Timeline for Calculating the impedance formula of an inductor
Current License: CC BY-SA 4.0
19 events
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| Sep 19, 2022 at 0:00 | history | tweeted | twitter.com/StackElectronix/status/1571650395998486530 | ||
| Sep 18, 2022 at 19:59 | history | became hot network question | |||
| Sep 18, 2022 at 18:57 | vote | accept | ozgun can | ||
| Sep 18, 2022 at 17:19 | answer | added | Andy aka | timeline score: 2 | |
| Sep 18, 2022 at 16:56 | comment | added | The Photon | @ozguncan, your circuit has no loss, therefore the initial transient will never die out and this circuit will never reach steady state. You could start out by doing the phaser analysis to find the steady state solution, and then choose an initial condition (a choice of \$i(0)\$) that is consistent with that solution. Or you could add a loss term (a series resistor) to the circuit and solve it in the time domain as you have done, and find the behavior after the transient decays...then take the limit as the resistor value goes to zero to find the steady state condition for your original circuit. | |
| Sep 18, 2022 at 13:37 | comment | added | ozgun can | @Neil_UK then, the only non-assumptionic way( unlike , by saying, current must be sinusoid function , thus taking its derivative. and then finding it) to find the impedances, is by using indefinite integral instead of the definite integral ? | |
| Sep 18, 2022 at 13:32 | comment | added | user319836 | @ozguncan: No, that will not work. The \$sin(-\infty)\$ is undefined. | |
| Sep 18, 2022 at 13:23 | history | edited | a concerned citizen | CC BY-SA 4.0 |
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| Sep 18, 2022 at 13:17 | comment | added | ozgun can | if so, is there anyway that you calculate the impedance from voltage in the definite integral form of inductor current, which is from time= minus infinity to present | |
| Sep 18, 2022 at 13:10 | comment | added | user319836 | Impedance is a property of a passive element that is independent of any signal. It exists whether or not the current and voltage are present. It is the transfer function for transforming current into voltage. The initial condition has nothing to do with impedance. | |
| Sep 18, 2022 at 13:05 | comment | added | Neil_UK | Zero! Zero current, or zero voltage? Work out both, and post the results. They are different, and have different transients. Remember impedance is what you get when the initial transient has died down. In a circuit with no resistors, there's no loss, and the initial transient never dies down, so it's going to be quite difficult to intuit what it is. As you have a simulation, put in a small amount of series R, and simulate it until the transient has finished. | |
| Sep 18, 2022 at 13:04 | answer | added | Amit M | timeline score: 1 | |
| Sep 18, 2022 at 13:03 | answer | added | user319836 | timeline score: 2 | |
| Sep 18, 2022 at 13:00 | comment | added | ozgun can | I assume initial conditions as zero, isn't that what makes it steady state? , if not, where should the adjustment be | |
| Sep 18, 2022 at 12:50 | comment | added | Neil_UK | Impedance is only defined in the steady state. Your expression seems to include the startup transient as well, with an arbitrary phase depending on where you take your t=0. | |
| Sep 18, 2022 at 12:43 | history | edited | ocrdu | CC BY-SA 4.0 |
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| Sep 18, 2022 at 12:02 | history | edited | ozgun can | CC BY-SA 4.0 |
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| S Sep 18, 2022 at 11:59 | review | First questions | |||
| Sep 18, 2022 at 13:04 | |||||
| S Sep 18, 2022 at 11:59 | history | asked | ozgun can | CC BY-SA 4.0 |