Two capacitance of 6(mu)F and 8(mu)F are connected in series. What additional capacitance must be connected in series with this combination to give a total of 3(mu)F
-
A.
3(mu)F -
B.
16(mu)F -
C.
24(mu)F -
D.
30(mu)F
Correct Answer: Option C
Explanation
3 = (frac{6 times 8 times c}{(6 times 8) + (6 times c) + (8 times c})
therefore, C = 24(mu)F