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. \( \left( \frac{n^2 - 1}{n^2} \right) R \)
• B. \( \left( \frac{n^2 + 1}{n^2 - 1} \right) R \)
• C. \( \left( \frac{n^2 - 1}{n} \right) R \)
• D. \( \frac{(n^2 + 1)R}{n^2} \)

Hint:

Remember: for \( n \) equal resistors in series, add normally. For parallel, take reciprocal sum. Always simplify using LCM when subtracting expressions.

Answer 1

The Correct Option is C

Let’s find the expressions for the resistors: - In series, total resistance is: \[ R_s = nR \] - In parallel, total resistance is: \[ \frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \cdots + \frac{1}{R} = \frac{n}{R} \Rightarrow R_p = \frac{R}{n} \] - Now compute the difference: \[ R_s - R_p = nR - \frac{R}{n} \] Take LCM: \[ R_s - R_p = \frac{n^2R - R}{n} = \frac{R(n^2 - 1)}{n} \] Thus, \[ R_s - R_p = \left( \frac{n^2 - 1}{n} \right) R \]