Chapter 20: Problem 96
Two capacitors are connected to a battery. The battery voltage is \(V=60.0 \mathrm{V}\) , and the capacitances are \(C_{1}=2.00 \mu \mathrm{F}\) and \(C_{2}=4.00 \mu \mathrm{F}\) . Determine the total energy stored by the two capacitors when they are wired \((\mathrm{a})\) in parallel and \((\mathrm{b})\) in series.
Short Answer
Step by step solution
Understanding the Parallel Configuration
Calculating Energy in Parallel Configuration
Understanding the Series Configuration
Calculating Energy in Series Configuration
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Capacitors
- When voltage is applied across the plates, electric charges accumulate on the plates.
- This creates an electric field that stores energy.
- The quantity of charge a capacitor can hold is directly proportional to its capacitance, which is measured in farads (F).
Energy Storage
- Energy increases with higher capacitance and greater voltage.
- The energy is directly proportional to the square of the voltage.
- For the same capacitance, doubling the voltage increases energy four-fold.
Series and Parallel Circuits
- Series connections generally result in a smaller equivalent capacitance.
- Parallel connections increase overall capacitance and energy storage capacity.
Equivalent Capacitance
- For parallel capacitors: Join up their capacitances directly since they combine their electric charge storage capacity, leading to a greater equivalent capacitance.
- For series capacitors: Calculate the reciprocal of the sum of the reciprocals of individual capacitances, resulting in a smaller equivalent capacitance.