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
Verified Correct Answer: C.
Physical Principles & Numerical Solution
3 = \(\frac{6 \times 8 \times c}{(6 \times 8) + (6 \times c) + (8 \times c}\)
therefore, C = 24\(\mu\)F
Source: Myschool Past Questions Archive (Record #28472)