To solve this problem, we need to analyze the two scenarios: when the resistors are connected in series and when they are connected in parallel.$\\$1. Series Connection:$\\$• Total resistance in series $R_{series} = nR + R$ (internal resistance)$\\$• Current drawn in series $I = \frac{E}{R_{series}} = \frac{E}{nR + R} \\$2. Parallel Connection:$\\$• Total resistance in parallel $R_{parallel} = \frac{R}{n} \\$• Total resistance including internal resistance $R_{total} = \frac{R}{n} + R \\$• Current drawn in parallel $I_{parallel} = \frac{E}{R_{total}} = \frac{E}{\frac{R}{n} + R} \\$According to the problem, the current in parallel is 10 times the current in series:$\\ I_{parallel} = 10I \\$Substitute the expressions for $I$ and $I_{parallel}: \\ \frac{E}{\frac{R}{n} + R} = 10 \times \frac{E}{nR + R} \\$Cancel $E$ from both sides:$\\ \frac{1}{\frac{R}{n} + R} = \frac{10}{nR + R} \\$Cross-multiply to solve for $n: \\ (nR + R) = 10 \left(\frac{R}{n} + R\right) \\ nR + R = \frac{10R}{n} + 10R \\$Multiply through by $n$ to eliminate the fraction:$\\ n^2R + nR = 10R + 10nR \\$Rearrange terms:$\\ n^2R + nR - 10nR - 10R = 0 \\ n^2R - 9nR - 10R = 0 \\$Factor out $R: \\ R(n^2 - 9n - 10) = 0 \\$Since $R \neq 0,$ solve the quadratic equation:$\\ n^2 - 9n - 10 = 0 \\$Use the quadratic formula:$\\ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\$Here, $a = 1, b = -9, c = -10: \\ n = \frac{-(-9) \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} \\ n = \frac{9 \pm \sqrt{81 + 40}}{2} \\ n = \frac{9 \pm \sqrt{121}}{2} \\ n = \frac{9 \pm 11}{2} \\$This gives two solutions:$\\ n = \frac{20}{2} = 10 \quad$and$\quad n = \frac{-2}{2} = -1 \\$Since $n$ must be positive, we have $n = 10. \\$Therefore, the value of $n$ is $10. \\$This corresponds to Option 3.