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

(D) In a series combination of capacitors,the charge on each capacitor is the same.
Let the equivalent capacitance be $C_{eq}$. For two capacitors in series,$C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$.
Given $C_1 = 10\,\mu F$,$C_2 = 20\,\mu F$,and $V = 30\,V$.
$C_{eq} = \frac{10 \times 20}{10 + 20} = \frac{200}{30} = \frac{20}{3}\,\mu F$.
The charge on each capacitor is $Q = C_{eq} \times V$.
$Q = \left( \frac{20}{3}\,\mu F \right) \times 30\,V = 200\,\mu C$.
Since they are in series,the charge on both capacitors is equal to $200\,\mu C$.

(C) When capacitors are connected in series,the equivalent capacitance $C_{eq}$ is given by the formula: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.
Substituting the given values: $\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{10} + \frac{1}{15}$.
Finding a common denominator $(30)$: $\frac{1}{C_{eq}} = \frac{10 + 3 + 2}{30} = \frac{15}{30} = \frac{1}{2}$.
Therefore,$C_{eq} = 2\,\mu F$.
The total charge $Q$ supplied by the source is $Q = C_{eq} \times V = 2\,\mu F \times 100\,V = 200\,\mu C$.
In a series combination,the charge on each capacitor is the same and is equal to the total charge supplied by the source.
Thus,the charge on the $15\,\mu F$ capacitor is $200\,\mu C$.

(B) From the circuit diagram,$C_2$ and $C_3$ are connected in parallel,and this parallel combination is in series with $C_1$.
$1$. For charge distribution: The total charge $Q$ flowing from the battery passes through $C_1$. This charge then splits into $Q_2$ and $Q_3$ across $C_2$ and $C_3$ respectively. Thus,$Q = Q_2 + Q_3$. Statement $II$ is correct.
$2$. For potential drop: Since $C_2$ and $C_3$ are in parallel,the potential drop across them is the same $(V_2 = V_3)$. However,the potential drop across $C_1$ $(V_1 = Q/C_1)$ is generally not equal to the potential drop across the parallel combination $(V_{23} = Q/(C_2+C_3))$. Thus,statement $III$ is incorrect.
$3$. For energy stored: Energy stored in a capacitor is $U = Q^2 / (2C)$. Since $Q_1 = Q$,$U_1 = Q^2 / (2C_1)$. The energy in the parallel combination is $U_{23} = Q_2^2 / (2C_2) + Q_3^2 / (2C_3)$. These are not generally equal. Thus,statement $I$ is incorrect.
Therefore,only statement $II$ is correct.

(C) In the given circuit,all three capacitors are connected in parallel between points $A$ and $B$.
By tracing the connections,we can see that one plate of each capacitor is connected to point $A$,and the other plate of each capacitor is connected to point $B$.
For capacitors connected in parallel,the equivalent capacitance $C_{eq}$ is given by the sum of individual capacitances:
$C_{eq} = C_1 + C_2 + C_3$
Given $C_1 = C_2 = C_3 = 1\,\mu F$,
$C_{eq} = 1\,\mu F + 1\,\mu F + 1\,\mu F = 3\,\mu F$.

(D) From the circuit diagram,we can observe that the three capacitors in the bottom branch are connected in series. The equivalent capacitance of these three capacitors is given by $\frac{1}{C_s} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C}$,which implies $C_s = \frac{C}{3}$.
This equivalent capacitor $C_s$ is connected in parallel with the top capacitor $C$. Therefore,the total equivalent capacitance between points $A$ and $B$ is $C_{eq} = C_s + C = \frac{C}{3} + C = \frac{4C}{3}$.

(B) $1$. The two $4 \, \mu F$ capacitors are connected in parallel. Their equivalent capacitance is $C_p = 4 \, \mu F + 4 \, \mu F = 8 \, \mu F$.
$2$. The $2 \, \mu F$ and $3 \, \mu F$ capacitors in the upper branch are in series. Their equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$,so $C_1 = \frac{6}{5} \, \mu F$.
$3$. Similarly,the $2 \, \mu F$ and $3 \, \mu F$ capacitors in the lower branch are in series. Their equivalent capacitance $C_2 = \frac{6}{5} \, \mu F$.
$4$. Now,the circuit consists of three capacitors $C_1$,$C_p$,and $C_2$ connected in series between $A$ and $B$.
$5$. The total equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_p} + \frac{1}{C_2} = \frac{5}{6} + \frac{1}{8} + \frac{5}{6}$.
$6$. $\frac{1}{C_{eq}} = \frac{10}{6} + \frac{1}{8} = \frac{5}{3} + \frac{1}{8} = \frac{40 + 3}{24} = \frac{43}{24}$.
$7$. Therefore,$C_{eq} = \frac{24}{43} \, \mu F$.

