[tex]\huge\star\underline\mathfrak\green{Question}[/tex]
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 ?
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[tex] \large\bf\underline {To \: find:-}[/tex]
[tex] \huge\bf\underline{Solution:-}[/tex]
[tex] \bf\underline{\red{Given:-}}[/tex]
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:-
[tex] \bf \red{R_{eq} = R_1 + R_2}[/tex]
[tex]\:\dashrightarrow \rm \: R_1 + R_2 = 9[/tex]
[tex]\:\dashrightarrow \rm \: R_1 = 9 - R_2.....(1)[/tex]
▶ When two resistors are connected in parallel
Then,
[tex] \bf \red{ \frac{1}{R_{eq} }= \dfrac{1}{R_1 }+ \dfrac{1}R_2}[/tex]
[tex] \dashrightarrow \rm \: \frac{1}{2 }= \dfrac{1}{R_1 }+ \dfrac{1}R_2[/tex]
[tex] \dashrightarrow \rm \: \frac{1}{2 }= \dfrac{R_2 + R_1}{R_1 .R_2}[/tex]
[tex] \dashrightarrow \rm \: 2= \dfrac{R_1 .R_2}{R_1 + R_2}[/tex]
[tex] \bf \: from \:eq. (1)[/tex]
[tex] \dashrightarrow \rm \: 2= \dfrac{(9 - R_2) .R_2}{(9 -R_2) + R_2}[/tex]
[tex] \dashrightarrow \rm \: 2= \dfrac{9 R_2 - {R_2 }^{2} }{9 }[/tex]
[tex] \dashrightarrow \rm \: {R_2 }^{2} - 9 R_2 + 18 = 0[/tex]
[tex] \dashrightarrow \rm \: {R_2 }^{2} - 6 R_2 - 3 R_2 + 18 = 0[/tex]
[tex] \dashrightarrow \rm \: {R_2 }(R_2- 6) - 3 ( R_2 - 6)= 0[/tex]
[tex] \dashrightarrow \rm \: (R_2 - 3) ( R_2 - 6)= 0[/tex]
[tex] \dashrightarrow \rm \: R_2 = 3 \: or \: R_2 = 6[/tex]
▶R1 = 9 - R2
▶R1 = 9 - 6
▶R1 = 3Ω
hence,
༒ the resistance of the component resistors is 3Ω and 6Ω
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Let the resistance be [tex]\sf R_1 \:and\: R_2[/tex]
series connection =
[tex]\sf \:9 = R_1 + R_2\\\sf R_1 =9- R_2[/tex]
Parallel connection =
[tex]\sfR_{eq}\frac{R_1R_2}{R_1+R_2}\\\sf 2 = \frac{(9-R_2)R_2}{9}\\\sf or\: R
^{2}_{2} −9R_2+18=0\\\sf R_2=6Ω \\\sf R_1 = 9 - 6 = 3Ω[/tex]