Question

I want the answer of this question step by step. A humble request not to give direct answers.​

Answer 1

$\large\underline{\underline{\rm{\red{Given:-}}}}$

• $\rm Net \ Resistance \ in \ Series \ combination = 80 \Omega$
• $\rm Net \ Resistance \ in \ parallel \ combination = 20 \Omega$

$\large\underline{\underline{\rm{\red{Solution:-}}}}$

Let the resistances are $\rm R_1, \ R_2$.

We need to find $\rm R_1, \ R_2$

Consider the series combination, use:

$\Large{\star} \red{\boxed{\green{R_s=R_1+ R_2}}}$

So , we get

$\rm :\longmapsto R_1 + R_2 = 80 \Omega$

$\rm :\longmapsto R_2 = 80 - R_1$

Now , consider the parallel combination, use;

$\Large{\star} \red{\boxed{\green{R_p = \left(\dfrac{1}{R_1} + \dfrac{1}{R_2}\right)^{-1}}}}$

Hence, you get:

$\rm :\longmapsto \dfrac{1}{R_1} + \dfrac{1}{R_2} = \dfrac{1}{20}\Omega$

$\rm :\longmapsto \dfrac{R_1+R_2}{R_1R_2} = \dfrac{1}{20}$

$\rm :\longmapsto \dfrac{R_1R_2}{R_1+R_2} = 20$

$\rm :\longmapsto \dfrac{R_1R_2}{80} = 20$

$\rm :\longmapsto R_1 R_2 = 1600$

Put the value of $\rm R_2$ to solve the equation in 1 variable

$\rm :\longmapsto R_1(80-R_1) = 1600$

$\rm :\longmapsto 80R_1 - R_1^2 = 1600$

$\rm :\longmapsto R_1^2 - 80R_1 + 1600 = 0$

$\rm :\longmapsto (R_1 - 40)^2 = 0$

${\boxed{\boxed{\rm :\longmapsto R_1 = 40 \Omega}}}$

$\boxed{\boxed{\rm :\longmapsto R_2 = 40 \Omega}}$