Question

When two resistance are joined in series,the equivalent resistance is 90ohm.when the same resistor are joined in parallel,the equivalent resistance is 20ohm.calculate the resistance of the two resistors.

Answer 1

Let the two resistances be $\tt R_1$ and $\tt R_2$ Ω respectively.

When connected in series , equivalent resistance is 90 Ω.  

$\therefore \rm R_s = R_1 + R_2$

$\rm 90 = R_1 + R_2$

$\rm R_2 = 90 - R_1$ ____(1)

When connected in parallel , the equivalent resistance is 20 Ω .

$\therefore \rm \dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$

$\implies \rm \dfrac{1}{20} = \dfrac{1}{R_1} + \dfrac{1}{90 - R_1}$

$\implies \rm \dfrac{1}{20} = \dfrac{90 - R_1 + R_1}{ R_1 ( 90 - R_1)}$

$\implies \rm \dfrac{1}{20} = \dfrac{90}{90\ R_1 - {R_1}^2}$

$\implies \rm 90\ R_1 - {R_1}^2 = 20 \times 90$

$\implies \rm 0 = {R_1}^2 - 90\ R_1 + 1800$

$\implies \rm 0 = {R_1}^2 - (60 + 30)R_1 + 1800$

$\implies \rm 0 = {R_1}^2 - 60\ R_1 - 30\ R_1 + 1800$

$\implies \rm 0 = R_1 ( R_1 - 60) - 30 ( R_1 - 60)$

$\implies \rm 0 = ( R_1 - 30) (R_1 - 60)$

$\implies \rm R_1 = 30 \Omega , 60 \Omega$

$\rm R_2 = 90 - 30 = 60 \Omega \\ \\ \rm Or \\ \\ \rm R_2 = 90 - 60 = 30 \Omega$

Therefore the two resistances are 30 Ω and 60 Ω .