are two resistors and re equivalent resistances in series and

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Solution

Q.
$R_{1} a n d R_{2}$ are two resistors and $r_{1} a n d r_{2}$ are equivalent resistances in series and parallel respectively, then $\frac{R_{1}}{R_{2}}$ :

1 $\frac{r_{1} r_{2}}{r_{1} + r_{2}}$

2 $\frac{r_{1} + r_{2}}{r_{1} r_{2}}$

3 $\frac{r_{1} + \sqrt{r_{1} - 4 r_{1} r_{2}}}{r_{1} + \sqrt{r_{1} + 4 r_{1} r_{2}}}$

4 $\frac{r_{1} + \sqrt{r_{1}^{2} - 4 r_{1} r_{2}}}{r_{1} - \sqrt{r_{1}^{2} - 4 r_{1} r_{2}}}$

Ans.
(4)

Hence $r_{1} = R_{1} + R_{2} (i) a n d r_{2} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} (i i) \Rightarrow R_{1} R_{2} = r_{2} r_{1} N o w, {(R_{1} - R_{2})}^{2} = {(R_{1} + R_{2})}^{2} - 4 R_{1} R_{2} = r_{1}^{2} - 4 r_{2} r_{1} R_{1} - R_{2} = \sqrt{r_{1}^{2} - 4 r_{2} r_{1}} \dots (i i i) a n d R_{1} + R_{2} = r_{1} \dots (i v) s o l v i n g e q u a t i o n (i i i) a n d (i v) f o r R_{1} a n d R_{2} . R_{1} = \frac{r_{1} + \sqrt{r_{1}^{2} - 4 r_{1} r_{2}}}{2} a n d R_{2} = \frac{r_{1} - \sqrt{r_{1}^{2} - 4 r_{1} r_{2}}}{2} \frac{R_{1}}{R_{2}} = \frac{r_{1} + \sqrt{r_{1}^{2} - 4 r_{1} r_{2}}}{r_{1} - \sqrt{r_{1}^{2} - 4 r_{1} r_{2}}}$

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