A 2.4 µF capacitor and a 3.9 µF capacitor are connected in parallel across a 230 V potential difference. Calculate the total energy in joules stored in the capacitors. Number i Units eTextbook and Media Save for Later Attempts: 0 of 2 used Submit Answer
A 2.4 µF capacitor and a 3.9 µF capacitor are connected in parallel across a 230 V potential difference. Calculate the total energy in joules stored in the capacitors. Number i Units eTextbook and Media Save for Later Attempts: 0 of 2 used Submit Answer
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)...
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![**Educational Content: Capacitors in Parallel**
---
**Problem Statement:**
A 2.4 µF capacitor and a 3.9 µF capacitor are connected in parallel across a 230 V potential difference. Calculate the total energy in joules stored in the capacitors.
---
**Input Fields:**
- **Number**: [ ]
- **Units**: [ (Dropdown Menu) ]
**Resources:**
- **eTextbook and Media**: [Link/Button]
**Interactive Features:**
- **Save for Later**: [Button]
- **Attempts: 0 of 2 used**
**Submission:**
- **Submit Answer**: [Button]
---
**Explanation for Energy Stored in Capacitors:**
Capacitors store electrical energy in an electric field between their plates. When connected in parallel, the total capacitance \(C_{total}\) is the sum of the individual capacitances:
\[ C_{total} = C_1 + C_2 \]
Where \(C_1\) and \(C_2\) are the capacitances of the individual capacitors.
The total energy \(E\) stored in the capacitors can be calculated using the formula:
\[ E = \frac{1}{2} C_{total} V^2 \]
Where \(V\) is the potential difference across the capacitors.
**Steps to Solve:**
1. Calculate the total capacitance:
\[ C_{total} = 2.4 \ \mu F + 3.9 \ \mu F = 6.3 \ \mu F \]
2. Convert the capacitance to farads (\(\mu F\) to F):
\[ C_{total} = 6.3 \times 10^{-6} \ F \]
3. Use the energy formula to find \(E\):
\[ E = \frac{1}{2} \times 6.3 \times 10^{-6} \times (230)^2 \]
4. Simplify the calculation:
\[ E = \frac{1}{2} \times 6.3 \times 10^{-6} \times 52900 \]
\[ E = 166.485 \times 10^{-6} \]
\[ E = 0.166485 \ Joules \]
**Answer:** The total energy stored in the capacitors is approximately](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fba35cc5f-959d-4f80-acdd-e9cc0bd31dfa%2Fb1c325c0-9b1f-43e1-9a80-35a2d12825a9%2F0dwftka_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Educational Content: Capacitors in Parallel**
---
**Problem Statement:**
A 2.4 µF capacitor and a 3.9 µF capacitor are connected in parallel across a 230 V potential difference. Calculate the total energy in joules stored in the capacitors.
---
**Input Fields:**
- **Number**: [ ]
- **Units**: [ (Dropdown Menu) ]
**Resources:**
- **eTextbook and Media**: [Link/Button]
**Interactive Features:**
- **Save for Later**: [Button]
- **Attempts: 0 of 2 used**
**Submission:**
- **Submit Answer**: [Button]
---
**Explanation for Energy Stored in Capacitors:**
Capacitors store electrical energy in an electric field between their plates. When connected in parallel, the total capacitance \(C_{total}\) is the sum of the individual capacitances:
\[ C_{total} = C_1 + C_2 \]
Where \(C_1\) and \(C_2\) are the capacitances of the individual capacitors.
The total energy \(E\) stored in the capacitors can be calculated using the formula:
\[ E = \frac{1}{2} C_{total} V^2 \]
Where \(V\) is the potential difference across the capacitors.
**Steps to Solve:**
1. Calculate the total capacitance:
\[ C_{total} = 2.4 \ \mu F + 3.9 \ \mu F = 6.3 \ \mu F \]
2. Convert the capacitance to farads (\(\mu F\) to F):
\[ C_{total} = 6.3 \times 10^{-6} \ F \]
3. Use the energy formula to find \(E\):
\[ E = \frac{1}{2} \times 6.3 \times 10^{-6} \times (230)^2 \]
4. Simplify the calculation:
\[ E = \frac{1}{2} \times 6.3 \times 10^{-6} \times 52900 \]
\[ E = 166.485 \times 10^{-6} \]
\[ E = 0.166485 \ Joules \]
**Answer:** The total energy stored in the capacitors is approximately
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