(A) The equivalent capacitance $C_{eq}$ for capacitors in series is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.
Substituting the values: $\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = \frac{6}{6} = 1\,\mu F$.
Thus,$C_{eq} = 1\,\mu F$.
The total charge $Q$ supplied by the battery is $Q = C_{eq} \times V = 1\,\mu F \times 24\,V = 24\,\mu C$.
In a series combination,the charge on each capacitor is the same.
Therefore,the potential difference $V'$ across the $6\,\mu F$ capacitor is $V' = \frac{Q}{C_3} = \frac{24\,\mu C}{6\,\mu F} = 4\,V$.

(B) When capacitors are connected in series,the equivalent capacitance $C_{eq}$ is given by the formula:
$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$
Given $C_1 = 1\,\mu F$ and $C_2 = 2\,\mu F$.
Substituting the values:
$\frac{1}{C_{eq}} = \frac{1}{1} + \frac{1}{2} = \frac{2+1}{2} = \frac{3}{2}$
Therefore,$C_{eq} = \frac{2}{3}\,\mu F \approx 0.67\,\mu F$.

(A) Let the capacitance of each identical capacitor be $C$.
When connected in series,the equivalent capacitance $C_s$ is given by $C_s = \frac{C}{2}$.
Given $C_s = 3\,\mu F$,so $\frac{C}{2} = 3$,which implies $C = 6\,\mu F$.
When connected in parallel,the equivalent capacitance $C_p$ is given by $C_p = C + C = 2C$.
Given $C_p = 12\,\mu F$,so $2C = 12$,which implies $C = 6\,\mu F$.
Thus,the capacitance of each capacitor is $6\,\mu F$.

(C) The two capacitors of $2 \mu F$ are connected in parallel. Their equivalent capacitance is $C_p = 2 \mu F + 2 \mu F = 4 \mu F$.
Now,the circuit consists of three capacitors of $4 \mu F$ each,connected in series between points $A$ and $B$.
The equivalent capacitance $C_{AB}$ for capacitors in series is given by $\frac{1}{C_{AB}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.
Substituting the values: $\frac{1}{C_{AB}} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$.
Therefore,$C_{AB} = \frac{4}{3} \mu F$.
Wait,re-evaluating the circuit diagram: The two $2 \mu F$ capacitors are in parallel,giving $4 \mu F$. This $4 \mu F$ is in series with the top $4 \mu F$ and bottom $4 \mu F$ capacitors. The total equivalent capacitance is $\frac{1}{C_{AB}} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$,so $C_{AB} = 1.33 \mu F$.
However,looking at the provided solution image,the circuit simplifies to three $4 \mu F$ capacitors in series. If the question implies a different configuration,based on the provided options,the intended answer is $C = 4/3 \mu F$. Given the options provided,there might be a discrepancy in the question's diagram or options. Assuming the standard interpretation of the provided image,the result is $4/3 \mu F$.

(C) The circuit consists of a capacitor in series with two parallel capacitors,all of which are in parallel with a fourth capacitor.
$1$. First,consider the two capacitors connected in parallel. Their equivalent capacitance is $C_p = C + C = 2C$.
$2$. This combination is in series with the third capacitor of capacitance $C$. The equivalent capacitance of this series branch is $C_s = \frac{C \times 2C}{C + 2C} = \frac{2C^2}{3C} = \frac{2}{3}C$.
$3$. Finally,this branch is in parallel with the fourth capacitor of capacitance $C$. The total effective capacitance between $A$ and $B$ is $C_{AB} = C_s + C = \frac{2}{3}C + C = \frac{5}{3}C$.

(A) The circuit consists of five capacitors,each of capacitance $C$.
Looking at the circuit,the two capacitors on the left side are in series,and their equivalent capacitance is $C_{s1} = \frac{C \times C}{C + C} = \frac{C}{2}$.
Similarly,the two capacitors on the right side are in series,and their equivalent capacitance is $C_{s2} = \frac{C \times C}{C + C} = \frac{C}{2}$.
Now,these two equivalent capacitors ($C_{s1}$ and $C_{s2}$) are in parallel with the central capacitor of capacitance $C$.
Therefore,the total effective capacitance between points $A$ and $B$ is $C_{AB} = C_{s1} + C_{s2} + C = \frac{C}{2} + \frac{C}{2} + C = C + C = 2\,C$.

(B) Let the three capacitors be $C_1 = C_2 = C_3 = 4\,\mu F$.
If we connect two capacitors in series,their equivalent capacitance $C_s$ is given by:
$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$
So,$C_s = 2\,\mu F$.
Now,if we connect this combination in parallel with the third capacitor $C_3$,the total effective capacitance $C_{eq}$ is:
$C_{eq} = C_s + C_3 = 2\,\mu F + 4\,\mu F = 6\,\mu F$.
Thus,connecting two capacitors in series and one in parallel gives an effective capacitance of $6\,\mu F$.

(A) To obtain the maximum capacitance,the capacitors must be connected in parallel.
In parallel combination,the equivalent capacitance is given by $C_{eq} = C_1 + C_2 + C_3$.
Since all capacitors have the same capacitance $C = 3\,\mu F$,we have $C_{max} = 3C = 3 \times 3\,\mu F = 9\,\mu F$.
To obtain the minimum capacitance,the capacitors must be connected in series.
In series combination,the equivalent capacitance is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.
Thus,$C_{min} = \frac{C}{3} = \frac{3\,\mu F}{3} = 1\,\mu F$.
Therefore,the maximum and minimum capacitances are $9\,\mu F$ and $1\,\mu F$ respectively.

