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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# In Fig. 25-31, a 20.0 V battery is connected across capacitors of capacitances ${{\mathbf{C}}}_{{\mathbf{1}}}{\mathbf{=}}{{\mathbf{C}}}_{{\mathbf{6}}}{\mathbf{=}}{\mathbf{3}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{\mu F}}{\mathbf{}}{\mathbf{and}}{\mathbf{}}{{\mathbf{C}}}_{{\mathbf{3}}}{\mathbf{=}}{{\mathbf{C}}}_{{\mathbf{5}}}{\mathbf{=}}{\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{{\mathbf{C}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{{\mathbf{C}}}_{{\mathbf{4}}}{\mathbf{=}}{\mathbf{4}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{\mu F}}$ What are (a) the equivalent capacitance ${{\mathbf{C}}}_{{\mathbf{eq}}}$ of the capacitors and (b) the charge stored by ${{\mathbf{C}}}_{{\mathbf{eq}}}$? What are (c) ${{\mathbf{V}}}_{{\mathbf{1}}}$ and (d) role="math" localid="1661748621904" ${{\mathbf{q}}}_{1}$ of capacitor 1, (e) role="math" localid="1661748675055" ${{\mathbf{V}}}_{{\mathbf{2}}}$ and (f) ${{\mathbf{q}}}_{{\mathbf{2}}}$of capacitor 2, and (g) ${{\mathbf{V}}}_{{\mathbf{3}}}$ and (h) ${{\mathbf{q}}}_{{\mathbf{3}}}$of capacitor 3?

a) The equivalent capacitance is ${\mathrm{C}}_{\mathrm{eq}}=3\mathrm{\mu F}$

b) The charge stored by role="math" localid="1661748823028" ${\mathrm{C}}_{\mathrm{eq}}\mathrm{is}\mathrm{q}=60\mathrm{\mu C}$

c) The value of ${\mathrm{V}}_{1}\mathrm{is}{\mathrm{V}}_{1}=10\mathrm{V}$

d) The charge of the capacitor $1\mathrm{is}{\mathrm{q}}_{1}=30\mathrm{\mu C}$

e) The value of ${\mathrm{V}}_{2}\mathrm{is}{\mathrm{V}}_{1}=10\mathrm{V}$

f) The charge of the capacitor $2\mathrm{is}{\mathrm{q}}_{2}=20\mathrm{\mu C}$

g) The value of ${\mathrm{V}}_{3}\mathrm{is}{\mathrm{V}}_{3}=5\mathrm{V}$

h) The charge ${\mathrm{q}}_{3}$ of the capacitor $2\mathrm{is}{\mathrm{q}}_{3}=20\mathrm{\mu C}$

See the step by step solution

## Step 1: Given

The potential of the battery is V = 20 V

${\mathrm{C}}_{1}={\mathrm{C}}_{6}=3\mathrm{\mu F}\phantom{\rule{0ex}{0ex}}{\mathrm{C}}_{3}={\mathrm{C}}_{5}=4\mathrm{\mu F}\phantom{\rule{0ex}{0ex}}{\mathrm{C}}_{2}={\mathrm{C}}_{4}=2\mathrm{\mu F}$

## Step 2: Determining the concept

Find the equivalent capacitance of the combination of different capacitors using the formula for equivalent capacitance connected in a series and in parallel. From the equivalent capacitance, find the charge stored by it. Find the potential difference and the charge stored across the capacitors using the formula for capacitance.

Formula:

q = CV

For parallel combination,${{\mathbf{C}}}_{{\mathbf{eq}}}{\mathbf{=}}\underset{\mathbf{j}\mathbf{=}\mathbf{1}}{\overset{\mathbf{n}}{\mathbf{\sum }{\mathbf{C}}_{\mathbf{j}}}}$

For series combination,$\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{eq}}}{\mathbf{=}}\underset{\mathbf{j}\mathbf{=}\mathbf{1}}{\overset{\mathbf{n}}{\mathbf{\sum }\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{j}}}}}$

Where C is capacitance, V is the potential difference, q is the charge on the capacitor

