#### Question

Find the capacitance of the combination shown in figure between A and B.

#### Solution

Capacitors 5 and 1 are in series.

Their equivalent capacitance, `C_(eq) = (C_1C_5)/(C_1+C_5)` =`(2 xx 2)/(2+2)` = `1 "uF"`

`therefore` `C_(eq) = 1`

Now, this capacitor system is parallel to capacitor 6. Thus, the equivalent capacitance becomes 1 + 1 = 2 μF

The above capacitor system is in series with capacitor 2. Thus, the equivalent capacitance become `(2 xx 2)/(2+2) = 1 "uF"`

The above capacitor system is in parallel with capacitor 7. Thus, the equivalent capacitance becomes 1 + 1 = 2 μF

The above capacitor system is in series with capacitor 3. Thus, the equivalent capacitance becomes `(2 xx 2)/(2+2) = 1 "uF"`

The above capacitor system is in parallel with capacitor 8. Thus, the equivalent capacitance becomes

1 + 1 = 2 μF

The above capacitor system is in series with capacitor 4. Thus, the equivalent capacitance becomes `(2 xx 2)/(2+2) = 1 "uF"`

Hence, the equivalent capacitance between points A and B of the given capacitor system is 1 μF.