Question

two resistance connected in parallel give resultant value of 2ohm , when connected in series value becomes 9ohm. calculate value of each resistance ​

Answer 1

Answer:

R1=60, R2=30

Explanation:

parallel R=R1.R2/R1+R2

and series R= R1+ R2

PUT they value and get answer

Answer 2

GiveN :-

• Two resistances when connected in parallel give resultant of 2Ω .
• When they are connected in series it gives resistance of 9Ω .

To FinD :-

• The value of each resistance .

AnsweR :-

Let us take :

$\sf \blue{First\: resistance\:be\:x.}$

$\sf \blue{Second\: resistance\:be\:y.}$

$\tt Now\:we\:know\:when\::-$

$\sf{\pink{\blue{\mapsto}When\: resistances\:are\: connected\:in\: series:}}$

$\boxed{\red{\bf{\pink{\dag}\:R_{net}=R_1+R_2+R_3+ . . +R_n}}}$

$\bf\qquad\qquad\qquad And,$

$\sf{\pink{\blue{\mapsto}When\: resistances\:are\: connected\:in\: parallel:}}$

$\boxed{\red{\bf{\pink{\dag}\:\dfrac{1}{R_{net}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+. . . +\dfrac{1}{R_n}}}}$

$\tt Let's\:use\:these\:one\:by\:one,$

$\tt{\orange{\mapsto}\green{Case\:1:In\: series}}$

$\sf:\implies R_{net}=R_1+R_2+R_3+. . . +R_n$

$\sf:\implies R_{net}=x+y$

$\sf:\implies 9\Omega=x+y$

$\underline{\boxed{\red{\tt\longmapsto x+y=9\Omega}}}$

___________________________________

$\tt{\orange{\mapsto}\green{Case\:2:In\: series}}$

$\sf:\implies \dfrac{1}{R_{net}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+. . .+\dfrac{1}{R_n}$

$\sf:\implies \dfrac{1}{R_{net}}=\dfrac{1}{x}+\dfrac{1}{y}$

$\sf:\implies \dfrac{1}{R_{net}}=\dfrac{x+y}{xy}$

$\sf:\implies \dfrac{1}{2\Omega} =\dfrac{x+y}{xy}$

$\sf:\implies 2(x+y)=xy$

$\sf:\implies 2(x+9-x)=x(9-x)$

$\sf:\implies 2\times9=9x-x^2$

$\sf:\implies 18=9x-x^2$

$\sf:\implies x^2-9x+18=0$

$\sf:\implies x^2-3x-6x+18=0$

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

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

$\underline{\boxed{\red{\frak{\longmapsto x\qquad=\qquad 3\Omega,6\Omega}}}}$

$\boxed{\green{\sf\pink{\dag}\:Hence\:two\: resistances\:are\:6\Omega\&\:3\Omega.}}$