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 C₁ to the capacitor C is:1:3

Correct Option: The correct option is 1:2.

Let us analyze the circuit shown in the diagram. The diagram displays a network of four capacitors: three capacitors in a series branch labeled $C_1=9\mathrm{\mu F}$, $C_2=9\mathrm{\mu F}$, and $C_3=9\mathrm{\mu F}$, and a fourth capacitor in a parallel branch labeled $C_4=6\mathrm{\mu F}$. The entire circuit is supplied by a DC source of $400V$.

Step 1: Calculate the equivalent capacitance of the series branch
The three capacitors $C_1$, $C_2$, and $C_3$ are connected in series with each other. The formula for the equivalent capacitance ($C_s$) of capacitors in series is:

$$\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$$

Substituting the values of $C_1=C_2=C_3=9\mathrm{\mu F}$ from the diagram:

$$\frac{1}{C_s}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9}=\frac{1}{3}{\mathrm{\mu F}}^{-1}$$

Taking the reciprocal, we get the equivalent capacitance of the series path:

$$C_s=3\mathrm{\mu F}$$

Step 2: Find the charge on capacitor $C_1$
For capacitors connected in series, the charge stored on each capacitor is identical and equals the charge supplied to that branch. Since the series combination is connected in parallel with the $400V$ voltage source, the total voltage across this series branch is $V=400V$.
The charge on $C_1$ is given by:

$$Q_1=C_s\times V$$

Substituting the values:

$$Q_1=3\mathrm{\mu F}\times 400V=1200\mathrm{\mu C}$$

Step 3: Find the charge on the fourth capacitor $C_4$
The capacitor $C_4$ is connected directly across the supply terminals, meaning it experiences the full potential difference of $V=400V$. Using the value $C_4=6\mathrm{\mu F}$ shown in the circuit diagram, the charge stored on it ($Q_4$) is:

$$Q_4=C_4\times V$$

Substituting the values:

$$Q_4=6\mathrm{\mu F}\times 400V=2400\mathrm{\mu C}$$

Step 4: Calculate the ratio of the charges
The ratio of the charge on the capacitor $C_1$ to the fourth capacitor (labeled $C_4$ in the diagram, referred to as $C$ in the question text) is:

$$\text{Ratio}=\frac{Q_1}{Q_4}=\frac{1200}{2400}=\frac{1}{2}$$

Therefore, the charge ratio is 1:2.