(C) For capacitors in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.
Substituting the values: $\frac{1}{C_{eq}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{8} = \frac{8+4+1}{8} = \frac{13}{8}$.
Thus,$C_{eq} = \frac{8}{13}\,\mu F$.
The total charge $Q$ supplied by the battery is $Q = C_{eq} \times V = \frac{8}{13} \times 13 = 8\,\mu C$.
In a series circuit,the charge on each capacitor is the same and equal to the total charge $Q$.
Therefore,the potential difference $V_2$ across the $2\,\mu F$ capacitor is $V_2 = \frac{Q}{C_2} = \frac{8\,\mu C}{2\,\mu F} = 4\,V$.

(D) The equivalent capacitance $C_{eq}$ of two capacitors in series is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$.
Thus,$C_{eq} = \frac{6}{5}\ \mu F$.
The total charge $Q$ supplied to the series combination is $Q = C_{eq} \times V_{total} = \frac{6}{5} \times 1000 = 1200\ \mu C$.
Since the capacitors are in series,the charge on each capacitor is the same,$Q = 1200\ \mu C$.
The potential difference across the $2\ \mu F$ capacitor is $V_1 = \frac{Q}{C_1} = \frac{1200}{2} = 600\ V$.
If the outer plate of the first capacitor is at $1000\ V$,the potential at the inner plate (which is shared with the second capacitor) is $V_{inner} = 1000\ V - V_1 = 1000 - 600 = 400\ V$.

(B) The given circuit consists of two groups of capacitors connected in series between points $A$ and $B$.
Each group consists of three identical capacitors connected in parallel.
Let $C$ be the capacitance of each individual capacitor.
For a parallel combination of three capacitors,the equivalent capacitance is $C_{eq} = C + C + C = 3C$.
Since each capacitor can withstand a maximum of $200\, V$,the entire parallel group of three capacitors can also withstand a maximum of $200\, V$.
The circuit effectively becomes two equivalent capacitors of $3C$ connected in series.
Let $V_1$ and $V_2$ be the voltages across the two groups. Since the capacitors in each group are identical,the total voltage $V_{AB}$ is divided equally across the two groups.
$V_{AB} = V_1 + V_2 = 200\, V + 200\, V = 400\, V$.
Thus,the maximum voltage that can be safely applied between $A$ and $B$ is $400\, V$.

(D) The given circuit consists of three plates. Let the middle plate be connected to terminal $A$ and the top and bottom plates be connected together to terminal $B$.
This arrangement forms two capacitors in parallel,each formed by the middle plate and one of the outer plates.
The capacitance of each capacitor is $C = \frac{\varepsilon_0 A}{d}$.
Since the two capacitors are in parallel,the effective capacitance is $C_{eq} = C + C = 2C$.
Substituting the value of $C$,we get $C_{eq} = \frac{2\varepsilon_0 A}{d}$.

(B) When capacitors are connected in series,the equivalent capacitance $C_{eq}$ is given by the formula:
$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$
Substituting the given values:
$\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$
$\frac{1}{C_{eq}} = \frac{3 + 2 + 1}{6} = \frac{6}{6} = 1$
Therefore,$C_{eq} = 1\,\mu F$.

(D) The given circuit is a Wheatstone bridge. We check the ratio of the capacitors in the arms: $\frac{C_1}{C_2} = \frac{3}{4}$ and $\frac{C_3}{C_4} = \frac{6}{8} = \frac{3}{4}$.
Since the ratios are equal,the bridge is balanced.
In a balanced Wheatstone bridge,the potential difference across the central capacitor $(2\,\mu F)$ is zero,so it can be removed from the circuit.
Now,the $3\,\mu F$ and $6\,\mu F$ capacitors are in series,and the $4\,\mu F$ and $8\,\mu F$ capacitors are in series.
The equivalent capacitance of the upper branch is $C_{up} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2\,\mu F$.
The equivalent capacitance of the lower branch is $C_{low} = \frac{4 \times 8}{4 + 8} = \frac{32}{12} = \frac{8}{3}\,\mu F$.
These two branches are in parallel,so the total effective capacitance is $C_{AB} = C_{up} + C_{low} = 2 + \frac{8}{3} = \frac{6 + 8}{3} = \frac{14}{3}\,\mu F$.

(D) In a series combination,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.
Substituting the values: $\frac{1}{C_{eq}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{6+3+2}{6} = \frac{11}{6}$.
Thus,$C_{eq} = \frac{6}{11}\ \mu F$.
The total charge $Q$ supplied by the battery is $Q = C_{eq} \times V = \frac{6}{11}\ \mu F \times 11\ V = 6\ \mu C$.
In a series circuit,the charge on each capacitor is the same. Therefore,the charge on the $1\ \mu F$ capacitor is $6\ \mu C$.
The potential difference $V_1$ across the $1\ \mu F$ capacitor is $V_1 = \frac{Q}{C_1} = \frac{6\ \mu C}{1\ \mu F} = 6\ V$.

