Given Information:
- Capacitors C1, C3, C4, and C5 have a capacitance of 4 µF each.
- Capacitor C2 has a capacitance of 10 µF.

To Find:
The effective capacitance between points A and B.

Explanation:

To find the effective capacitance between points A and B, we need to analyze the given circuit. Let's break down the steps to find the solution.

Step 1: Identifying the Capacitors in Series and Parallel:

- Capacitors C1, C2, and C3 are in series.
- Capacitors C4 and C5 are in parallel.

Step 2: Simplifying the Circuit:

Series Combination:
When capacitors are connected in series, the reciprocal of the effective capacitance is equal to the sum of the reciprocals of the individual capacitances.

1/C = 1/C1 + 1/C2 + 1/C3

Substituting the values:
1/C = 1/4 + 1/10 + 1/4
1/C = 5/20 + 2/20 + 5/20
1/C = 12/20
C = 20/12
C = 5/3 µF

Parallel Combination:
When capacitors are connected in parallel, the effective capacitance is equal to the sum of the individual capacitances.

C_parallel = C4 + C5
C_parallel = 4 µF + 4 µF
C_parallel = 8 µF

Step 3: Finding the Effective Capacitance:
Now that we have simplified the circuit, we can combine the remaining capacitors.

The simplified circuit consists of two capacitors in series (C1, C2, and C3) and one capacitor in parallel (C_parallel).

The capacitors in series can be added to get the equivalent capacitance:
C_series = C1 + C2 + C3
C_series = 4 µF + 10 µF + 4 µF
C_series = 18 µF

Finally, the effective capacitance between points A and B is given by the sum of the capacitances in series and the capacitance in parallel:
Effective Capacitance = C_series + C_parallel
Effective Capacitance = 18 µF + 8 µF
Effective Capacitance = 26 µF

Conclusion:
The effective capacitance between points A and B in the given circuit is 26 µF.