1. Circuit Analysis

The circuit consists of a series of resistors $R, 2R, 3R, \dots, 100R$ connected in a closed loop. An ideal battery is connected across two points in this loop, effectively splitting the loop into two parallel branches.

• Branch 1 (Upper): Contains resistors from $R$ to $nR$.
• Branch 2 (Lower): Contains resistors from $(n+1)R$ to $100R$.

2. Resistance Calculation

First, we calculate the total resistance of the entire series loop: $$ R_{total} = \sum_{k=1}^{100} kR = R \frac{100(101)}{2} = 5050R $$ Let $R_1$ be the resistance of the first branch (resistors $R$ to $nR$): $$ R_1 = \sum_{k=1}^{n} kR = \frac{n(n+1)}{2}R $$ Let $R_2$ be the resistance of the second branch: $$ R_2 = R_{total} – R_1 $$

3. Minimizing Current

The two branches are in parallel. The equivalent resistance $R_{eq}$ is: $$ R_{eq} = \frac{R_1 R_2}{R_1 + R_2} $$ The current drawn from the battery is $I = V / R_{eq}$. To minimize current, we must maximize $R_{eq}$.
Since the sum $R_1 + R_2 = R_{total}$ is constant ($5050R$), the product $R_1 R_2$ (and thus $R_{eq}$) is maximized when $R_1 = R_2$. $$ R_1 \approx \frac{R_{total}}{2} = \frac{5050R}{2} = 2525R $$ Substituting the formula for $R_1$: $$ \frac{n(n+1)}{2} \approx 2525 $$ $$ n(n+1) \approx 5050 $$

4. Solving for n

We look for an integer $n$ such that $n(n+1)$ is closest to 5050. Approximating $n^2 \approx 5050$, we get $n \approx \sqrt{5050} \approx 71.06$. We check the integers $n=70$ and $n=71$.

• For $n=70$: $n(n+1) = 70 \times 71 = 4970$ (Difference from 5050 is 80)
• For $n=71$: $n(n+1) = 71 \times 72 = 5112$ (Difference from 5050 is 62)

Since 5112 is closer to 5050 than 4970 is, $n=71$ provides a resistance closer to the ideal midpoint, yielding the maximum equivalent resistance and minimum current.