Question

Two resistances when connected in parallel give a equivalent resistance of 2ohms, when
same resistances were connected in series the resultant resistance became 9ohms.calculate
value of each resistance.​

Answer 1

Answer:

• The Resistances are 6 ohms & 3 ohms Respectively.

Given:

1. Two resistances when connected in parallel give a equivalent resistance of 2 ohms.
2. Resistances were connected in series the resultant resistance became 9 ohms.

Explanation:

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From First Statement:-

Two resistances when connected in parallel give a equivalent resistance of 2 ohms.

We Know,

$\large\bigstar \: {\boxed{\tt \dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}}}$

$\bold{Here}\begin{cases}\text{R Denotes Effective Resistance} \\ \sf{R_1} \: \text{Denotes Resistance} \\ \sf{R_1} \: \text{Denotes Resistance}\end{cases}$

Now,

$\large{\boxed{\tt \dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}}}$

$\large{\tt \longmapsto \dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}}$

$\large{\tt \longmapsto \dfrac{1}{2} = \dfrac{1}{R_1} + \dfrac{1}{R_2} \: ----(1)}$

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From Second Statement:-

Resistances were connected in series the resultant resistance became 9ohms.

We Know,

$\large\bigstar \: {\boxed{\tt R = R_1 + R_2}}$

$\bold{Here}\begin{cases}\text{R Denotes Effective Resistance} \\ \sf{R_1} \: \text{Denotes Resistance} \\ \sf{R_1} \: \text{Denotes Resistance}\end{cases}$

Now,

$\large{\boxed{\tt R = R_1 + R_2}}$

$\large{\tt \longmapsto R = R_1 + R_2}$

$\large{\tt \longmapsto 9 = R_1 + R_2}$

Now

$\large{\tt \longmapsto R_1 = 9 - R_2}$

Substituting the Above equation in Equation (1).

$\large{\tt \longmapsto \dfrac{1}{2} = \dfrac{1}{R_1} + \dfrac{1}{R_2} \: ----(1)}$

$\large{\tt \longmapsto \dfrac{1}{2} = \dfrac{1}{ 9 - R_2} + \dfrac{1}{R_2}}$

$\large{\tt \longmapsto \dfrac{1}{2} = \dfrac{R_2 + (9 - R_2)}{( 9 - R_2) \times R_2}}$

$\large{\tt \longmapsto \dfrac{1}{2} = \dfrac{\cancel{R_2} + 9 - \cancel{R_2}}{( 9 - R_2) \times R_2}}$

$\large{\tt \longmapsto \dfrac{1}{2} = \dfrac{9}{( 9 - R_2) \times R_2}}$

$\large{\tt \longmapsto \dfrac{1}{2} = \dfrac{9}{( 9R_2 - R_2^2)}}$

Cross- Multiplying,

$\large{\tt \longmapsto 9R_2 - R_2^2 = 9 \times 2}$

$\large{\tt \longmapsto 9R_2 - R_2^2 = 18}$

Multiplying with " - " on Both sides, Results in.

$\large{\tt \longmapsto - 9R_2 + R_2^2 = - 18}$

$\large{\tt \longmapsto R_2^2 - 9 R_2 + 18 = 0}$

By Splitting the middle term.

$\large{\tt \longmapsto R_2^2 - 6R_2 - 3R_2 + 18 = 0}$

$\large{\tt \longmapsto R_2(R_2 - 6) - 3(R_2 - 6) = 0}$

$\large{\tt \longmapsto (R_2 - 6) (R_2 - 3) = 0}$

$\large\longmapsto {\underline{\boxed{\tt R_2 = 6\: (or) \; 3 \: \Omega}}}$

Now,

When

⟼ R₂ = 6 Ω then R₂ = 3 Ω

⟼ R₂ = 3 Ω then R₁ = 6 Ω.

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