Timeline for Transient analysis of inductor circuit and back EMF
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
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| Aug 25, 2023 at 15:32 | answer | added | user347520 | timeline score: 1 | |
| Jun 23, 2021 at 18:02 | answer | added | SystemTheory | timeline score: 0 | |
| Jun 23, 2021 at 15:35 | comment | added | G36 | Can you explain to me why do you think that "mother nature" will overshoot so badly and allowed the current to jumps and next to compensate it the emf to jump well beyond supply voltage? Why can't "mother nature" knows that the current through the inductor ( for a DC input voltage) needs to rise at a steady rate determined by the amount of inductance and the amount of supply voltage? | |
| Jun 22, 2021 at 12:33 | comment | added | G36 | Everyone understands the capacitors and that he stores his energy in form of a voltage (in electric field). But we have a problem with the inductor because the induction is charged via current and store energy in a magnetic field. And the energy is always conserved, energy cannot simply appear or vanish. Since energy is a function of current for any inductance, the amount of current through a fixed inductor’s coil cannot vary unless energy is added to or removed from that inductance. And this is why we can't have a sudden change in the inductor current. | |
| Jun 21, 2021 at 19:12 | comment | added | Matt | But there is a very brief sudden change in the current that generates the induced voltage \$V_L\$ so we get \$V_L=L\frac{di}{dt}\$ right? | |
| Jun 21, 2021 at 15:33 | comment | added | G36 | There will be no current "spike" because at T0 we applying (imposed across the inductor) the voltage from a voltage source. Thus , V_L = V_S, and at this instant of time, the current will start to flow so the di/dt is satisfied. And we are using this in DC to DC converters. | |
| Jun 21, 2021 at 12:29 | answer | added | sarthak | timeline score: 0 | |
| Jun 21, 2021 at 12:06 | answer | added | tobalt | timeline score: 1 | |
| Jun 21, 2021 at 7:39 | comment | added | Matt | Maybe I quote your answer from your other thread. "At the beginning (at time 0+) we apply +10V across the inductor by doing this we are attempting to cause a sudden change in the current. The induced voltage now steps in and tries to keep the current down to its initial value (0A)". What I was trying to ask is that does the sudden change in current actually happened? If so, does the current go up high so when the induced voltage steps in, the current drops? Thanks | |
| Jun 21, 2021 at 7:33 | answer | added | Bruce Abbott | timeline score: 2 | |
| Jun 21, 2021 at 7:05 | comment | added | G36 | What do you mean by saying "current actually shoot up before the back EMF steps in". And maybe this will help you allaboutcircuits.com/textbook/direct-current/chpt-15/… | |
| Jun 21, 2021 at 6:04 | comment | added | Matt | hi, I've read your answer on the above thread. I did get some insights from it but I still have something that I am not so sure with that I hope you could clarify for me. First, when we apply +V across the inductor to cause a sudden change in current, does the current actually shoot up before the back EMF steps in or only the \$\frac{di}{dt}\$ goes up? Thanks! | |
| Jun 21, 2021 at 4:56 | comment | added | G36 | electronics.stackexchange.com/questions/470171/… | |
| Jun 21, 2021 at 4:33 | history | edited | Matt | CC BY-SA 4.0 |
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| Jun 21, 2021 at 4:26 | history | edited | Matt | CC BY-SA 4.0 |
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| Jun 21, 2021 at 4:21 | history | edited | Matt | CC BY-SA 4.0 |
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| Jun 21, 2021 at 3:26 | history | edited | Matt | CC BY-SA 4.0 |
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| Jun 21, 2021 at 3:14 | history | edited | Matt | CC BY-SA 4.0 |
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| Jun 21, 2021 at 3:09 | history | edited | Matt | CC BY-SA 4.0 |
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| Jun 21, 2021 at 0:43 | answer | added | ErikR | timeline score: 0 | |
| Jun 21, 2021 at 0:00 | history | asked | Matt | CC BY-SA 4.0 |