Equivalent Capacitance of Capacitor connected in Series and Parallel Questions in English

(D) For capacitors connected in series,the equivalent capacitance $C_{eq}$ is given by:
$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$
$\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} + \frac{1}{6} = \frac{2+1+1}{6} = \frac{4}{6} = \frac{2}{3} \mu F^{-1}$
So,$C_{eq} = 1.5 \mu F$.
The total charge $q$ flowing through the circuit is:
$q = C_{eq} \times V = 1.5 \mu F \times 120 V = 180 \mu C$.
In a series circuit,the charge on each capacitor is the same. Therefore,the charge on the $3 \mu F$ capacitor is $q = 180 \mu C$.
The potential difference $V_1$ across the $3 \mu F$ capacitor is:
$V_1 = \frac{q}{C_1} = \frac{180 \mu C}{3 \mu F} = 60 V$.

(D) Let the potential difference across the capacitor $C$ (connected between $X$ and $Y$) be $V_{XY} = 60 \ V$.
The circuit consists of a capacitor $C$ in parallel with a series combination of capacitors $2C$,$C$,and $2C$.
Since the capacitors $2C$,$C$,and $2C$ are in series,the same charge $q$ flows through them.
The potential difference across the series branch is $V_{XY} = 60 \ V$.
The equivalent capacitance of the series branch is given by:
$\frac{1}{C_{eq}} = \frac{1}{2C} + \frac{1}{C} + \frac{1}{2C} = \frac{1+2+1}{2C} = \frac{4}{2C} = \frac{2}{C}$
So,$C_{eq} = \frac{C}{2}$.
The charge $q$ on each capacitor in the series branch is:
$q = C_{eq} \times V_{XY} = \frac{C}{2} \times 60 \ V = 30C$.
The potential difference between points $M$ and $N$ is the potential difference across the capacitor $C$ located between them.
$V_{MN} = \frac{q}{C} = \frac{30C}{C} = 30 \ V$.

(D) Let the capacitance of each capacitor be $C = 3 \mu F$.
Looking at the circuit,we can identify the arrangement of capacitors.
There are two capacitors in parallel in the middle branch. Their equivalent capacitance is $C_p = C + C = 2C = 2 \times 3 = 6 \mu F$.
This combination is in series with another capacitor $C$ in that same branch. The equivalent capacitance of this branch is $C_s = \frac{C \times C_p}{C + C_p} = \frac{C \times 2C}{C + 2C} = \frac{2C^2}{3C} = \frac{2}{3}C = \frac{2}{3} \times 3 = 2 \mu F$.
Finally,this branch is in parallel with the top capacitor $C$. The total effective capacitance between $A$ and $B$ is $C_{eq} = C + C_s = 3 + 2 = 5 \mu F$.

(B) Let the capacitance of each capacitor be $C$.
When $4$ capacitors are connected in series,the equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{4}{C}$,which implies $C_1 = \frac{C}{4}$.
When $4$ capacitors are connected in parallel,the equivalent capacitance $C_2$ is given by $C_2 = C + C + C + C = 4C$.
Now,the ratio $\frac{C_1}{C_2}$ is calculated as $\frac{C/4}{4C} = \frac{1}{16}$.

(D) Let the potential at $A$ be $V_A$ and at $B$ be $V_B$.
By analyzing the circuit,we see that the wire connecting the node between $C_1$ and $C_2$ to the node between $C_2$ and $C_3$ effectively short-circuits $C_2$.
Thus,the circuit simplifies to $C_1$ and $C_3$ in series,which are in parallel with $C_4$.
However,looking closely at the diagram,$C_1$ and $C_2$ are in series,and this combination is in parallel with $C_4$.
Actually,the circuit shows $C_1$ and $C_2$ in series,and this combination is in parallel with $C_4$. The capacitor $C_3$ is in series with the parallel combination of $(C_1, C_2)$ and $C_4$.
Let $C_{12} = \frac{C_1 C_2}{C_1 + C_2} = \frac{1 \times 1}{1 + 1} = 0.5\text{ }\mu\text{F}$.
Now,$C_{12}$ is in parallel with $C_4$,so $C_{p} = C_{12} + C_4 = 0.5 + 2 = 2.5\text{ }\mu\text{F}$.
Finally,$C_p$ is in series with $C_3$,so $C_{eq} = \frac{C_p C_3}{C_p + C_3} = \frac{2.5 \times 1}{2.5 + 1} = \frac{2.5}{3.5} = \frac{25}{35} = \frac{5}{7}\text{ }\mu\text{F}$.
Wait,re-evaluating the diagram: The wire connects the node after $C_1$ to the node after $C_2$. This means $C_2$ is shorted. The circuit is $C_1$ in series with $C_3$,and $C_4$ is in parallel with the combination of $C_1$ and $C_3$.
$C_{13} = \frac{1 \times 1}{1 + 1} = 0.5\text{ }\mu\text{F}$.
$C_{eq} = C_{13} + C_4 = 0.5 + 2 = 2.5 = 5/2\text{ }\mu\text{F}$.