(B) Let the given capacitor have capacitance $C = 8\,\mu F$ and voltage rating $V = 250\,V$. The required composite capacitor has capacitance $C' = 16\,\mu F$ and voltage rating $V' = 1000\,V$.
Suppose $m$ rows of capacitors are connected in parallel,and each row contains $n$ capacitors connected in series.
The potential difference across each capacitor in a series row is $V = \frac{V'}{n}$.
Substituting the values: $250 = \frac{1000}{n} \implies n = 4$.
The equivalent capacitance of the network is $C' = \frac{mC}{n}$.
Substituting the values: $16 = \frac{m \times 8}{4} \implies 16 = 2m \implies m = 8$.
The total number of capacitors required is $N = n \times m = 4 \times 8 = 32$.
Alternatively,using the formula: $N = \frac{C'}{C} \times \left( \frac{V'}{V} \right)^2 = \frac{16}{8} \times \left( \frac{1000}{250} \right)^2 = 2 \times (4)^2 = 2 \times 16 = 32$.

(B) The circuit consists of an infinite number of rows connected in parallel.
Each row contains a certain number of capacitors connected in series.
- The first row has $1$ capacitor of $1\,\mu F$. Its equivalent capacitance is $C_1 = 1\,\mu F$.
- The second row has $2$ capacitors in series,each of $1\,\mu F$. Its equivalent capacitance is $C_2 = \frac{1}{1+1} = \frac{1}{2}\,\mu F$.
- The third row has $4$ capacitors in series,each of $1\,\mu F$. Its equivalent capacitance is $C_3 = \frac{1}{1+1+1+1} = \frac{1}{4}\,\mu F$.
- The $n^{th}$ row has $2^{n-1}$ capacitors in series. Its equivalent capacitance is $C_n = \frac{1}{2^{n-1}}\,\mu F$.
Since all these rows are connected in parallel,the total equivalent capacitance $C_{eq}$ is the sum of the capacitances of each row:
$C_{eq} = C_1 + C_2 + C_3 + ... = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...$
This is an infinite geometric progression with the first term $a = 1$ and common ratio $r = \frac{1}{2}$.
The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$.
Substituting the values: $C_{eq} = \frac{1}{1 - 1/2} = \frac{1}{1/2} = 2\,\mu F$.

(A) Let the equivalent capacitance of the ladder between points $A$ and $B$ be $C_{eq}$.
If we add one more section to the ladder,the equivalent capacitance remains unchanged because the ladder is infinite or effectively behaves as such for the condition to hold.
Looking at the circuit,the first section consists of a $2\,\mu F$ capacitor in parallel with the rest of the ladder,which is in series with a $4\,\mu F$ capacitor.
Thus,$C_{eq} = 2 + \frac{4 \times C_{eq}}{4 + C_{eq}}$.
Solving for $C_{eq}$:
$C_{eq} - 2 = \frac{4 C_{eq}}{4 + C_{eq}}$
$(C_{eq} - 2)(4 + C_{eq}) = 4 C_{eq}$
$4 C_{eq} + C_{eq}^2 - 8 - 2 C_{eq} = 4 C_{eq}$
$C_{eq}^2 - 2 C_{eq} - 8 = 0$
$(C_{eq} - 4)(C_{eq} + 2) = 0$
Since capacitance cannot be negative,$C_{eq} = 4\,\mu F$.
For the ladder to be independent of the number of sections,the terminating capacitor $C$ must be equal to this equivalent capacitance $C_{eq}$.
Therefore,$C = 4\,\mu F$.

(C) In a series combination,the charge $Q$ on each capacitor is the same.
First,calculate the maximum charge each capacitor can hold:
$(Q_1)_{max} = C_1 V_1 = 1\ \mu F \times 6\ kV = 6\ \mu C$.
$(Q_2)_{max} = C_2 V_2 = 3\ \mu F \times 4\ kV = 12\ \mu C$.
Since the capacitors are in series,the maximum charge the system can hold is limited by the capacitor with the smaller maximum charge,which is $6\ \mu C$.
When the system holds a charge of $Q = 6\ \mu C$,the potential difference across $C_1$ is $V_1 = 6\ kV$ and the potential difference across $C_2$ is $V_2 = Q / C_2 = 6\ \mu C / 3\ \mu F = 2\ kV$.
The total maximum voltage the series combination can withstand is $V_{total} = V_1 + V_2 = 6\ kV + 2\ kV = 8\ kV$.

