### 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 µFThe 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: \[ C_{eq2} = 2.22 \mu F + 6.00 \mu F = 8.22 \mu F \]

Answered: Image /qna-images/answer/87cf3cc8-8e4e-44a9-b924-3890c01fb2f8.jpg

(a) Find the equivalent capacitance of the group of capacitors shown in Figure. (b) Find the charge on 7 uF capacitor if the potential difference of the battery is 90 V. 5.00 HF 3.00 µF 2.00 µF 4.00 µF 3.00 0 µF 6.00 HF 7.00 µF

(a) Find the equivalent capacitance of the group of capacitors shown in Figure. (b) Find the charge on 7 uF capacitor if the potential difference of the battery is 90 V. 5.00 HF 3.00 µF 2.00 µF 4.00 µF 3.00 0 µF 6.00 HF 7.00 µF

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

Q: Two capacitors, C₁ = 4.00 μF and C₂ = 10.0 μF, are connected in parallel, and the resulting…

A: a) 14.0μFb) V1 = 9.00V, V2 = 9.00Vc) Q1 = 36.0μC, Q2 = 90.0μCExplanation:Please see the attached…

Q: Two capacitors, C1 = 4.53 ?F and C2 = 11.5 ?F, are connected in parallel, and the resulting…

A: since the capacitors are connected in parallel Ceq=C1+C2 =4.53+11.5 =16.03F Answer:…

Q: Consider the following. (LetC₁ = 25.60 μF and C₂ = 19.60 μF.) 6.00 με C₁ C₂ C₁ + 9.00 V (a) Find the…

Q: In the figure the battery has a potential difference of V = 10.0 V and the five capacitors each have…

Q: Each capacitor shown below has a capacitance of 5.0 uF. Initially, all the capacitors were…

Q: Two capacitors, C₁ = 4.44 μF and C₂ = 11.6 μF, are connected in parallel, and the resulting…

A: Given: C1=4.44 μFC2=11.6 μF The connection of the capacitors is in parallel. The supply voltage is 9…

Q: Two capacitors (C, = 8.00 µF and C, = 13.0 µF) are now connected in series and to a 9.00-V battery.…

A: Given: C1=8 μFC2=13 μF The connection of the capacitors is in series. The supply voltage is 9 V.

Q: Two capacitors, C1 = 6.00 µF and C2 = 11.0 µF, are connected in parallel, and the resulting…

Q: Two capacitors C, = 3.00 ÂµF and C2 = 14.0 ÂµF are connected in series to a 9.00 V battery. (a) Find…

Q: In the figure the battery has a potential difference of V = 10.0 V and the five capacitors each have…

Q: Two capacitors (C₁ = 4.00 μF and C₂ = 14.0 μF) are now connected in series and to a 9.00-V battery…

A: We are given 2 capacitors. We are given their values. We are also given the voltage of battery. We…

Q: Four capacitors are connected as shown in the figure below. (CE 14.0 UE 3.00 uF 20,.0 aF 6.00 pF (a)…

A: (a) The series combination of capacitor is, 1Cs=114+13Cs=14×33+14=4217=2.47 μF The parallel…

Q: How many 1.01 µF capacitors must be connected in parallel to store a charge of 1.36 C with a…

Q: the figure a 15 V battery is connected across capacitors of capacitances C1 = e (a) the equivalent…

A: Given:- A circuit with the capacitance is given C1 = C6 = 3 microfarad C3 = C5 = 1.5C2 = 1.5 C4…

Q: - In the figure a 27 V battery is connected across capacitors of capacitances C₁ C6 = 4.5 µF and C3…

Q: Problem 4: The battery has a potential difference of 10 V and each capacitor has a capacitance of C…

Q: Two capacitors, C1 = 4.00 µF and C2 = 13.0 µF, are connected in parallel, and the resulting…

A: We have:Part a:The equivalent capacitance of the capacitors connected in parallel is given by;

Q: Part (f) Express the energy stored in a capacitor in terms of capacitance C and the potential…

Q: Two capacitors C1 = 7.4 µF, C2 individually to V1 = 17.0 V, V2 = 3.5 V. The two = 14.7 µF are…

A: Given values :C1 = 7.4μF C2 = 14.7μF V1 = 17 V V2 = 3.5 V

Q: Two capacitors, C, = 6.00 µF and C, = 13.0 µF, are connected in parallel, and the resulting…

A: Given data: Two capacitors are connected in parallel C1 = 6.00 μF C2 = 13.0 μF Battery voltage (V)…

Q: Consider the following. (LetC, = 30.60 µF and C, = 24.60 µF.) 6.00 µF C2 + 9.00 V (a) Find the…

A: Given values: Voltage applied, V=9.00 V Capacitance : C1=30.60 μFC2=24.60 μF

Q: In the figure the battery has a potential difference of V = 10.0 V and the five capacitors each have…

Q: Problem 4: Three capacitors are connected as shown in the figure. C1 = 7.8 µF, C2 = 9.2 µF, C3 = 2.6…

