91Ó°ÊÓ

The capacitance $C_1$androle="math" localid="1661337756030" $C_2$with$C_1>C_2$ is given.

Using Eq.25-1, we can find the amount of charge on individual capacitors. Then using 25-20 and 25-19, we can find the equivalent capacitance when they are connected in series and parallel respectively. Again using Eq.25-1, we can find the amount of charge stored.

Formulae:

The charge within the plates of the capacitor, q=CV …(i)

If capacitors are in series, the equivalent capacitance $C_{eq}$is given by,

$\frac{1}{C_{eq}}=\underset{}{∑}\frac{1}{C}$ …(¾±¾±)

If capacitors are in parallel, the equivalent capacitance $C_{eq}$is given by,

role="math" localid="1661337878259" $C_{eq}=\stackrel{}{∑C}$ …(¾±¾±¾±)

First, let’s assume that the capacitors 1 and 2 are connected individually.

Thus, the amount of charge on the capacitors using equation (i) are given as:

$q_1=C_{1}V$ …(¾±±¹)

and

$q_2=C_{2}V$ …(±¹)

Here, we see that the potential difference is same as that of the battery.

Now, let’s assume that the capacitors1 and 2 are in series.

Then, the equivalent capacitance of the capacitors using equation (ii) is given as follows:

$C_{eq}=\frac{C_1{C}_2}{C_1+C_2}$

Now, the charge of the equivalent capacitance is given using equation (i) as follows: localid="1661338068588" $q_s=\left(\frac{C_1{C}_2}{C_1+C_2}\right)V$ …(±¹¾±)

Now, let’s assume that capacitor 1 and 2in parallel.

Then, the equivalent capacitance of the capacitors using equation (iii) is given by:

$C_{eq}=C_1+C_2$

Now, the charge of the equivalent capacitance is given using equation (i) as follows:

$q_p=\left(C_1+C_2\right)V$ …(±¹¾±¾±)

Since,$C_1>C_2$

Thus, comparing Eq.(iv), (v), (vi) and (vii), we get

$q_p>q_1>q_2>q_s$

Therefore, the ranking of the arrangements according to the amount of charge stored is $\mathrm{Parallel}\mathrm{capacitance}>\mathrm{Indvidual}\mathrm{capacitance}>\mathrm{Series}\mathrm{capacitance}$.