(C) In a series combination of capacitors,the charge $Q$ on each capacitor is the same.
The potential difference across a capacitor is given by $V = \frac{Q}{C}$.
From the graph,the potential drop across the first capacitor $(C_1)$ is greater than the potential drop across the second capacitor $(C_2)$.
Let $V_1$ be the potential difference across $C_1$ and $V_2$ be the potential difference across $C_2$.
Since $V_1 > V_2$ and $Q = C_1 V_1 = C_2 V_2$,we have $\frac{C_1}{C_2} = \frac{V_2}{V_1}$.
Since $V_1 > V_2$,it follows that $\frac{V_2}{V_1} < 1$,which implies $\frac{C_1}{C_2} < 1$,or ${C_1} < {C_2}$.

(A) The energy stored in a capacitor is given by $U = \frac{1}{2} C_{eq} V^2$,where $C_{eq}$ is the equivalent capacitance and $V$ is the potential difference across the combination.
Since the potential difference $V$ is constant,the stored energy $U$ is directly proportional to the equivalent capacitance $C_{eq}$ $(U \propto C_{eq})$.
To maximize the stored energy,we must maximize the equivalent capacitance $C_{eq}$.
For a given set of capacitors,the equivalent capacitance is maximum when they are connected in parallel $(C_{eq} = C_1 + C_2 + C_3)$.
Therefore,all capacitors should be connected in parallel to maximize the stored energy.

(D) Let the capacitance of each capacitor be $C = 1 \ \mu\text{F}$.
If two capacitors are connected in series,their equivalent capacitance is $C_s = \frac{C \times C}{C + C} = \frac{C}{2} = 0.5 \ \mu\text{F}$.
If this combination is connected in parallel with the third capacitor,the total equivalent capacitance is $C_{\text{eff}} = C_s + C = 0.5 \ \mu\text{F} + 1 \ \mu\text{F} = 1.5 \ \mu\text{F}$.
Therefore,the third capacitor is connected in parallel with the series combination of the other two capacitors.

(D) The given circuit is a balanced Wheatstone bridge.
In a Wheatstone bridge,if the ratio of capacitors in opposite arms is equal,the central capacitor $(C_5 = 20 \ \mu F)$ carries no charge and can be removed.
Here,$\frac{C_1}{C_3} = \frac{6 \ \mu F}{6 \ \mu F} = 1$ and $\frac{C_2}{C_4} = \frac{6 \ \mu F}{6 \ \mu F} = 1$.
Since the ratios are equal,the bridge is balanced.
Removing $C_5$,the circuit simplifies to two parallel branches:
Branch $1$: $C_1$ and $C_2$ are in series. $C_{s1} = \frac{6 \times 6}{6 + 6} = 3 \ \mu F$.
Branch $2$: $C_3$ and $C_4$ are in series. $C_{s2} = \frac{6 \times 6}{6 + 6} = 3 \ \mu F$.
Total effective capacitance $C_{eq} = C_{s1} + C_{s2} = 3 + 3 = 6 \ \mu F$.

(C) Since the capacitors $C_1$ and $C_2$ are connected in series,the charge $Q$ on both capacitors must be the same.
Let $V_D$ be the potential at point $D$.
The charge on capacitor $C_1$ is $Q = C_1(V_1 - V_D)$.
The charge on capacitor $C_2$ is $Q = C_2(V_D - V_2)$.
Equating the charges,we get: $C_1(V_1 - V_D) = C_2(V_D - V_2)$.
Expanding the terms: $C_1V_1 - C_1V_D = C_2V_D - C_2V_2$.
Rearranging to solve for $V_D$: $C_1V_1 + C_2V_2 = C_1V_D + C_2V_D$.
$C_1V_1 + C_2V_2 = V_D(C_1 + C_2)$.
Therefore,$V_D = \frac{C_1V_1 + C_2V_2}{C_1 + C_2}$.

(A) For capacitors connected in parallel, the equivalent capacitance $C_{eq}$ is the sum of individual capacitances.
$C_{eq} = C_1 + C_2 + C_3 + \dots + C_n$
Given the values are $C, 2C, 4C, \dots, \infty$, the equivalent capacitance is:
$C_{eq} = C + 2C + 4C + \dots + \infty$
$C_{eq} = C(1 + 2 + 4 + \dots + \infty)$
The series $(1 + 2 + 4 + \dots)$ is a geometric progression with a common ratio $r = 2$. Since $r > 1$, the sum of this infinite series diverges to infinity.
Therefore, $C_{eq} = C \times \infty = \infty$.

(A) The circuit consists of four parallel branches connected between points $A$ and $B$.
$1$. The top branch has two capacitors of capacitance $C$ in series,so their equivalent capacitance is $C_{1} = \frac{C \times C}{C + C} = \frac{C}{2}$.
$2$. The bottom branch also has two capacitors of capacitance $C$ in series,so its equivalent capacitance is $C_{4} = \frac{C}{2}$.
$3$. The middle two branches form a balanced Wheatstone bridge structure. The capacitors in the middle branch (between nodes $D$ and $E$) do not carry any charge because the potential at $D$ and $E$ is the same. Thus,the central capacitor can be ignored.
$4$. For the second branch,the series combination of two $C$ capacitors gives $\frac{C}{2}$. Similarly,for the third branch,the series combination gives $\frac{C}{2}$.
$5$. Since all four branches are in parallel,the total equivalent capacitance is $C_{AB} = \frac{C}{2} + \frac{C}{2} + \frac{C}{2} + \frac{C}{2} = 2C$.
Therefore,the value is $2$.

