A

To find the capacitance between points A and B in the circuit systems shown in the diagrams:
Step 1: Identify the arrangement of capacitors in each circuit. In the first circuit (a), we have capacitors arranged in series and parallel.
Step 2: For capacitors in series, the equivalent capacitance (C_eq) can be calculated using the formula:
$$ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + ... $$
In this case, we have three capacitors (10μF, 30μF, and 20μF) in series, and the preceding 5μF and 10μF capacitors in parallel.
Step 3: First, calculate the equivalent capacitance of the series combination:
$$ \frac{1}{C_{eq}^{series}} = \frac{1}{10} + \frac{1}{30} + \frac{1}{20} $$
After calculating this, you can find the total capacitance by factoring in the parallel capacitors with 5μF and 10μF.
Step 4: Finally, sum up the equivalent capacitances for a final equivalent capacitance between points A and B.
Therefore, the answer is Option A, corresponding to the first diagram.