Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

In Fig. 25-31, a 20.0 V battery is connected across capacitors of capacitances $C_{1} = C_{6} = 3.00 μF and C_{3} = C_{5} = 2.00 C_{2} = 2.00 C_{4} = 4.00 μF$ What are (a) the equivalent capacitance $C_{eq}$ of the capacitors and (b) the charge stored by $C_{eq}$ ? What are (c) $V_{1}$ and (d) role="math" localid="1661748621904" $q_{1}$ of capacitor 1, (e) role="math" localid="1661748675055" $V_{2}$ and (f) $q_{2}$ of capacitor 2, and (g) $V_{3}$ and (h) $q_{3}$ of capacitor 3?

a) The equivalent capacitance is $C_{eq} = 3 μF$

b) The charge stored by role="math" localid="1661748823028" $C_{eq} is q = 60 μC$

c) The value of $V_{1} is V_{1} = 10 V$

d) The charge of the capacitor $1 is q_{1} = 30 μC$

e) The value of $V_{2} is V_{1} = 10 V$

f) The charge of the capacitor $2 is q_{2} = 20 μC$

g) The value of $V_{3} is V_{3} = 5 V$

h) The charge $q_{3}$ of the capacitor $2 is q_{3} = 20 μC$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given

The potential of the battery is V = 20 V

$C_{1} = C_{6} = 3 μF C_{3} = C_{5} = 4 μF C_{2} = C_{4} = 2 μF$

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, $C_{eq} = {\sum C_{j}}_{j = 1}^{n}$

For series combination, $\frac{1}{C_{eq}} = {\sum \frac{1}{C_{j}}}_{j = 1}^{n}$

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 $C_{3} and C_{5}$ connected them in a series. Find the equivalent capacitance for this combination by using the formula,

$\frac{1}{C_{eq}} = {\sum \frac{1}{C_{j}}}_{j = 1}^{n} \frac{1}{C_{35}} = \frac{1}{C_{3}} + \frac{1}{C_{5}} = \frac{C_{5} + C_{3}}{C_{3} C_{5}} = \frac{4 μF + 4 μF}{4 μF \times 4 μF} = \frac{1}{2 μF} C_{35} = 2 μF$

Thus, equivalent capacitance is $2 μF$ .

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

$C_{eq} = {\sum C_{j}}_{j = 1}^{n} C = C_{35} + C_{2} + C_{4} = 2 μF + 2 μF + 2 μF = 6 μF$

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

$C' = C_{1} + C_{6} = 3 μF + 3 μF = 6 μF$ p;1

Thus, the equivalent capacitance of the circuit is,

$C_{eq} = \frac{CC'}{C + C'} = \frac{6 μF \times 6 μF}{6 μF + 6 μF} = \frac{36 μF}{12} = 3 μF$

Hence, the equivalent capacitance is $C_{eq} = 3 μF$

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,

$q = C_{eq} V = 3 μF \times 20 V = (3 \times 10^{- 6} F) \times 20 V = 60 \times 10^{- 6} (\frac{1 μC}{10^{- 6} C}) = 60 μC$

Hence, the charge stored by $C_{eq} is q = 60 μC$

Step 5: (c) Determining the value of V1

The potential difference across $C_{1}$ is given by,

$V_{1} = \frac{CV}{C + C'} = \frac{6 μF \times 20 V}{6 μF + 6 μF} = \frac{120 V}{12} = 10 V$

Hence, the value of potential across $C_{1} is V_{1} = 10 V$

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

The charge carried by $C_{1}$ is,

$q_{1} = C_{1} V_{1} = 3 μF \times 10 V = 30 μC$

Hence, the charge of the capacitor 1 is $q_{1} = 30 μC$

Step 7: (e) Determining the value of V2

The potential difference across $C_{2}$ can be calculated as,

$V = V_{1} + V_{2} V_{2} = V - V_{1} = 20 V - 10 V V = 10 V$

Hence, the value of $V_{2} is V_{2} = 10 V$

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

The charge carried by $C_{2}$ is,

$q_{2} = C_{2} V_{2} = 2 μF \times 10 V = 20 μC$

Hence, the charge $q_{2}$ of the capacitor 2 is $q_{2} = 20 μC$

Step 9: (g) Determining the value of V3

Since the potential difference $V_{2}$ is divided equally between $C_{3} and C_{5}$ . The potential difference across $C_{3}$ is,

$V_{3} = \frac{V_{2}}{2} = \frac{10 V}{2} = 5 V$

Hence, the value of $V_{3} is V_{3} = 5 V$

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

The charge carried by $q_{3}$ is,

$q_{3} = C_{3} V_{3} = 4 μF \times 5 V = 20 μC$

Hence, the charge $q_{3}$ of the capacitor 3 is $q_{3} = 20 μC$

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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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Given

Step 2: Determining the concept

Step 3: (a) Determining the equivalent capacitance

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

Step 5: (c) Determining the value of V1

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

Step 7: (e) Determining the value of V2

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

Step 9: (g) Determining the value of V3

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

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Given

Step 2: Determining the concept

Step 3: (a) Determining the equivalent capacitance

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

Step 5: (c) Determining the value of V1

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

Step 7: (e) Determining the value of V2

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

Step 9: (g) Determining the value of V3

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

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