Q: 24Je are in the of capacitors are C-1.SuF, C 3.SF, C 2.OuF, Ce 2SuF, C 3.0aF and C-4.0µF. The…

A: C1=1.5uF, C2= 3.5uF, C3= 2.0uF, C4=2.5uF, C5= 3.0uF C6=4.0uF voltage V = 12.0V

Q: capacitors are connected in series between the terminals of a 350-V battery. If their capacitances…

Q: The figure below shows three capacitors with capacitances CA = 1.00 μF, CB = 1.20 μF, and Cc = 3.30…

A: CA = 1 uf CB = 1.20 uf Cc = 3.30 uf V = 6 V

Concept explainers

Dielectric Constant Of Water

Water constitutes about 70% of earth. Some important distinguishing properties of water are high molar concentration, small dissociation constant and high dielectric constant.

Electrostatic Potential and Capacitance

An electrostatic force is a force caused by stationary electric charges /fields. The electrostatic force is caused by the transfer of electrons in conducting materials. Coulomb’s law determines the amount of force between two stationary, charged particles. The electric force is the force which acts between two stationary charges. It is also called Coulomb force.

Question

100%

### 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:
\[
C_{eq2} = 2.22 \mu F + 6.00 \mu F = 8.22 \mu F
\]

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step 3

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 3 steps with 3 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions

Find the following. (In the figure, use C, = 31.00 µF and C, = 25.00 µF.) 6.00 µF C2 µF CµF 9.00 V (a) the equivalent capacitance of the capacitors in the figure above uF (b) the charge on each capacitor on the right 31.00-µF capacitor on the left 31.00-µF capacitor μC on the 25.00-µF capacitor on the 6.00-µF capacitor
In the figure the battery has a potential difference of V = 10.0 V and the five capacitors each have a capacitance of 9.00 uF. What is the charge on (a) capacitor 1 and (b) capacitor 2? FF (a) Number i (b) Number i Units Units
The capacitance of a parallel-plate capacitor can be increased by: O increasing the charge O decreasing the charge O increasing the plate separation O decreasing the plate separation O decreasing the plate area
Find the following. (In the figure use C1 = 37.80 µF and C2 = 31.80 µF.) (a) the equivalent capacitance of the capacitors in the figure above µF(b) the charge on each capacitor on the right 37.80 µF capacitor µC on the left 37.80 µF capacitor µC on the 31.80 µF capacitor µC on the 6.00 µF capacitor µC (c) the potential difference across each capacitor on the right 37.80 µF capacitor V on the left 37.80 µF capacitor V on the 31.80 µF capacitor V on the 6.00 µF capacitor V
Find the equivalent capacitance Ceq of the combination of capacitors shown in the figure, where C = 0.825 µF, %3D C2 = 10.0 µF, and C3 = 1.25 µF. %3D %3D C, Cea HF %3D C, Question Credit: OpenStax College Physics
In the figure the battery potential difference V is 12.0V and each of the seven capacitors has capacitance 16.0 µF. What is the charge on (a) capacitor 1 and (b) capacitor 2? (a) Number i (b) Number Units Units C₂
Where a 2.7 V battery is connected to a network of capacitors with C1 = 54 F, C2 = 63 F, and C3 = 45 F. What is the equivalent capacitance connected to the battery? (in F) 162 F 17.7 F 36.0 F 80.2 F
Consider the following. (LetC, = 28.60 µF and C2 = 22.60 µF.) 6.00 μΕ 9.00 V (a) Find the equivalent capacitance of the capacitors in the figure. uF (b) Find the charge on each capacitor. on the right 28.60 uF capacitor Juc on the left 28.60 uF capacitor on the 22.60 uF capacitor UC on the 6.00 uF capacitor UC (c) Find the potential difference across each capacitor. on the right 28.60 uF capacitor on the left 28.60 µF capacitor on the 22.60 uF capacitor on the 6.00 uF capacitor > >
An arrangement of capacitors is shown in the figure below. (a) If C = 7.20 ✕ 10−5 F, what is the equivalent capacitance between points a and b?F(b) A battery with a potential difference of 21.00 V is connected to a capacitor with the equivalent capacitance. What is the energy stored by this capacitor? J
Two capacitors, C₁ = 6.00 μF and C₂ = 13.0 μF, are connected in parallel, and the resulting combination is connected to a 9.00-V battery. (a) Find the equivalent capacitance of the combination. UF (b) Find the potential difference across each capacitor. V₁1 V₂ = = (c) Find the charge stored on each capacitor. Q₁ UC μC = Q₂ V V =
A 1.00-?F capacitor is charged by being connected across a 9.0-V battery. It is then disconnected from the battery and connected across an uncharged 8.00-?F capacitor. Determine the resulting charge on each capacitor.
3.00 uF 6.00 uF 10. For the system of capacitors shown in Figure, find: (a) the equivalent capacitance of the system, (b) the voltage on each capacitor 2.00 μF 4.00 μF 90.0 V