The equivalent resistance of a series combination of two resistors is $s$ . When they are connected in parallel, the equivalent resistance is $p$ . If $s = n p$ , then the maximum value for $n$ is ______. (Round off to the nearest integer)

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Solution

Step 1: Given data
The equivalent resistance of a series combination is $s$ .
The equivalent resistance of a parallel combination is $p$
The relationship between the equivalent resistance of series and parallel combinations is
$s = n p$
Step 2: Formula used
Suppose the two are resistors are $R_{1}$ and $R_{2}$ .
The equivalent resistance of two resistors $R_{1}$ and $R_{2}$ when they are in series combination are-
$s = R_{1} + R_{2}$
When parallel connected then-
$\frac{1}{p} = \frac{1}{R_{1}} + \frac{1}{R_{2}} p = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$
Step 3: Compute the maximum value for n
Substitute the known values in the relation $s = n p$ ,
$R_{1} + R_{2} = n \times \frac{R_{1} R_{2}}{R_{1} + R_{2}}$
Multiply both sides by $(R_{1} + R_{2})$ ,
${(R_{1} + R_{2})}^{2} = n R_{1} R_{2} R_{1}^{2} + R_{2}^{2} + (2 - n) R_{1} R_{2} = 0$
Divide the whole equation by ${R_{2}}^{2}$
$\frac{{R_{1}}^{2}}{{R_{2}}^{2}} + 1 + (2 - n) \frac{R_{1}}{R_{2}}$
Let $\frac{R_{1}}{R_{2}} = k$
Now simplifying further,
$k^{2} + 1 + (2 - n) k = 0 k^{2} + (2 - n) k + 1 = 0 . . . . . . (1)$
On comparing equation (1) with $a x^{2} + b x + c = 0$
$a = 1, b = (2 - n), c = 1$
Use the Dharacharya rule to compute a discriminant,
$D \geq 0$
For the maximum value of $n$ , substitute $D = 0$ ,
$D = 0 b^{2} - 4 a c = 0 {(2 - n)}^{2} - 4 = 0 n^{2} - 4 n + 4 - 4 = 0 n^{2} = 4 n n = 4$
Hence, the maximum value for the $n$ is $4$ .

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