Three capacitors each of capacitance 18\(\mu F\) are connected in series. Calculate the effective capacitance of the capacitors
- 54.00\(\mu F\)
- 18.00\(\mu F\)
-
6.00\(\mu F\)
- 0.17\(\mu F\)
The correct answer is: C
Explanation
\(\frac{1}{c} = \frac{1}{18} + \frac{1}{18} + \frac{1}{18} = \frac{3}{18}\)c = 6\(\mu F\)