Question

Two resistances when connected in parallel give resultant value of 2 ohm when connected in series the value become 9 ohm calculate the value of each resistance

Answer 1

Let the Resistances be R1 and R2 then,
A/q
R1+R2=9
And
1/R1+1/R2=2
Now You can easily solve it using linear equations in two variables
If u want me to do that too then reply in the comment section.

Answer 2

✦QUESTION✦

Two resistances when connected in parallel give resultant value of 2 ohm;when connected in series the value becomes 9 ohms. calculate the value of each resistance.

✦CONCEPT✦

•Here the concept of using of the formula of resistance(R) in parallel and series has been clarified to find the other required value of resistance .Moreover we also used factorization method to find the required value of resistance

★CALCULATION★

We know that two resistance are in parallel and hence

$\sf\blue{ \dfrac{1}{R_P} = \dfrac{1}{R_1} +\dfrac{1}{R_2}}$

$\sf \implies R_P= \dfrac{R_1 R_2}{R_1+R_2}$

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Given that ,

$\sf R_P=2 \Omega$

$\therefore \sf 2= \dfrac{R_1R_2}{R_1+R_2}$

$\sf \longrightarrow 2(R_1+R_2)=R_1 R_2 .....(1)$

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⟾ Now same resistance are in series

$\sf\red{ R_S =R_1 +R_2}$

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Given that

$\sf R_S =9\Omega$

$\therefore \sf 9=R_1+R_2......(2)$

From (1) and (2) ,we get

$:: \implies \sf R_1 R_2 =18$

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Again using (2),we have

$\sf R_2 =9-R_1$

$\therefore \sf R_1 (9-R_1)=18$

$:: \implies \sf R_{1}^{2}-9R_{1}+18=0$

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Factorization above equation

$\sf::\implies R_{1}^{2}- 6R_1-3R_1+18 = 0$

$:: \implies\sf (R_1-6)(R_1-3)=0$

$\sf Either \: R_1=6 \Omega\:or\:R_1=3 \Omega$

$\bf \green{R_2 =3 \Omega }\: or \pink{ \: R_2=6 \Omega }$

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$\sf\pink\bigstar Thus \: two \: resistance \: are \: \bf 3 \Omega\sf \: and \bf \: 6 \Omega.$

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