A 2.0 µF capacitor and a 4.0 µF capacitor are connected in parallel across a 300 V potential difference. Calculate the total energy stored in the capacitors.
A 2.0 µF capacitor and a 4.0 µF capacitor are connected in parallel across a 300 V potential difference. Calculate the total energy stored in the capacitors.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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![### Capacitors in Parallel and Energy Storage
**Problem Statement:**
A 2.0 µF capacitor and a 4.0 µF capacitor are connected in parallel across a 300 V potential difference. Calculate the total energy stored in the capacitors.
---
**Solution:**
To solve this problem, we need to follow these steps:
1. **Calculate the total capacitance:**
In a parallel connection, the total capacitance \( C_t \) is the sum of the individual capacitances:
\[
C_t = C_1 + C_2
\]
Where:
- \( C_1 = 2.0 \, \mu\text{F} \)
- \( C_2 = 4.0 \, \mu\text{F} \)
So,
\[
C_t = 2.0 \, \mu\text{F} + 4.0 \, \mu\text{F} = 6.0 \, \mu\text{F}
\]
2. **Calculate the energy stored in each capacitor:**
The energy \( U \) stored in a capacitor is given by the formula:
\[
U = \frac{1}{2} C V^2
\]
For \( C_1 \):
\[
U_1 = \frac{1}{2} \times 2.0 \, \mu\text{F} \times (300 \, V)^2 = \frac{1}{2} \times 2.0 \times 10^{-6} \, F \times 90000 \, V^2
\]
\[
U_1 = \frac{1}{2} \times 0.18 \, J = 0.09 \, J
\]
For \( C_2 \):
\[
U_2 = \frac{1}{2} \times 4.0 \, \mu\text{F} \times (300 \, V)^2 = \frac{1}{2} \times 4.0 \times 10^{-6} \, F \times 90000 \, V^2
\]
\[
U_2 = \frac{1}{2} \times 0.36](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fba35cc5f-959d-4f80-acdd-e9cc0bd31dfa%2F63569234-f6c8-485e-8a6c-8cdcc22961c2%2F4ljgqo8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Capacitors in Parallel and Energy Storage
**Problem Statement:**
A 2.0 µF capacitor and a 4.0 µF capacitor are connected in parallel across a 300 V potential difference. Calculate the total energy stored in the capacitors.
---
**Solution:**
To solve this problem, we need to follow these steps:
1. **Calculate the total capacitance:**
In a parallel connection, the total capacitance \( C_t \) is the sum of the individual capacitances:
\[
C_t = C_1 + C_2
\]
Where:
- \( C_1 = 2.0 \, \mu\text{F} \)
- \( C_2 = 4.0 \, \mu\text{F} \)
So,
\[
C_t = 2.0 \, \mu\text{F} + 4.0 \, \mu\text{F} = 6.0 \, \mu\text{F}
\]
2. **Calculate the energy stored in each capacitor:**
The energy \( U \) stored in a capacitor is given by the formula:
\[
U = \frac{1}{2} C V^2
\]
For \( C_1 \):
\[
U_1 = \frac{1}{2} \times 2.0 \, \mu\text{F} \times (300 \, V)^2 = \frac{1}{2} \times 2.0 \times 10^{-6} \, F \times 90000 \, V^2
\]
\[
U_1 = \frac{1}{2} \times 0.18 \, J = 0.09 \, J
\]
For \( C_2 \):
\[
U_2 = \frac{1}{2} \times 4.0 \, \mu\text{F} \times (300 \, V)^2 = \frac{1}{2} \times 4.0 \times 10^{-6} \, F \times 90000 \, V^2
\]
\[
U_2 = \frac{1}{2} \times 0.36
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