(B) First,calculate the equivalent capacitance of $C_1$ and $C_2$ in series:
$C_{12} = \frac{C_1 C_2}{C_1 + C_2} = \frac{2 \times 6}{2 + 6} = \frac{12}{8} = 1.5 \,\mu F$.
Next,calculate the total equivalent capacitance $C_{eq}$ of the circuit,where $C_{12}$ is in parallel with $C_3$:
$C_{eq} = C_{12} + C_3 = 1.5 \,\mu F + 4 \,\mu F = 5.5 \,\mu F = 5.5 \times 10^{-6} \,F$.
The energy supplied by the battery is given by the formula $E = \frac{1}{2} C_{eq} V^2$,where $V = 2 \,V$:
$E = \frac{1}{2} \times (5.5 \times 10^{-6}) \times (2)^2$
$E = \frac{1}{2} \times 5.5 \times 10^{-6} \times 4$
$E = 11 \times 10^{-6} \,J$.

(A) For capacitors in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.
Substituting the values: $\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = \frac{6}{6} = 1\ \mu F$.
The total charge $Q$ supplied by the battery is $Q = C_{eq} \times V = 1\ \mu F \times 24\ V = 24\ \mu C$.
In a series circuit,the charge on each capacitor is the same,so $Q = 24\ \mu C$ for the $6\ \mu F$ capacitor.
The potential difference $V'$ across the $6\ \mu F$ capacitor is $V' = \frac{Q}{C} = \frac{24\ \mu C}{6\ \mu F} = 4\ V$.

(C) When capacitors are connected in series,the equivalent capacitance $C_{eq}$ is given by the formula: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.
Since $C_1 = C_2 = C_3 = C$,we have $\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C}$.
Therefore,$C_{eq} = \frac{C}{3}$.
For capacitors in series,the total breakdown voltage $V_{total}$ is the sum of the individual breakdown voltages of each capacitor: $V_{total} = V_1 + V_2 + V_3$.
Given $V_1 = V_2 = V_3 = V$,we get $V_{total} = V + V + V = 3V$.
Thus,the equivalent capacitance is $C/3$ and the equivalent breakdown voltage is $3V$.

(C) For capacitors connected in series,the equivalent capacitance $C_{eq}$ is given by the formula:
$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ... + \infty$
Substituting the given values:
$\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{4C} + \frac{1}{16C} + ... + \infty$
$\frac{1}{C_{eq}} = \frac{1}{C} \left( 1 + \frac{1}{4} + \frac{1}{16} + ... + \infty \right)$
The term in the bracket is an infinite geometric progression ($G$.$P$.) with the first term $a = 1$ and common ratio $r = \frac{1}{4}$.
The sum of an infinite $G$.$P$. is given by $S_{\infty} = \frac{a}{1 - r}$.
$S_{\infty} = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.
Therefore,$\frac{1}{C_{eq}} = \frac{1}{C} \times \frac{4}{3} = \frac{4}{3C}$.
Thus,$C_{eq} = \frac{3}{4}C = 0.75C$.

(D) For $n$ capacitors each of capacitance $C$,the minimum capacitance is obtained when they are connected in series: $C_{min} = \frac{C}{n} = \frac{6 \ \mu F}{3} = 2 \ \mu F$.
The maximum capacitance is obtained when they are connected in parallel: $C_{max} = n \times C = 3 \times 6 \ \mu F = 18 \ \mu F$.
Therefore,the minimum and maximum capacitances are $2 \ \mu F$ and $18 \ \mu F$ respectively.

(B) The capacitors are connected in parallel to the battery of $E$ volts.
Since the potential difference $V$ across both capacitors is the same $(V = E)$,the charge $Q$ on each capacitor is given by $Q = CV$.
Therefore,the ratio of the charges is:
$\frac{Q_1}{Q_2} = \frac{C_1 V}{C_2 V} = \frac{C_1}{C_2}$
Given $C_1 = 4 \mu F$ and $C_2 = 2 \mu F$:
$\frac{Q_1}{Q_2} = \frac{4 \mu F}{2 \mu F} = \frac{2}{1}$
Thus,the ratio is $2 : 1$.

(B) $1$. The capacitors $6\, \mu F$ and $3\, \mu F$ are connected in parallel. Their equivalent capacitance is $C_p = 6\, \mu F + 3\, \mu F = 9\, \mu F$.
$2$. Now,this equivalent capacitor $C_p = 9\, \mu F$ is in series with the $9\, \mu F$ capacitor.
$3$. The total capacitance $C_{eq}$ of the circuit is given by $\frac{1}{C_{eq}} = \frac{1}{9} + \frac{1}{9} = \frac{2}{9}$,so $C_{eq} = 4.5\, \mu F$.
$4$. The total charge $Q$ supplied by the battery is $Q = C_{eq} \times V = 4.5\, \mu F \times 10\, V = 45\, \mu C$.
$5$. Since the $9\, \mu F$ capacitor is in series with the parallel combination,the same charge $Q = 45\, \mu C$ flows through it.
$6$. The potential difference $V_9$ across the $9\, \mu F$ capacitor is $V_9 = \frac{Q}{C} = \frac{45\, \mu C}{9\, \mu F} = 5\, V$.

