Three capacitors of 4F each are connected in such a way that net capacitance oftheir combination is 6F. It is possible if:A. All the three in series ⃝B. All the three in parallel ⃝C. Two in series and one in parallel ⃝D. Two in parallel and one in series

The answer is C. Two in series and one in parallel.

Here's the step-by-step explanation:

1. When capacitors are connected in series, the total or equivalent capacitance (Ct) is given by the reciprocal of the sum of the reciprocals of their individual capacitances (C1, C2, ... , Cn). So, if two 4µF capacitors are connected in series, their total capacitance would be:

   1/Ct = 1/C1 + 1/C2 1/Ct = 1/4 + 1/4 1/Ct = 1/2 Ct = 2µF

2. When capacitors are connected in parallel, the total capacitance is simply the sum of their individual capacitances. So, if a 4µF capacitor is connected in parallel with the 2µF total capacitance from the series connection, their total capacitance would be:

   Ct = C1 + C2 Ct = 2µF + 4µF Ct = 6µF

So, the net capacitance of their combination is 6µF if two capacitors are connected in series and one in parallel.