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Step-by-Step Solution
Step 1: Identify the equivalent resistances in series and parallel
Let the two resistors have resistances R_1 and R_2 . Their equivalent resistance when connected in series is:
s = R_1 + R_2.
When the same two resistors are connected in parallel, their equivalent resistance is:
p = \frac{R_1 R_2}{R_1 + R_2}.
Step 2: Set up the condition s = np
We are given that the series equivalent s is n times the parallel equivalent p . Hence,
s = np \quad\Longrightarrow\quad R_1 + R_2 = n \left( \frac{R_1 R_2}{R_1 + R_2} \right).
Step 3: Rearrange into a quadratic form
Multiply both sides by (R_1 + R_2) :
(R_1 + R_2)^2 = n R_1 R_2.
Expand (R_1 + R_2)^2 :
R_1^2 + 2R_1R_2 + R_2^2 = n R_1 R_2.
Rearrange terms:
R_1^2 + (2 - n)R_1 R_2 + R_2^2 = 0.
Step 4: Apply the condition for real (physical) solutions
For R_1 and R_2 to be physically real and positive, the discriminant of the quadratic in R_1 (treating R_2 as a constant) must be non-negative:
\text{Discriminant} \; \Delta = b^2 - 4ac \ge 0.
In the quadratic form R_1^2 + (2 - n)R_1 R_2 + R_2^2 = 0 , we identify:
\[
a = 1,\quad b = (2 - n)R_2,\quad c = R_2^2.
\]
So,
\Delta = \bigl[(2 - n)R_2\bigr]^2 - 4\times 1 \times (R_2^2) \ge 0.
Substitute and simplify:
(2 - n)^2 R_2^2 - 4 R_2^2 \ge 0 \quad\Longrightarrow\quad R_2^2 \bigl[(2 - n)^2 - 4\bigr] \ge 0.
Since R_2^2 > 0 , we need:
(2 - n)^2 - 4 \ge 0 \quad\Longrightarrow\quad (2 - n)^2 \ge 4.
Step 5: Solve the inequality for n
From (2 - n)^2 \ge 4 , we get
\[
2 - n \ge 2 \quad \text{or} \quad 2 - n \le -2.
\]
Solving each:
1) If 2 - n \ge 2 , then n \le 0 .
2) If 2 - n \le -2 , then n \ge 4 .
Since n must be positive (it relates the ratio of series to parallel resistances), the acceptable range is n \ge 4 .
Step 6: Conclude the minimum value of n
Therefore, the smallest integer value satisfying n \ge 4 is:
\boxed{4}.