(B) In a series combination,the charge $Q$ on each capacitor is the same.
First,calculate the equivalent capacitance $C_{eq}$ for the two capacitors in series:
$C_{eq} = \frac{C_1 \times C_2}{C_1 + C_2} = \frac{2.0 \times 8.0}{2.0 + 8.0} \mu F = \frac{16}{10} \mu F = 1.6 \mu F$.
The total charge $Q$ supplied by the source is given by $Q = C_{eq} \times V$.
$Q = 1.6 \times 10^{-6}\ F \times 300\ V = 480 \times 10^{-6}\ C = 4.8 \times 10^{-4}\ C$.
Since the capacitors are in series,the charge on the $2.0\ \mu F$ capacitor is the same as the total charge $Q$,which is $4.8 \times 10^{-4}\ C$.

(C) $1$. Observe the circuit carefully. The two $1\ \mu F$ capacitors connected to point $A$ are in parallel with each other.
$2$. The equivalent capacitance of these two parallel capacitors is $C_p = 1\ \mu F + 1\ \mu F = 2\ \mu F$.
$3$. Now,this $C_p$ is in series with the other $1\ \mu F$ capacitor. The equivalent capacitance of this series branch is $C_s = \frac{2 \times 1}{2 + 1} = \frac{2}{3}\ \mu F$.
$4$. Finally,this $C_s$ is in parallel with the $2\ \mu F$ capacitor connected directly between $A$ and $B$.
$5$. The total equivalent capacitance is $C_{AB} = C_s + 2\ \mu F = \frac{2}{3} + 2 = \frac{2 + 6}{3} = \frac{8}{3}\ \mu F$.

(A) In a series connection,the charge $Q$ on each capacitor is the same.
The energy stored in a capacitor is given by $U = \frac{Q^2}{2C}$.
Since $Q$ is constant for both capacitors in series,the energy stored is inversely proportional to the capacitance: $U \propto \frac{1}{C}$.
Therefore,the ratio of energy stored in the first capacitor $(U_1)$ to the second capacitor $(U_2)$ is:
$\frac{U_1}{U_2} = \frac{C_2}{C_1} = \frac{0.3 \ \mu F}{0.6 \ \mu F} = \frac{1}{2} = 0.5$.

(C) From the circuit diagram,$C_4$ is connected directly across the battery of potential $V$. Therefore,the charge on $C_4$ is $Q_4 = C_4 V = (4C)V = 4CV$.
The upper branch consists of $C_3$,$C_2$,and $C_1$ in series. The equivalent capacitance $C_{eq}$ of this branch is given by $\frac{1}{C_{eq}} = \frac{1}{C_3} + \frac{1}{C_2} + \frac{1}{C_1} = \frac{1}{3C} + \frac{1}{2C} + \frac{1}{C} = \frac{2+3+6}{6C} = \frac{11}{6C}$.
Thus,$C_{eq} = \frac{6C}{11}$.
The charge on this branch is $Q_{branch} = C_{eq} V = \frac{6CV}{11}$.
Since $C_2$ is in series in this branch,the charge on $C_2$ is $Q_2 = Q_{branch} = \frac{6CV}{11}$.
The ratio of charges is $\frac{Q_2}{Q_4} = \frac{6CV/11}{4CV} = \frac{6}{11 \times 4} = \frac{6}{44} = \frac{3}{22}$.

(A) By analyzing the circuit diagram,we can see that all three capacitors $(C_1, C_2, C_3)$ are connected in parallel between points $a$ and $b$.
Each capacitor has a value of $3 \ \mu F$.
For capacitors connected in parallel,the equivalent capacitance is given by the sum of individual capacitances:
$C_{eq} = C_1 + C_2 + C_3$
$C_{eq} = 3 \ \mu F + 3 \ \mu F + 3 \ \mu F = 9 \ \mu F$.
Therefore,the equivalent capacitance between points $a$ and $b$ is $9 \ \mu F$.

(B) The circuit consists of two parallel branches connected to a battery of potential $V$.
Branch $A$ contains capacitors $C_1$,$C_2$,and $C_3$ in series. The equivalent capacitance of this branch is given by:
$\frac{1}{C_A} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{C} + \frac{1}{2C} + \frac{1}{3C} = \frac{6+3+2}{6C} = \frac{11}{6C}$
Thus,$C_A = \frac{6C}{11}$.
The charge on branch $A$ is $Q_A = C_A V = \frac{6CV}{11}$. Since $C_1, C_2, C_3$ are in series,the charge on each is the same,so $Q_{C2} = Q_A = \frac{6CV}{11}$.
Branch $B$ contains only capacitor $C_4$ in parallel with branch $A$. The potential across $C_4$ is $V$,so the charge on $C_4$ is $Q_{C4} = C_4 V = 4CV$.
The ratio of charges on $C_2$ and $C_4$ is:
$\frac{Q_{C2}}{Q_{C4}} = \frac{6CV/11}{4CV} = \frac{6}{44} = \frac{3}{22}$.

