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. `(R)/(n)`
B. `(R)/(n^(2))`
C. `nR`
D. `n^(2)R`

Answer

To solve the problem, we need to find the equivalent resistance when n similar conductors, each with resistance R, are connected in series.

### Step-by-Step Solution:

1. **Understanding Series Connection**:
- In a series connection, the total or equivalent resistance (R_eq) is the sum of the individual resistances.

2. **Identifying the Individual Resistances**:
- Given that there are n conductors and each has a resistance of R, we can denote the resistances as R1, R2, ..., Rn, where each Ri = R.

3. **Applying the Formula for Series Resistance**:
- The formula for the equivalent resistance in series is:
\[ R_{eq} = R_1 + R_2 + R_3 + ... + R_n \]
- Since all resistances are equal (R), we can write:
\[ R_{eq} = R + R + R + ... + R \quad (n \text{ times}) \]

4. **Calculating the Total Resistance**:
- This can be simplified to:
\[ R_{eq} = n \times R \]

5. **Final Result**:
- Therefore, when n similar conductors each of resistance R are connected in series, the resultant resistance is:
\[ R_{eq} = nR \]

### Conclusion:
The equivalent resistance when n similar conductors of resistance R are connected in series is \( nR \).