Question

Two capacitors with capacitance of 2uf and 0.4 uf respectively are connected in series

Answer 1

Answer:

Capacitance of two capacitors are given by;

C(1) = 2 μF

C(2) = 0.4 μF

Since the two capacitors are arranged in series combination, so their capacitance will be;

1/C(total) = 1/C(1) + 1/C(2)

Substituting the values;

1/C(total) = [1/2+ 1/0.4] μF = 3 μF

C(total) = 0.334 μF

The total capacitance of the combination is 0.334 μF.