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
- 3\(\mu\)F
- 16\(\mu\)F
-
24\(\mu\)F
- 30\(\mu\)F
The correct answer is: C
Explanation
3 = \(\frac{6 \times 8 \times c}{(6 \times 8) + (6 \times c) + (8 \times c}\)therefore, C = 24\(\mu\)F