First, a set of $$n$$ equal resistors of 10 $$\Omega$$ each are connected in series to a battery of E.M.F. 20 V and internal resistance 10 $$\Omega$$. A current $$I$$ is observed to flow. Then, the $$n$$ resistors are connected in parallel to the same battery. It is observed that the current is increased 20 times, then the value of $$n$$ is _________.

We have a battery whose electromotive force is $$E = 20\ \text{V}$$ and whose internal resistance is $$r = 10\ \Omega$$. Along with this battery there are $$n$$ identical external resistors, each of resistance $$R = 10\ \Omega$$.

First the $$n$$ resistors are attached in series. The rule for series combination is that all resistances add directly, so the total external resistance becomes

$$R_{\text{series}} = R + R + \dots + R \; (n\ \text{times}) = nR = n(10\ \Omega) = 10n\ \Omega.$$

The total resistance of the complete circuit in this situation must also include the battery’s own resistance, therefore

$$R_{\text{total,1}} = r + R_{\text{series}} = 10 + 10n\ \Omega.$$

Ohm’s law states that the current in a closed circuit is the emf divided by the total resistance, that is

$$I = \frac{E}{R_{\text{total}}}.$$

Applying this to the first arrangement gives

$$I = \frac{20}{10 + 10n} = \frac{20}{10(1 + n)} = \frac{2}{1 + n}\ \text{A}.$$

Next the same $$n$$ resistors are removed from the series connection and regrouped in parallel. For resistors in parallel the reciprocals add, so with identical resistors we can write

$$\frac{1}{R_{\text{parallel}}} = \frac{1}{R} + \frac{1}{R} + \dots + \frac{1}{R}\; (n\ \text{times}) = \frac{n}{R},$$

which implies

$$R_{\text{parallel}} = \frac{R}{n} = \frac{10}{n}\ \Omega.$$

Now the total resistance of the entire circuit in the parallel arrangement becomes

$$R_{\text{total,2}} = r + R_{\text{parallel}} = 10 + \frac{10}{n}\ \Omega.$$

Using Ohm’s law again, the new current (call it $$I'$$) is

$$I' = \frac{E}{R_{\text{total,2}}} = \frac{20}{10 + \frac{10}{n}}.$$

We are told that this second current is twenty times the first current, i.e.

$$I' = 20\,I.$$

Substituting the explicit expressions for $$I'$$ and $$I$$:

$$\frac{20}{10 + \dfrac{10}{n}} = 20 \left(\frac{2}{1 + n}\right).$$

To clear the complex fraction on the left we rewrite its denominator:

$$10 + \frac{10}{n} = 10\left(1 + \frac{1}{n}\right) = 10\left(\frac{n + 1}{n}\right) = \frac{10(n + 1)}{n}.$$

Hence

$$\frac{20}{\dfrac{10(n + 1)}{n}} = 20 \left(\frac{2}{1 + n}\right).$$

The fraction on the left simplifies as follows:

$$\frac{20}{\dfrac{10(n + 1)}{n}} = 20 \times \frac{n}{10(n + 1)} = \frac{2n}{n + 1}.$$

Therefore the equality of currents becomes

$$\frac{2n}{n + 1} = \frac{40}{n + 1}.$$

Multiplying both sides by the common denominator $$(n + 1)$$ eliminates it:

$$2n = 40.$$

Dividing by 2 gives us the number of resistors,

$$n = 20.$$

So, the answer is $$20$$.