CUET Preparation Today
Three capacitors each of capacitance 9 pF are connected in series. The capacitance of the combination would be: |
3 pF 1 / 3 pF 6 pF 27 pF |
3 pF |
The correct answer is Option (1) → 3 pF
$\frac{1}{C_{eq}}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}$ [In series] $\frac{1}{C_{eq}}=\frac{3}{9}=\frac{1}{3}$ $C_{eq}=3pF$ |
