Three capacitors each of capacitance 18\(\mu F\) are connected in series. Calculate the effective capacitance of the capacitors
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Correct Answer: Option C
Explanation:
\(\frac{1}{c} = \frac{1}{18} + \frac{1}{18} + \frac{1}{18} = \frac{3}{18}\)
c = 6\(\mu F\)
\(\frac{1}{c} = \frac{1}{18} + \frac{1}{18} + \frac{1}{18} = \frac{3}{18}\)
c = 6\(\mu F\)