Question

Find the resistance of R1 and R2.​

Answer 1

Question :-

In resistor R$_1$ and R $_2$where connected in series will give 16Ω and when connected in parallel gives 3Ω. Find the resistance of R$_1$ and R$_2$

⟶ Given :-

R$_1$ + R$_2$ = 16Ω

$\sf\dfrac{1}{R_1} + \dfrac{1}{ R_2} = \dfrac{ R_2 \times R_1}{R_1+ R_2} =3 \Omega$

⟶ To find:-

The resistance of R$_1$ and R$_2$

⟶ Solution :-

R$_1$ + R$_2$ = 16Ω

$\sf R_2 = 16 - R_2 \: \: \: \: \: \: \: \: .....(1)$

$\sf\dfrac{ R_2 \times R_1}{R_1+ R_2} =3 \Omega$

$\sf R_2 \times R_1 = 3(R_1+ R_2) \: \: \: \: .....(2)$

Put eq (1) in eq (2)

$\sf16 - R_1 \times R_1 = 3(R_1+ 16 - R_1)$

$\sf R_1(16 - R_1) = 3( \: \cancel {R_1}+ 16 \: \cancel{ - R_1})$

$\sf16 R_1 - R_1^{2} = 48$

$\sf R_1 ^{2} - 16 R_1 + 48 = 0$

$\sf R_1 ^{2} - 12 R_1 - 4R_1 + 48 = 0$

$\sf R_1(R_1 - 12) - 4(R_1 - 12) = 0$

$\sf(R_1 - 4)(R_1 - 12) = 0$

Hence, $\sf R_1$= 4 or $\sf R_1$= 12

Thus if $\sf R_1$ = 4 then $\sf R_2$= 12

lf $\sf R_1$ = 12 then $\sf R_2$= 4