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:
Step 1 (Formula / Governing Law & Given Data): [0.25 Mark]
For $n$ resistors each of resistance $R$:
- Equivalent resistance in series: $R_s = nR$
- Equivalent resistance in parallel: $\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \dots + \frac{1}{R} \text{ (n times)} \implies R_p = \frac{R}{n}$
Step 2 (Step-by-Step Mathematical Working): [0.5 Mark]
We need to find the value of $(R_s - R_p)$. Substituting the expressions derived in Step 1:
$$R_s - R_p = nR - \frac{R}{n}$$Taking $R$ as a common factor:
$$R_s - R_p = R \left( n - \frac{1}{n} \right)$$Finding the common denominator inside the bracket:
$$R_s - R_p = R \left( \frac{n^2 - 1}{n} \right)$$Correction Note: Upon reviewing the provided options, the standard algebraic simplification leads to $\left(\frac{n^2 - 1}{n}\right)R$.
Step 3 (Final Concluded Answer): [0.25 Mark]
The final expression is:
$$R_s - R_p = \left( \frac{n^2 - 1}{n} \right)R$$Correct Option: (C)
💡 EL10 Concept Insight: Think of series as stacking resistors to make the path longer (increasing resistance) and parallel as adding extra lanes to a road to make it easier for current to flow (decreasing resistance); subtracting the "easy" path from the "long" path gives you the difference in total resistance.