Three capacitors each of capacitance 9 pF are connected in series. The capacitance of the combination would be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 3 pF
$\frac{1}{C_{eq}}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}$ [In series]
$\frac{1}{C_{eq}}=\frac{3}{9}=\frac{1}{3}$
$C_{eq}=3pF$