B

Step 1: Analyze the circuit. We have a series-parallel circuit with capacitors and resistors.
Step 2: Calculate the equivalent capacitance for the capacitors connected to the battery (C1 = 3 µF and C2 = 3 µF in parallel):
$$ C_{eq1} = C1 + C2 = 3 + 3 = 6 \mu F $$
Step 3: The 1 µF capacitors are in series with the equivalent capacitance calculated from the previous step. The formula for capacitors in series is:
$$ \frac{1}{C_{eq2}} = \frac{1}{C_{eq1}} + \frac{1}{1} $$
Therefore,
$$ C_{eq2} = \frac{6 \cdot 1}{6 + 1} = \frac{6}{7} \mu F $$
Step 4: Now calculate the voltage across the equivalent capacitor and the resistors using the voltage divider rule:
The total resistance is 50 Ohms (30Ω + 20Ω). The voltage across the resistors can be calculated as follows:
$$ V_{AC} = 100 \cdot \frac{20}{30 + 20} = 100 \cdot \frac{20}{50} = 40 V $$
Step 5: The potential difference between points A and B is 40 V.
Step 6: Calculate the potential difference between B and C: The voltage drop across the second branch (1 µF) can be calculated because the voltage across the components is the same, leading us to conclude that the potential difference between B and C is 60 V.
Therefore, the answers are 40V (A to B) and 60V (B to C). Thus, the correct answer option is B.