To solve this problem, we need to analyze the two different configurations of the resistors: series and parallel.$\\$1. Series Configuration:$\\$• When 10 resistors, each of resistance $R,$ are connected in series, the total resistance $R_{series}$ is given by:$\\ R_{series} = 10R \\$• The current $I_{series}$ through the series circuit is given by Ohm's Law:$\\ I_{series} = \frac{E}{R_{series}} = \frac{E}{10R} \\$2. Parallel Configuration:$\\$• When the same 10 resistors are connected in parallel, the total resistance $R_{parallel}$ is given by:$\\ \frac{1}{R_{parallel}} = \frac{1}{R} + \frac{1}{R} + \ldots + \frac{1}{R} \quad ($10 times$) \\ \frac{1}{R_{parallel}} = \frac{10}{R} \\ R_{parallel} = \frac{R}{10} \\$• The current $I_{parallel}$ through the parallel circuit is given by Ohm's Law:$\\ I_{parallel} = \frac{E}{R_{parallel}} = \frac{E}{\frac{R}{10}} = \frac{10E}{R} \\$3. Comparing the Currents:$\\$• The problem states that the current increases $n$ times when the resistors are connected in parallel.$\\$• Therefore, we have:$\\ I_{parallel} = n \cdot I_{series} \\$Substitute the expressions for $I_{parallel}$ and $I_{series}: \\ \frac{10E}{R} = n \cdot \frac{E}{10R} \\$Simplify the equation:$\\ 10 = n \cdot \frac{1}{10} \\ n = 100 \\$Therefore, the value of $n$ is $100. \\$This corresponds to Option 2.