Timeline for Pure inductor as load in inverter
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 20, 2017 at 17:48 | comment | added | Andy aka | @AnshKumar see my answer here: electronics.stackexchange.com/questions/281398/… | |
| Jan 20, 2017 at 17:33 | comment | added | Vedanshu | If the current starts from 0, in a time period the current will end to 0, making the inverter operation useless because we will be getting dc current and I(peak) = 20 which is wrong. I(peak) = 10. | |
| Jan 20, 2017 at 17:32 | comment | added | calcium3000 | Good call! Forgot to mention that. | |
| Jan 20, 2017 at 17:31 | history | edited | calcium3000 | CC BY-SA 3.0 |
thanks to Neil's comment for the reminder
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| Jan 20, 2017 at 17:29 | comment | added | Neil_UK | ... and the current will start from zero | |
| Jan 20, 2017 at 17:29 | history | answered | calcium3000 | CC BY-SA 3.0 |