A network of three capacitors each 9 µF connected in series and the fourth capacitor of 6 µF is supplied 400 V as shown in figure. The ratio of charge on the capacitor $C_1$ to the capacitor $C_4$ is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1 : 2
As $C_1,C_2$ and $C_3$ are in series -
$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$
$=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{2+2+2}{18}$
$⇒C_{eq}=3μF$
Charge on $C_1$, $Q_1=C'×V$
$=3×4×10^{-6}=12×10^{-6}C$
Charge on $C_4$, $Q_4=C×V$
$=6×4×10^{-6}=24×10^{-6}C$
$∴\frac{Q_1}{Q_4}=\frac{12×10^{-6}}{24×10^{-6}}=\frac{1}{2}$