Question

Two resistors of resistances R(1) = 100 $\: \pm \:$ 3 ohm and R(2) = 200 $\: \pm \:$ 4 ohm are connected (a) in series,(b) in Parrellel.Find the equivalent resistance of the (a) series combination, (b) parallel combination.Use for (a) the relation R= R(1)+R(2) . and for (b)

$\frac{1}{ r_{1} } + \frac{1}{ r_{2} }$
and
$\frac{ \triangle \: R }{ {R }^{2} } = \frac{ \triangle \: R_{1}}{ {R _{1} {}^{2} } } + \frac{ \triangle R_{2}}{ R_{2} {}^{2} }$
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Answer 1

$\underline{\huge{Answer:-}}$

Explanation:

$\: \large\bold{(a) \: \:The \: equivalent \: resistance\: of \: series\: combination.} \:$

$\large \mathcal{R = R _{1} + R_{2} = (100 \pm3) \: ohm \: + (200 \pm4) \: ohm } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large \mathcal{ =300 \pm7 \: ohm. }$

$\: \large\bold{(b) \: \: The \: equivalent \: resistance\: of \: parallel \: combination.} \:$

$\large \mathcal{ {R}^{'} = \frac{ R_{1} R_{2} }{ R_{1} + R _{2} } = \frac{200}{3} = 66.7 \: ohm }$

$\large \bold{Then , \: from} \: \: \large \mathcal{ \frac{1}{R'} = \frac{1}{ R_{1} } + \frac{1}{ R_{2} } }$

$\large \bold{we \: get, }$

$\large \mathcal{ \frac{ \triangle \: R'}{ {R'}^{2} } = \frac{ \triangle \: R_{1} }{ R_{1} ^{2} } + \frac{ \triangle \: R _{2} }{ R_{2} {}^{2} } }$

$\large \mathcal{ \triangle \: R' = (R' {}^{2}) \frac{ \triangle \: R _{1} }{R _{1} {}^{2} } + ( {R'}^{2} ) \frac{ \triangle R_{2} }{ R_{2} {}^{2} } }$

$\large \mathcal{( \frac{66.7}{100} ){}^{2}3 + ( \frac{66.7}{200} ) {}^{2} 4 } \\ \\ \: = \large \fbox{1.8}$

$\large \mathcal{Then , \: R' \: = 66.7 \pm1.8 \: ohm}$

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(Here , ∆R is expressed as 1.8 instead of 2 to keep in confirmity with the rules of significant figures.)

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Answer 2

Answer:

$the \: answer \: is \: 1.8$