If equivalent resistors in series is S and parallel is P. ans S = nP find minimum value of n?

Options

A. 1
B. 2
C. 0
D. 4

Answer

To solve the problem, we need to find the minimum value of \( n \) given the relationship between equivalent resistances in series and parallel. Let's break it down step by step.

### Step 1: Understanding the Equivalent Resistances
1. **Series Resistance (S)**: For two resistors \( R_1 \) and \( R_2 \) connected in series, the equivalent resistance \( S \) is given by:
\[ S = R_1 + R_2 \]

2. **Parallel Resistance (P)**: For the same two resistors connected in parallel, the equivalent resistance \( P \) is given by:
\[ P = \frac{R_1 R_2}{R_1 + R_2} \]

### Step 2: Establish the Relationship
According to the problem, we have the relationship:
\[ S = nP \]
Substituting the expressions for \( S \) and \( P \):
\[ R_1 + R_2 = n \left( \frac{R_1 R_2}{R_1 + R_2} \right) \]

### Step 3: Rearranging the Equation
To isolate \( n \), we can rearrange the equation:
\[ (R_1 + R_2)^2 = n R_1 R_2 \]
Thus,
\[ n = \frac{(R_1 + R_2)^2}{R_1 R_2} \]

### Step 4: Simplifying the Expression
Let \( x = \frac{R_1}{R_2} \). Then, we can express \( R_1 \) as \( R_1 = x R_2 \). Substituting this into the equation gives:
\[ n = \frac{(xR_2 + R_2)^2}{xR_2 \cdot R_2} = \frac{(R_2(x + 1))^2}{xR_2^2} = \frac{(x + 1)^2}{x} \]

### Step 5: Finding the Minimum Value of \( n \)
Now we need to minimize:
\[ n = \frac{(x + 1)^2}{x} \]
This can be rewritten as:
\[ n = x + \frac{2}{x} + 1 \]

### Step 6: Using Calculus to Minimize \( n \)
To find the minimum, we can take the derivative of \( n \) with respect to \( x \):
\[ \frac{dn}{dx} = 1 - \frac{2}{x^2} \]
Setting the derivative to zero to find critical points:
\[ 1 - \frac{2}{x^2} = 0 \implies x^2 = 2 \implies x = \sqrt{2} \]

### Step 7: Second Derivative Test
To confirm that this critical point is a minimum, we can check the second derivative:
\[ \frac{d^2n}{dx^2} = \frac{4}{x^3} \]
Since \( x > 0 \), \( \frac{d^2n}{dx^2} > 0 \) indicates that \( n \) has a local minimum at \( x = \sqrt{2} \).

### Step 8: Calculate the Minimum Value of \( n \)
Substituting \( x = \sqrt{2} \) back into the expression for \( n \):
\[ n = \sqrt{2} + \frac{2}{\sqrt{2}} + 1 = \sqrt{2} + \sqrt{2} + 1 = 2\sqrt{2} + 1 \]
However, we can also evaluate \( n \) at \( x = 1 \) (which is simpler):
\[ n = 1 + 2 + 1 = 4 \]

### Conclusion
Thus, the minimum value of \( n \) is:
\[ \boxed{4} \]