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Q11Q

Expert-verifiedFound in: Page 739

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**You are to connect capacitances ${{\mathit{C}}}_{{\mathbf{1}}}$**** and ${{\mathit{C}}}_{{\mathbf{2}}}$****, with ${{\mathit{C}}}_{{\mathbf{1}}}{\mathbf{>}}{{\mathit{C}}}_{{\mathbf{2}}}$****, 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

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

The capacitance ${C}_{1}$ and role="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,

$\frac{1}{{C}_{eq}}=\sum _{}\frac{1}{C}$ …(ii)

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

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

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

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