Question

Two resistors of resistances R1 and R2, with R2>R1, are connected to a voltage source with voltage V0. When the resistors are connected in series, the current is Is. When the resistors are connected in parallel, the current Ip from the source is equal to 10Is.
Let r be the ratio R1/R2. Find r. Round your answer to the nearest thousandth.

Answer 1

Answer

The Value of  r  = 0.127

Explanation:

The mathematical representation of the two resistors connected in series is

                               $R_T = R_1 +R_2$

 And from Ohm law

                           $I_s =\frac{ V}{R_T}$

                            $I_s = \frac{V_0}{R_1 +R_2} ---(1)$

The mathematical representation of the two resistors connected in parallel  is

                    $R_T = \frac{1}{R_1} +\frac{1}{R_2}$

                          $= \frac{R_1 R_2}{R_1 +R_2}$

From the question $I_p =10I_s$

          =>                 $I_p =10I_s = \frac{V_0 }{\frac{R_1R_2}{R_1 +R_2} } = \frac{V_0 (R_1 +R_2)}{R_1 R_2}---(2)$

     Dividing equation 2 with equation 1

       =>                 $\frac{10I_s}{I_s} =\frac{\frac{V_0 (R_1 +R_2)}{R_1 R_2}}{\frac{V_0}{R_1 +R_2}}$

                                  $10 = \frac{(R_1+R_2)^2}{R_1 R_2}----(3)$

We are told that    $r = \frac{R_1}{R_2} \ \ \ \ \ = > R_1 = rR_2$

From equation 3  

                            $10 = \frac{(1-r)^2}{r}$

$=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1+r^2 + 2r = 10r$

$=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r^2 -8r+1 = 0$

Using the quadratic formula

                             $r =\frac{-b\pm \sqrt{(b^2 - 4ac)} }{2a}$

        a = 1  b = -8 c =1  

                              $= \frac{8 \pm\sqrt{((-8)^2- (4*1*1))} }{2*1}$

                               $r= \frac{8+ \sqrt{60} }{2} \ or \ r = \frac{8 - \sqrt{60} }{2}$

                              $r = \ 7.87\ or \ r \ = \ 0.127$

Now  r =  0.127 because it is the least value among the obtained values