When all 10 resistors (\( R \)) are connected in series, the maximum resistance is:
\[ R_{\text{max}} = 10R = 10 \times 10 = 100 \, \Omega \]
When all 10 resistors (\( R \)) are connected in parallel, the minimum resistance is:
\[ R_{\text{min}} = \frac{R}{10} = \frac{10}{10} = 1 \, \Omega \]
The ratio is given by:
\[ \frac{R_{\text{max}}}{R_{\text{min}}} = \frac{100}{1} = 100 \]
From the above calculations:
\[ R_{\text{min}} = 1 \, \Omega \]
A wire of resistance $ R $ is bent into a triangular pyramid as shown in the figure, with each segment having the same length. The resistance between points $ A $ and $ B $ is $ \frac{R}{n} $. The value of $ n $ is:

In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by: