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ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
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![### Problem Statement
(a) Find the equivalent capacitance of the group of capacitors shown in the figure.
(b) Find the charge on the 7 µF capacitor if the potential difference of the battery is 90 V.
### Explanation
#### Capacitor Network Diagram:
The diagram indicates a network of capacitors connected in a mixed series-parallel combination. The capacitors and their respective capacitances are as follows:
- 5.00 µF
- 4.00 µF
- 6.00 µF
- 3.00 µF
- 2.00 µF
- 3.00 µF
- 7.00 µF
The battery connected to the network is 90 V.
##### Detailed Capacitor Network Configuration:
1. The 5.00 µF and 4.00 µF capacitors are in series.
2. The result of their series combination is parallel with the 6.00 µF capacitor.
3. This parallel combination is in series with 3.00 µF.
4. This resultant is in parallel with both 2.00 µF and a combination of 3.00 µF and 7.00 µF capacitors in series.
### Solution Approach:
**Finding Equivalent Capacitance (a):**
1. **Series Combination of Capacitors:**
- Capacitors in series: \( \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n} \)
For series combination of 5.00 µF and 4.00 µF:
\[
\frac{1}{C_{eq1}} = \frac{1}{5.00} + \frac{1}{4.00}
\]
\[
C_{eq1} \approx 2.22 \mu F
\]
2. **Parallel Combination of Capacitors:**
- Capacitors in parallel: \( C_p = C_1 + C_2 + \ldots + C_n \)
Parallel combination of \( C_{eq1} \) and 6.00 µF:
\[
C_{eq2} = 2.22 \mu F + 6.00 \mu F = 8.22 \mu F
\]](https://content.bartleby.com/qna-images/question/b6f89a01-a0d2-4d6f-ab48-f2d74370b2c9/87cf3cc8-8e4e-44a9-b924-3890c01fb2f8/qbgzaqv_processed.jpeg)
Transcribed Image Text:### Problem Statement
(a) Find the equivalent capacitance of the group of capacitors shown in the figure.
(b) Find the charge on the 7 µF capacitor if the potential difference of the battery is 90 V.
### Explanation
#### Capacitor Network Diagram:
The diagram indicates a network of capacitors connected in a mixed series-parallel combination. The capacitors and their respective capacitances are as follows:
- 5.00 µF
- 4.00 µF
- 6.00 µF
- 3.00 µF
- 2.00 µF
- 3.00 µF
- 7.00 µF
The battery connected to the network is 90 V.
##### Detailed Capacitor Network Configuration:
1. The 5.00 µF and 4.00 µF capacitors are in series.
2. The result of their series combination is parallel with the 6.00 µF capacitor.
3. This parallel combination is in series with 3.00 µF.
4. This resultant is in parallel with both 2.00 µF and a combination of 3.00 µF and 7.00 µF capacitors in series.
### Solution Approach:
**Finding Equivalent Capacitance (a):**
1. **Series Combination of Capacitors:**
- Capacitors in series: \( \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n} \)
For series combination of 5.00 µF and 4.00 µF:
\[
\frac{1}{C_{eq1}} = \frac{1}{5.00} + \frac{1}{4.00}
\]
\[
C_{eq1} \approx 2.22 \mu F
\]
2. **Parallel Combination of Capacitors:**
- Capacitors in parallel: \( C_p = C_1 + C_2 + \ldots + C_n \)
Parallel combination of \( C_{eq1} \) and 6.00 µF:
\[
C_{eq2} = 2.22 \mu F + 6.00 \mu F = 8.22 \mu F
\]
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