## Step 3: (a) Determining the equivalent capacitance

To find the equivalent capacitance, first, consider the capacitors${\mathrm{C}}_{3}\mathrm{and}{\mathrm{C}}_{5}$ connected them in a series. Find the equivalent capacitance for this combination by using the formula,

$\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\underset{\mathrm{j}=1}{\overset{\mathrm{n}}{\sum \frac{1}{{\mathrm{C}}_{\mathrm{j}}}}}\phantom{\rule{0ex}{0ex}}\frac{1}{{\mathrm{C}}_{35}}=\frac{1}{{\mathrm{C}}_{3}}+\frac{1}{{\mathrm{C}}_{5}}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{C}}_{5}+{\mathrm{C}}_{3}}{{\mathrm{C}}_{3}{\mathrm{C}}_{5}}\phantom{\rule{0ex}{0ex}}=\frac{4\mathrm{\mu F}+4\mathrm{\mu F}}{4\mathrm{\mu F}×4\mathrm{\mu F}}\phantom{\rule{0ex}{0ex}}=\frac{1}{2\mathrm{\mu F}}\phantom{\rule{0ex}{0ex}}{\mathrm{C}}_{35}=2\mathrm{\mu F}$

Thus, equivalent capacitance is $2\mathrm{\mu F}$.

This combination is then connected in parallel with ${\mathrm{C}}_{2}\mathrm{and}{\mathrm{C}}_{4}$. The resulting equivalent capacitance can be found by using the formula,

${\mathrm{C}}_{\mathrm{eq}}=\underset{\mathrm{j}=1}{\overset{\mathrm{n}}{\sum {\mathrm{C}}_{\mathrm{j}}}}\phantom{\rule{0ex}{0ex}}\mathrm{C}={\mathrm{C}}_{35}+{\mathrm{C}}_{2}+{\mathrm{C}}_{4}\phantom{\rule{0ex}{0ex}}=2\mathrm{\mu F}+2\mathrm{\mu F}+2\mathrm{\mu F}\phantom{\rule{0ex}{0ex}}=6\mathrm{\mu F}$

This is now in a series with another combination that consists of capacitors ${\mathrm{C}}_{1}\mathrm{and}{\mathrm{C}}_{6}$ in parallel. Find the equivalent capacitance for ${\mathrm{C}}_{1}\mathrm{and}{\mathrm{C}}_{6}$.

$\mathrm{C}\text{'}={\mathrm{C}}_{1}+{\mathrm{C}}_{6}\phantom{\rule{0ex}{0ex}}=3\mathrm{\mu F}+3\mathrm{\mu F}\phantom{\rule{0ex}{0ex}}=6\mathrm{\mu F}$p;1

Thus, the equivalent capacitance of the circuit is,

${\mathrm{C}}_{\mathrm{eq}}=\frac{\mathrm{CC}\text{'}}{\mathrm{C}+\mathrm{C}\text{'}}\phantom{\rule{0ex}{0ex}}=\frac{6\mathrm{\mu F}×6\mathrm{\mu F}}{6\mathrm{\mu F}+6\mathrm{\mu F}}\phantom{\rule{0ex}{0ex}}=\frac{36\mathrm{\mu F}}{12}\phantom{\rule{0ex}{0ex}}=3\mathrm{\mu F}$

Hence, the equivalent capacitance is ${\mathrm{C}}_{\mathrm{eq}}=3\mathrm{\mu F}$

## Step 4: (b) Determining the charge stored by Ceq

The potential difference supplied by the battery is 20 V, then the total charge stored by the equivalent capacitor is,

$\mathrm{q}={\mathrm{C}}_{\mathrm{eq}}\mathrm{V}\phantom{\rule{0ex}{0ex}}=3\mathrm{\mu F}×20\mathrm{V}\phantom{\rule{0ex}{0ex}}=\left(3×{10}^{-6}\mathrm{F}\right)×20\mathrm{V}\phantom{\rule{0ex}{0ex}}=60×{10}^{-6}\left(\frac{1\mathrm{\mu C}}{{10}^{-6}\mathrm{C}}\right)\phantom{\rule{0ex}{0ex}}=60\mathrm{\mu C}$

