Question

2 resistances when connected in series gives value of 2ohm and when connected in parallel becomes 9 on. calculate value is each resistance​

Answer 1

Correct Question :-  

 Two resistance when connected in series gives value of 9 ohm and when

 connected in parallel gives value of 2 ohm .

GIVEN :-

• The effective  resistance when two resistances are                          

        connected in parallel = $\sf 2 \ \Omega$

• The effective resistance when two resistances are

        connected in series = $\sf 9 \ \Omega$

TO FIND :-

• Value of each resistance

SOLUTION :-

  Let  $\sf R_1$ and  $\sf R_2$  be the two resistance .

   Effective resistance when resistors are connected in parallel ,

            $\bf \dfrac{1}{R} = \dfrac{1}{R_1}+\dfrac{1}{R_2}$

         $\hookrightarrow \sf \dfrac{1}{2} = \dfrac{1}{R_1}+\dfrac{1}{R_2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \.........................(1)$

   Effective resistance when resistors are connected in series ,

          $\bf R= R_1+R_2$

         $\hookrightarrow \sf R_1+R_2 = 9\\\\\hookrightarrow R_2 = 9 - R_1$

    Substitute the value of  $\sf R_2$ in Eq(1) ,

     

      $\hookrightarrow \sf \dfrac{1}{2} = \dfrac{1}{R_1} + \dfrac{1}{(9-R_1)} \\\\\hookrightarrow \dfrac{1}{2}=\dfrac{9-R_1+R_1}{R_1(9-R_1)}}\\\\\hookrightarrow \dfrac{1}{2} = \dfrac{9}{9R_1-R_1^2}\\\\\hookrightarrow 9R_1-R_1^2 = 18 \\\\\hookrightarrow R_1^2 -9R_1+18 = 0\\\\\hookrightarrow R_1^2-6R_1-3R_1+18 =0 \\\\\hookrightarrow R_1(R_1-6)-3(R_1-6) = 0 \\\\\hookrightarrow (R_1-6)(R_1-3) = 0 \\\\\hookrightarrow \bf R_1 = 6 \ , \ R_2 = 3$

 When  $\sf R_1 = 6 \ \Omega$ ,  $\sf R_2 = 9-6 = 3 \ \Omega$

 When  $\sf R_1 = 3 \ \Omega \ , \ R_2 = 9-3 = 6 \ \Omega$