If n resistors, each of resistance r ohms, are connected in parallel, the equivalent resistance is R. What would be the equivalent resistance if they were connected in series?

When $n$ resistors, each of resistance $r \Omega$, are connected in parallel, the equivalent resistance $R$ is given by the formula:

$$\frac{1}{R} = \frac{1}{r} + \frac{1}{r} + \dots + \frac{1}{r} \quad (\text{n times})$$ $$\frac{1}{R} = \frac{n}{r}$$

From this equation, we can express $r$ in terms of $R$ and $n$:

$$r = nR$$

If the same $n$ resistors, each of resistance $r \Omega$, were connected in series, the equivalent resistance $R_{series}$ would be:

$$R_{series} = r + r + \dots + r \quad (\text{n times})$$ $$R_{series} = n \times r$$

Substitute $r = nR$ into the equation for $R_{series}$:

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

The equivalent resistance of the combination if connected in series would be $n^2R$.

Comparing this with the given options:
[A] $n^2R$
[B] $R/n^2$
[C] $R/n$
[D] $nR$

The calculated resistance matches option [A].

The final answer is $\text{A}$.