\frac{1}{R_2}[/tex]
[tex]= \frac{R_1 R_2}{R_1 +R_2}[/tex]
From the question [tex]I_p =10I_s[/tex]
=> [tex]I_p =10I_s = \frac{V_0 }{\frac{R_1R_2}{R_1 +R_2} } = \frac{V_0 (R_1 +R_2)}{R_1 R_2}---(2)[/tex]
Dividing equation 2 with equation 1
=> [tex]\frac{10I_s}{I_s} =\frac{\frac{V_0 (R_1 +R_2)}{R_1 R_2}}{\frac{V_0}{R_1 +R_2}}[/tex]
[tex]10 = \frac{(R_1+R_2)^2}{R_1 R_2}----(3)[/tex]
We are told that [tex]r = \frac{R_1}{R_2} \ \ \ \ \ = > R_1 = rR_2[/tex]
From equation 3
[tex]10 = \frac{(1-r)^2}{r}[/tex]
[tex]=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1+r^2 + 2r = 10r[/tex]
[tex]=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r^2 -8r+1 = 0[/tex]
Using the quadratic formula
[tex]r =\frac{-b\pm \sqrt{(b^2 - 4ac)} }{2a}[/tex]
a = 1 b = -8 c =1
[tex]= \frac{8 \pm\sqrt{((-8)^2- (4*1*1))} }{2*1}[/tex]
[tex]r= \frac{8+ \sqrt{60} }{2} \ or \ r = \frac{8 - \sqrt{60} }{2}[/tex]
[tex]r = \ 7.87\ or \ r \ = \ 0.127[/tex]
Now r = 0.127 because it is the least value among the obtained values
