The resultant resistance value of n resistance each of r ohms and connected is series is X. When those n , resistance are connected in parallel , the resultant values is .

Options

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

Answer

To solve the problem, we need to find the resultant resistance when \( n \) resistors, each of resistance \( r \) ohms, are connected in series and then in parallel.

### Step-by-Step Solution:

1. **Calculate the Resistance in Series:**
When \( n \) resistors are connected in series, the total or resultant resistance \( R_s \) is given by the formula:
\[ R_s = r_1 + r_2 + r_3 + \ldots + r_n = n \cdot r \]
Here, since all resistors have the same resistance \( r \), we can simplify this to:
\[ R_s = n \cdot r \]
Given that the resultant resistance in series is \( X \), we can equate:
\[ X = n \cdot r \quad \text{(1)} \]

2. **Calculate the Resistance in Parallel:**
When the same \( n \) resistors are connected in parallel, the total or resultant resistance \( R_p \) is given by the formula:
\[ \frac{1}{R_p} = \frac{1}{r_1} + \frac{1}{r_2} + \ldots + \frac{1}{r_n} = \frac{n}{r} \]
Since all resistors have the same resistance \( r \), we can write:
\[ \frac{1}{R_p} = \frac{n}{r} \]
Therefore, the resultant resistance \( R_p \) can be calculated as:
\[ R_p = \frac{r}{n} \quad \text{(2)} \]

3. **Substituting for \( r \):**
From equation (1), we can express \( r \) in terms of \( X \):
\[ r = \frac{X}{n} \quad \text{(3)} \]
Now, substitute this value of \( r \) into equation (2):
\[ R_p = \frac{\frac{X}{n}}{n} = \frac{X}{n^2} \]

4. **Final Result:**
Thus, the resultant resistance when the \( n \) resistors are connected in parallel is:
\[ R_p = \frac{X}{n^2} \]

### Summary:
The resultant resistance when \( n \) resistors of \( r \) ohms are connected in series is \( X \), and when connected in parallel, the resultant resistance is:
\[ R_p = \frac{X}{n^2} \]