Final answer:
The total capacitance of the circuit consisting of two 8MFD capacitors connected in series is 4 MFD.
Step-by-step explanation:
In a series circuit, the total capacitance is given by the reciprocal of the sum of the reciprocals of the individual capacitances. In this case, with two capacitors connected in series, the total capacitance (C_total) is:
C_total = 1/(1/C1 + 1/C2)
Substituting C1 = C2 = 8 MFD (microfarads), we get:
C_total = 1/(1/8 + 1/8) = 1/(1/4) = 4 MFD
Therefore, the total capacitance of the circuit is 4 MFD.