Question

① Two resistor when connected in parallel give resulant value of 2 ohm when connected in series the value become 9 ohm calculate the value resistance
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Answer 1

➩Given:

Two resistors when connected in parallel gives resultant value of 2 ohms.

The same resistors when connected in series gives resultant value of 9 ohms.

➩Solution:

Let the two resistances be R₁and R₂

If connected in series, then

R₁ + R₂ = 9

R₂ = 9 - R₁

When connected in parallel, then

$\frac{1}{R_{1}} + \frac{1}{ R₂ } = \frac{1}{2}$

$\frac{1}{R_{1}} + \frac{1}{ 9 - R_{1}} = \frac{1}{2}$

$\frac{9 - {R_{1}} + {R_{1}}}{ {R_{1}} (9 - {R_{1}})} = \frac{1}{2}$

$\frac{9}{9 {R_{1}} - {R_{1}} ^{2} } = \frac{1}{2}$

Now

Cross multiplication

$9 {R_{1}} - {R_{1}} ^{2} = 18$

$- {R_{1}} ^{2} + 9 {R_{1}}= 18$

$- {R_{1}} ^{2} + 9 {R_{1}} - 18 = 0$

Taking - (minus) as common, we get

${R_{1}} ^{2} - 9 {R_{1}} + 18 = 0$

Now we are left with a Quadratic Equation , we need to solve it .

${R_{1}} ^{2} - 9 {R_{1}} + 18 = 0$

$( {R_{1}} - 3)( {R_{1}} - 6)= 0$

$\sf So, {R_ { 1 } } = 3Ω,6 Ω$

When ${R_ { 1 } } = 3Ω \: \: , {R_ { 2 } } = 9-3=6Ω$

When ${R_ { 1 } } = 6Ω \: \:, {R_ { 2 } } = 9-6=3Ω$