Question

Resistance $n$, each of $r$ ohm, when connected in parallel give an equivalent resistance of $R$ ohm. If these resistances were connected in series, the combination would have a resistance in ohms, equal to

• A. $n^2 R$
• B. $R/ n^2$
• C. $R/n$
• D. $n R $

Answer 1

The Correct Option is A

To solve this question, we need to understand and use the concepts of resistors connected in series and parallel. Let's break the problem down step-by-step:

1. When resistances are connected in parallel, the reciprocal of the total (or equivalent) resistance \( R \) is the sum of the reciprocals of each individual resistance. If each resistance is \( r \) ohms and there are \( n \) such resistances in parallel, the formula is given by:

\(\frac{1}{R} = \frac{1}{r} + \frac{1}{r} + \ldots + \frac{1}{r}\) (n times)

Resulting in:

\(\frac{1}{R} = \frac{n}{r}\)

2. The equivalent resistance \( R \) for parallel configuration can be found by:

\(R = \frac{r}{n}\)

3. Now, if these \( n \) resistances are connected in series, then the total resistance \( R_{\text{series}} \) is simply the sum of individual resistances. Hence, it is given by:

\(R_{\text{series}} = n \times r\)

4. We derived that \( r = n \times R \) from the equivalent resistance of the parallel connection:

\(R = \frac{r}{n} \Rightarrow r = n \times R\)

5. Substituting \( r = n \times R \) back into the equation for series resistance:

\(R_{\text{series}} = n \times (n \times R) = n^2 \times R\)

6. Therefore, the total resistance when the resistors are connected in series is:

\(n^2 R\)

Conclusion: When the resistances are connected in series, the equivalent resistance is \( n^2 R \). Thus, the correct answer is:

n^2 R