Question

Two resistances when connected in parallel give resultant value of 2 ohm when connected in series give 9 ohm.Calculate the value of each resistance​

Answer 1

$\huge\underline{\underline{\bf\bigstar ANSWER \bigstar}}$

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

➤ When connencted in series ,

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

➤ When connected in parallel ,

       $\hookrightarrow\sf \dfrac{1}{R_1} + \dfrac{1}{R_2} = \dfrac{1}{2} \\\\\hookrightarrow \dfrac{R_1R_2}{R_1+R_2} = 2 \\\\\hookrightarrow \dfrac{R_1(9-R_2)}{9} = 2\\\\\hookrightarrow 9R_1-(R_1)^2 = 18 \\\\\hookrightarrow (R_1)^2 - 9R_1 + 18 = 0 \\\\\hookrightarrow (R_1)^2-3R_1-6R_1+18 = 0 \\\\\hookrightarrow R_1(R_1-3)-6(R_1-3)=0\\\\\hookrightarrow (R_1-3)(R_1-6) = 0$

 From this we get ,

       $\sf R_1 = 3 \ ohm \ , 6 \ ohm$

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$\sf If \ R_1= 3 \ ohm , \ R_2 = 9-3 = 6 \ ohm\\\\If \ R_2 = 6 \ ohm \ , \ R_2 = 9-6 = 3 \ ohm$

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