Hence, the charge stored by ${\mathrm{C}}_{\mathrm{eq}}\mathrm{is}\mathrm{q}=60\mathrm{\mu C}$

## Step 5: (c) Determining the value of V1

The potential difference across ${\mathrm{C}}_{1}$ is given by,

${\mathrm{V}}_{1}=\frac{\mathrm{CV}}{\mathrm{C}+\mathrm{C}\text{'}}\phantom{\rule{0ex}{0ex}}=\frac{6\mathrm{\mu F}×20\mathrm{V}}{6\mathrm{\mu F}+6\mathrm{\mu F}}\phantom{\rule{0ex}{0ex}}=\frac{120\mathrm{V}}{12}\phantom{\rule{0ex}{0ex}}=10\mathrm{V}$

Hence, the value of potential across ${\mathrm{C}}_{1}\mathrm{is}{\mathrm{V}}_{1}=10\mathrm{V}$

## Step 6: (d) Determining the charge  q1 of the capacitor 1

The charge carried by ${\mathrm{C}}_{1}$ is,

${\mathrm{q}}_{1}={\mathrm{C}}_{1}{\mathrm{V}}_{1}\phantom{\rule{0ex}{0ex}}=3\mathrm{\mu F}×10\mathrm{V}\phantom{\rule{0ex}{0ex}}=30\mathrm{\mu C}$

Hence, the charge of the capacitor 1 is ${\mathrm{q}}_{1}=30\mathrm{\mu C}$

## Step 7: (e) Determining the value of V2

The potential difference across ${\mathrm{C}}_{2}$ can be calculated as,

$\mathrm{V}={\mathrm{V}}_{1}+{\mathrm{V}}_{2}\phantom{\rule{0ex}{0ex}}{\mathrm{V}}_{2}=\mathrm{V}-{\mathrm{V}}_{1}\phantom{\rule{0ex}{0ex}}=20\mathrm{V}-10\mathrm{V}\mathrm{V}\phantom{\rule{0ex}{0ex}}=10\mathrm{V}$

Hence, the value of ${V}_{2}\mathrm{is}{\mathrm{V}}_{2}=10\mathrm{V}$

## Step 8: (f) Determining the charge q2 of the capacitor

The charge carried by ${\mathrm{C}}_{2}$ is,

${\mathrm{q}}_{2}={\mathrm{C}}_{2}{\mathrm{V}}_{2}\phantom{\rule{0ex}{0ex}}=2\mathrm{\mu F}×10\mathrm{V}\phantom{\rule{0ex}{0ex}}=20\mathrm{\mu C}$

Hence, the charge ${\mathrm{q}}_{2}$ of the capacitor 2 is ${\mathrm{q}}_{2}=20\mathrm{\mu C}$

## Step 9: (g) Determining the value of V3

Since the potential difference ${\mathrm{V}}_{2}$ is divided equally between ${\mathrm{C}}_{3}\mathrm{and}{\mathrm{C}}_{5}$. The potential difference across ${\mathrm{C}}_{3}$ is,

${\mathrm{V}}_{3}=\frac{{\mathrm{V}}_{2}}{2}\phantom{\rule{0ex}{0ex}}=\frac{10\mathrm{V}}{2}\phantom{\rule{0ex}{0ex}}=5\mathrm{V}$

Hence, the value of ${\mathrm{V}}_{3}\mathrm{is}{\mathrm{V}}_{3}=5\mathrm{V}$

## Step 10: (h) Determining the charge q3 of the capacitor 3

The charge carried by ${\mathrm{q}}_{3}$ is,

${\mathrm{q}}_{3}={\mathrm{C}}_{3}{\mathrm{V}}_{3}\phantom{\rule{0ex}{0ex}}=4\mathrm{\mu F}×5\mathrm{V}\phantom{\rule{0ex}{0ex}}=20\mathrm{\mu C}$

Hence, the charge ${q}_{3}$ of the capacitor 3 is ${\mathrm{q}}_{3}=20\mathrm{\mu C}$

Therefore, find the equivalent capacitance using the formula for equivalent capacitance connected in series and parallel. The potential difference and the charge stored across the capacitors can be found using the formula for capacitance.

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