There are n similar conductors each of resistance R . The resultant resistance comes out to be x when connected in parallel. If they are connected in series, the resistance comes out to be

Options

A. `x//n^(2)`
B. `n^(2)x`
C. `x//n`
D. nx

Answer

To solve the problem, we need to find the resultant resistance when 'n' similar conductors, each with resistance 'R', are connected in series, given that the resultant resistance when connected in parallel is 'X'.

### Step-by-Step Solution:

1. **Understanding Parallel Resistance**:
When 'n' resistors of resistance 'R' are connected in parallel, the formula for the equivalent resistance \( R_p \) is given by:
\[ \frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \ldots + \frac{1}{R} \quad (n \text{ times}) \]
This simplifies to:
\[ \frac{1}{R_p} = \frac{n}{R} \]
Therefore, the equivalent resistance \( R_p \) can be expressed as:
\[ R_p = \frac{R}{n} \]
Given that this resultant resistance is equal to \( X \), we have:
\[ X = \frac{R}{n} \]

2. **Finding Resistance in Terms of X**:
From the equation \( X = \frac{R}{n} \), we can rearrange it to find \( R \):
\[ R = nX \]

3. **Understanding Series Resistance**:
When the same 'n' resistors are connected in series, the equivalent resistance \( R_s \) is simply the sum of all the resistances:
\[ R_s = R + R + \ldots + R \quad (n \text{ times}) = nR \]

4. **Substituting R in Series Resistance**:
Now, substituting the value of \( R \) from the previous step into the series resistance formula:
\[ R_s = n(nX) = n^2X \]

5. **Final Result**:
Therefore, the resultant resistance when 'n' similar conductors each of resistance 'R' are connected in series is:
\[ R_s = n^2X \]

### Conclusion:
The resultant resistance when the conductors are connected in series is \( n^2X \).