Timeline for Calculating charge time of a large capacitor from a boost with a limited power source
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 3, 2019 at 14:49 | vote | accept | J. Jones | ||
| May 31, 2019 at 22:33 | comment | added | TimWescott | Your confusion confuses me. It's simple conservation of energy. 1st-year college physics, if not high-school physics. Power is the rate of energy change. Energy is energy. Time = amount / rate. So time = (energy change) / (energy rate of change). | |
| May 31, 2019 at 21:56 | comment | added | J. Jones | What formula are you using to come up with this expression that the energy needed divided by the total energy equals time? | |
| May 30, 2019 at 19:04 | comment | added | TimWescott | 2.5W = 2.5J/s. Conservation of energy says that if you're going to end up with 135mJ, you have to wait for 135mJ/2.5W (or 2.5J/s) to get it. | |
| May 30, 2019 at 18:44 | comment | added | J. Jones | Can you elaborate the comment regarding 135mJ divided by 2.5J/s? How is that the lower limit on the capacitor charge time? | |
| May 30, 2019 at 18:18 | history | answered | TimWescott | CC BY-SA 4.0 |