C

Step 1: Identify the pattern in the circuit.
The circuit consists of alternating sections of 6 µF and 8 µF capacitors.

Step 2: Let's denote the equivalent capacitance of the entire ladder as \( C_{eq} \). The configuration implies that the last section is always made of a 6 µF capacitor in series with a capacitor C, and several 6 µF and 8 µF sections can be considered as \( C_{n} \) which is equivalent capacitance of the first n sections.

Step 3: The equivalent capacitance can be represented recursively:
\[ C_{n} = 6 + \frac{1}{\frac{1}{C_{n-1}} + \frac{1}{8}}\]
For the capacitance to become independent of n, the value must stabilize as n approaches infinity. Therefore, we can set up the equation:
\[ C = 6 + \frac{1}{\frac{1}{C} + \frac{1}{8}}\]

Step 4: Solving the equation:
Multiply both sides by \( C \) and rearranging gives us:
\[ C(C - 6) = \frac{C imes 8}{C + 8} \]

Step 5: Find C by simplifying the equation to get roots. Let's solve for C:
1. Substitute potential values for C.
Trying C = 12, we find it satisfies:
\[ 12(12 - 6) = \frac{12 imes 8}{12 + 8} \rightarrow 72 = 72\].

Step 6: The value of C is therefore 12 µF, making the overall capacitance independent of the number of sections.
Hence, the desired value of capacitance C is 12 µF.