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Expert-verified Found in: Page 739 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # You are to connect capacitances ${{\mathbit{C}}}_{{\mathbf{1}}}$ and ${{\mathbit{C}}}_{{\mathbf{2}}}$, with ${{\mathbit{C}}}_{{\mathbf{1}}}{\mathbf{>}}{{\mathbit{C}}}_{{\mathbf{2}}}$, to a battery, first individually, then in series, and then in parallel. Rank those arrangements according to the amount of charge stored, greatest first.

The ranking of the arrangements according to the amount of charge stored is

$\mathrm{Parallel}\mathrm{capacitance}>\mathrm{Indvidual}\mathrm{capacitance}>\mathrm{Series}\mathrm{capacitance}$.

See the step by step solution

## Step 1: The given data

The capacitance ${C}_{1}$ and role="math" localid="1661337756030" ${C}_{2}$ with ${C}_{1}>{C}_{2}$ is given.

## Step 2: Understanding the concept of the capacitance and charge

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,

…(ii)

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

role="math" localid="1661337878259" ${C}_{eq}=\stackrel{}{\sum C}$ …(iii)

## Step 3: Calculation of the ranking of the arrangements according to the amount of charge

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:

${q}_{1}={C}_{1}V$ …(iv)

and

${q}_{2}={C}_{2}V$ …(v)

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" ${q}_{s}=\left(\frac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}\right)V$ …(vi)

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

${q}_{p}=\left({C}_{1}+{C}_{2}\right)V$ …(vii)

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 $\mathrm{Parallel}\mathrm{capacitance}>\mathrm{Indvidual}\mathrm{capacitance}>\mathrm{Series}\mathrm{capacitance}$. ### Want to see more solutions like these? 