CBSE 2025, Set 2

Question

If $R_s$ and $R_p$ are the equivalent resistances of $n$ resistors, each of value $R$, in series and parallel combinations respectively, then the value of $(R_s - R_p)$ is : A). $(\frac{n^2 - 1}{n^2})R$ B). $(\frac{n^2 + 1}{n^2 - 1})R$ C). $(\frac{n^2 - 1}{n})R$ D). $\frac{(n^2 + 1)R}{n^2}$

Edvaya · Accepted Answer

The correct answer is $(\frac{n^2 - 1}{n})R$. When $n$ identical resistors, each of resistance $R$, are connected in **series**, their equivalent resistance $R_s$ is:
$$R_s = R + R + ... 	ext{ (n times)} = nR$$

When they are connected in **parallel**, their equivalent resistance $R_p$ is:
$$\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + ... 	ext{ (n times)} = \frac{n}{R}$$
$$R_p = \frac{R}{n}$$

We need to find the value of $(R_s - R_p)$:
$$R_s - R_p = nR - \frac{R}{n}$$
$$R_s - R_p = R \left( n - \frac{1}{n} ight)$$
$$R_s - R_p = R \left( \frac{n^2 - 1}{n} ight) = \boxed{\left( \frac{n^2 - 1}{n} ight)R}$$

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