Question

$\huge\star\underline\mathfrak\green{Question}$



Two resistors are connected (a) in series (b) in parallel. The equivalent resistance in the two cases are 9 ohm and 2 ohm respectively. Then what will be the resistance of the component resistors ?


Give answer with proper Explanation... Wrong answer will be reported. ​

Answer 1

$\large\bf\underline {To \: find:-}$

• we need to find the resistance of component resistors .

$\huge\bf\underline{Solution:-}$

$\bf\underline{\red{Given:-}}$

Two resistors are connected (a) in series (b) in parallel. The equivalent resistance in the two cases are 9 ohm and 2 ohm respectively.

Let

Let two resistors be R1 and R2

▶when two resistors are connected in series:-

$\bf \red{R_{eq} = R_1 + R_2}$

$\:\dashrightarrow \rm \: R_1 + R_2 = 9$

$\:\dashrightarrow \rm \: R_1 = 9 - R_2.....(1)$

▶ When two resistors are connected in parallel

Then,

$\bf \red{ \frac{1}{R_{eq} }= \dfrac{1}{R_1 }+ \dfrac{1}R_2}$

$\dashrightarrow \rm \: \frac{1}{2 }= \dfrac{1}{R_1 }+ \dfrac{1}R_2$

$\dashrightarrow \rm \: \frac{1}{2 }= \dfrac{R_2 + R_1}{R_1 .R_2}$

$\dashrightarrow \rm \: 2= \dfrac{R_1 .R_2}{R_1 + R_2}$

$\bf \: from \:eq. (1)$

$\dashrightarrow \rm \: 2= \dfrac{(9 - R_2) .R_2}{(9 -R_2) + R_2}$

$\dashrightarrow \rm \: 2= \dfrac{9 R_2 - {R_2 }^{2} }{9 }$

$\dashrightarrow \rm \: {R_2 }^{2} - 9 R_2 + 18 = 0$

$\dashrightarrow \rm \: {R_2 }^{2} - 6 R_2 - 3 R_2 + 18 = 0$

$\dashrightarrow \rm \: {R_2 }(R_2- 6) - 3 ( R_2 - 6)= 0$

$\dashrightarrow \rm \: (R_2 - 3) ( R_2 - 6)= 0$

$\dashrightarrow \rm \: R_2 = 3 \: or \: R_2 = 6$

• R2 = 6Ω then,

▶R1 = 9 - R2

▶R1 = 9 - 6

▶R1 = 3Ω

hence,

༒ the resistance of the component resistors is 3Ω and 6Ω

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Answer 2

Let the resistance be $\sf R_1 \:and\: R_2$

series connection =

$\sf \:9 = R_1 + R_2\\\sf R_1 =9- R_2$

Parallel connection =

$\sfR_{eq}\frac{R_1R_2}{R_1+R_2}\\\sf 2 = \frac{(9-R_2)R_2}{9}\\\sf or\: R </p><p>^{2}_{2} −9R_2+18=0\\\sf R_2=6Ω \\\sf R_1 = 9 - 6 = 3Ω$