(D) The given circuit is a balanced Wheatstone bridge.
In a balanced Wheatstone bridge,the central capacitor (between $B$ and $D$) does not contribute to the charge flow,so it can be removed.
Now,the two upper capacitors of $6\ \mu F$ each are in series,giving an equivalent capacitance of $C_1 = (6 \times 6) / (6 + 6) = 3\ \mu F$.
Similarly,the two lower capacitors of $6\ \mu F$ each are in series,giving an equivalent capacitance of $C_2 = (6 \times 6) / (6 + 6) = 3\ \mu F$.
These two combinations ($C_1$ and $C_2$) are in parallel with each other.
Therefore,the total effective capacitance is $C_{eq} = C_1 + C_2 = 3\ \mu F + 3\ \mu F = 6\ \mu F$.

(C) The given circuit can be simplified as follows:
$1$. The two capacitors of $4\, \mu F$ and $2\, \mu F$ are connected in parallel between point $P$ and point $B$. Their equivalent capacitance is $C_{eq} = 4\, \mu F + 2\, \mu F = 6\, \mu F$.
$2$. Now, the circuit consists of a $3\, \mu F$ capacitor in series with a $6\, \mu F$ capacitor, connected between $A$ and $B$.
$3$. Since the capacitors are in series, the charge $Q$ on both must be the same.
$4$. Using $Q = CV$, we have $C_1(V_A - V_P) = C_{eq}(V_P - V_B)$.
$5$. Substituting the values: $3\, \mu F \times (1200 - V_P) = 6\, \mu F \times (V_P - 0)$.
$6$. $3(1200 - V_P) = 6V_P \implies 1200 - V_P = 2V_P \implies 3V_P = 1200$.
$7$. Therefore, $V_P = 400\, V$.

(D) For capacitors in parallel,the equivalent capacitance is $C_{eq} = C_1 + C_2 + C_3$.
Given $C_1 + C_2 + C_3 = 12$ $... (i)$
Given $C_1 \cdot C_2 \cdot C_3 = 48$ $... (ii)$
Given $C_1 + C_2 = 6$ $... (iii)$
Substituting $(iii)$ into $(i)$: $6 + C_3 = 12 \implies C_3 = 6$ units.
Substituting $C_3 = 6$ into $(ii)$: $C_1 \cdot C_2 \cdot 6 = 48 \implies C_1 \cdot C_2 = 8$.
We know $(C_1 - C_2)^2 = (C_1 + C_2)^2 - 4C_1 C_2$.
$(C_1 - C_2)^2 = (6)^2 - 4(8) = 36 - 32 = 4$.
Therefore,$C_1 - C_2 = 2$ $... (iv)$.
Solving $(iii)$ and $(iv)$: $2C_1 = 8 \implies C_1 = 4$ and $C_2 = 2$.
Thus,the capacitances are $C_1 = 4, C_2 = 2, C_3 = 6$.

(C) $1$. The two $2 \, \mu F$ capacitors are in parallel. Their equivalent capacitance is $C_p = 2 \, \mu F + 2 \, \mu F = 4 \, \mu F$.
$2$. Now,the circuit consists of three $4 \, \mu F$ capacitors in series (the original top $4 \, \mu F$,the original left $4 \, \mu F$,and the original bottom $4 \, \mu F$ are not in series,let's re-examine the diagram).
$3$. Looking at the diagram: The two $2 \, \mu F$ capacitors are in parallel,giving $4 \, \mu F$. This $4 \, \mu F$ is in series with the top $4 \, \mu F$ capacitor,giving $C_{top} = (4 \times 4) / (4 + 4) = 2 \, \mu F$.
$4$. Similarly,the bottom $4 \, \mu F$ capacitor is in series with the left $4 \, \mu F$ capacitor,giving $C_{bottom} = (4 \times 4) / (4 + 4) = 2 \, \mu F$.
$5$. Finally,these two branches ($C_{top}$ and $C_{bottom}$) are in parallel between $A$ and $B$. Thus,$C_{AB} = 2 \, \mu F + 2 \, \mu F = 4 \, \mu F$.

(C) From the circuit diagram,capacitor $C_1$ is in series with the parallel combination of $C_2$ and $C_3$.
For capacitors in series,the charge $Q$ remains the same. Thus,the charge on $C_1$ is equal to the total charge flowing through the parallel combination: $Q_1 = Q_2 + Q_3$.
For capacitors in parallel,the potential difference $V$ across them is the same. Thus,$V_2 = V_3$.
The total potential $V$ of the battery is the sum of the potential across $C_1$ and the potential across the parallel combination: $V = V_1 + V_2$ (or $V = V_1 + V_3$).
Comparing this with the given options,option $C$ correctly describes the charge distribution and the potential relationship.