Question

The resistance of two conductors joined in series is 8 ohm and in parallel is 1.5 ohm. Show that the resistance are 6 ohm and 2 ohm respectively.​

Answer 1

Hey Pretty Stranger!

Let Resistance A be $\sf\: R_{A}$ and Resistance B be $\sf\: R_{B}$

It is mentioned that $\sf\: R_{A}\: + R{B}$ = 8 Ω... i)

→ $\sf\: R_{B}$ = (8 - $\sf\: R_{A}$ )... ii)

$\sf \: \dfrac{1}{R_A} \: + \dfrac{1}{R_B} = 1.5$

Simplifying :

$\sf \dfrac {R_B+ R_A}{R_A R_B} = \dfrac{1}{1.5}$

From Equation i)

$\sf \dfrac {8}{R_A R_B} = \dfrac{1}{1.5}$

$\sf {R_A R_B} = 12$

From Equation ii)

$\sf {R_A(8 -R_A) } = 12$

$\sf {8R_A -(R_A) ^{2} } = 12$

$\sf(R_A) ^{2} = 6 \: or \: 2$

$\therefore$ Two resistors are 6 and 2 ohm respectively.

Proved! ;)

Answer 2

Explanation:

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