Discussion
A 3000 F supercapacitor charged to 2.7 V stores about
Answer: C
Correct answer: 3.0 Wh
The energy in a capacitor is
\[ E = \tfrac{1}{2}CV^2 = 0.5 \times 3000 \times 2.7^2 = 0.5 \times 3000 \times 7.29 \approx 10\,935 \ \mathrm{J} \]\[ \frac{10\,935}{3600} \approx 3.0 \ \mathrm{Wh} \]Three watt hours from a device the size of a large drink can shows the limitation: a lithium ion cell of the same volume holds perhaps thirty times as much. What the supercapacitor offers instead is the ability to deliver it in a second and to do so a million times.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development