Three capacitors of 4F each are connected in such a way that net capacitance oftheir combination is 6F. It is possible if

The given problem can be solved by understanding the concept of series and parallel combination of capacitors.

1. In a parallel combination of capacitors, the total capacitance (Ct) is the sum of the individual capacitances (C1, C2, C3, ...). So, Ct = C1 + C2 + C3 + ...

2. In a series combination of capacitors, the total capacitance (Ct) is given by the reciprocal of the sum of the reciprocals of the individual capacitances. So, 1/Ct = 1/C1 + 1/C2 + 1/C3 + ...

Given that we have three capacitors each of 4µF and the total capacitance is 6µF, we can achieve this by connecting two capacitors in parallel and one in series.

Step-by-step solution:

Step 1: Connect two capacitors in parallel. The total capacitance (Cp) of this combination is Cp = C1 + C2 = 4µF + 4µF = 8µF.

Step 2: Connect this parallel combination (Cp = 8µF) in series with the third capacitor (C3 = 4µF). The total capacitance (Ct) of this combination is given by 1/Ct = 1/Cp + 1/C3 = 1/8 + 1/4 = 1/8 + 2/8 = 3/8.

Therefore, Ct = 8/3 µF = 2.67 µF.

This does not match the given total capacitance of 6µF. Therefore, it is not possible to achieve a total capacitance of 6µF with three 4µF capacitors.