Real inductor I/V relationship - Page 1
Forum section: electronics / projects

author: ElectronicsGeek
author_rank: Contributor
timestamp: July 12, 2017, 05:50:32 pm
content: The I/V relationship for an inductor is usually given by v = L di/dt, however, the voltage induced by the inductor is actually opposite the applied voltage, so why isn't it v = -Ldi/dt instead ?_____+ L - _______||+V In this circuit we have a voltage source V, and an inductor L. Look at the polarity of the inductor. If v = Ldi/dt, then the voltage will be against the- source i the current is increasing, and with the source if the current is decreasing. In this way the inductor has the correct polarity. Is this why it's| Ldi/dt rather than -Ldi/dt ?|----

author: Benta
author_rank: Super Contributor
timestamp: July 12, 2017, 06:32:14 pm
content: You have to differentiate between applied voltage and induced voltage. Your equation is for applied voltage.

author: free_electron
author_rank: Super Contributor
timestamp: July 12, 2017, 06:53:52 pm
content: what about the series resistance ? inter winding capacitance ? resistance drift due to heating ?how come those aspects are never considered or explained ?

author: ElectronicsGeek
author_rank: Contributor
timestamp: July 12, 2017, 07:41:27 pm
content: Quote from: Benta on July 12, 2017, 06:32:14 pmAs you can see the induced voltage is the applied voltage in this case as there is nothing between it and the L

author: suicidaleggroll
author_rank: Super Contributor
timestamp: July 12, 2017, 08:26:46 pm
content: Quote from: ElectronicsGeek on July 12, 2017, 07:41:27 pmThere's also nothing between the other side of the L and the - on the applied voltage. It's an incomplete circuit, there is no current, and the entire inductor is at the same voltage as the + terminal on the source. If you'd like to analyze what's happening in the circuit, you need to complete it first.

author: ElectronicsGeek
author_rank: Contributor
timestamp: July 12, 2017, 08:53:59 pm
content: Quote from: suicidaleggroll on July 12, 2017, 08:26:46 pmJust assume there is a ground there. is there a current now?

author: rfeecs
author_rank: Frequent Contributor
timestamp: July 12, 2017, 09:14:21 pm
content: Quote from: ElectronicsGeek on July 12, 2017, 05:50:32 pmThis is simply a convention:https://en.wikipedia.org/wiki/Passive_sign_conventionDon't overthink it.

author: Benta
author_rank: Super Contributor
timestamp: July 12, 2017, 09:15:18 pm
content: Your question is meaningless, as u (not v) is constant from an ideal source and will take precedent.The only result you can get from this is ut/L = i, which will yield a constant current rise if u is a constant DC voltage.Setting up an equation to solve for u if u is already given is nonsense.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: July 12, 2017, 09:17:56 pm
content: Quote from: ElectronicsGeek on July 12, 2017, 08:53:59 pmSo the voltage source is shorted across the inductor? I'm assuming it's a DC source?The instant you connect it there will be no current, applied voltage and induced voltage on the inductor are equal and opposite.A fraction of a second later, current will start to flow. Applied voltage will be slightly higher than the induced voltage on the inductor, and as a result you'll get some current flowing through the inductor, I = (Vapplied - Vinduced)/ESR, where ESR is the sum of the source, inductor, and wiring resistance.As time goes on, Vinduced will continue to drop until it approaches 0, at which point I = Vapplied/ESR

author: ElectronicsGeek
author_rank: Contributor
timestamp: July 12, 2017, 09:25:28 pm
content: Quote from: Benta on July 12, 2017, 09:15:18 pmIt's an Ac source not a DC source. Are you crazy ?

author: ElectronicsGeek
author_rank: Contributor
timestamp: July 12, 2017, 09:28:46 pm
content: Its an AC source as DC clearly would be nonsense here folks.It seems I am the only one who knows this stuff round here folks

author: uwezi
author_rank: Unknown
timestamp: July 12, 2017, 09:37:42 pm
content: Quote from: free_electron on July 12, 2017, 06:53:52 pmDefinenever.Of course all these additional, parasitic parameters are important, explained and used. However, the resulting equation system is extremely non-linear. Most of the time you will see that the real device is split/replaced by a new model, which is a series and parallel combination of the main inductance and all the parasitics as individual devices. Most of the time you will try to use devices in your circuits which are dominated by the main component, however, for high or low frequencies, large currents etc the importance can and will shift...In LTspice for example the following equivalent circuit is used:http://ltwiki.org/LTspiceHelp/LTspiceHelp/L_Inductor.htm

author: uwezi
author_rank: Unknown
timestamp: July 12, 2017, 09:51:21 pm
content: Quote from: ElectronicsGeek on July 12, 2017, 09:28:46 pmIf you are the only one, then what is your question? Connecting an inductor to an AC voltage source like this (assuming the open end of the inductor is connected to ground) will force the voltage over the inductor to be identical to the voltage of the voltage source. The current through the inductor will be lagging after the voltage, which you can see when you integrate the equation once to solve for the current.If you start at a point without any current through the inductor and apply a positive step voltage, the current through the inductor will flow from the positive of the voltage source. The current will increase linearly with time under the step.If you apply a sine wave AC voltage then the current through an ideal inductor will lag 90 degrees or 1/4 of a period after the voltage waveform, i.e. when the sine voltage reaches the positive maximum value the current will be on the rising edge just passing through zero.v(t) = L di(t)/dtintegr v(t) dt = L i(t) + consti(t) = integr v(t) dt / L = integr sin(wt) dt /L = - cos(wt)/L

author: mikerj
author_rank: Super Contributor
timestamp: July 12, 2017, 09:53:48 pm
content: Quote from: ElectronicsGeek on July 12, 2017, 09:28:46 pmNonsense. The equation in the OP post simply relates voltage to rate of change of current which could be either DC or AC.

author: Benta
author_rank: Super Contributor
timestamp: July 12, 2017, 10:29:09 pm
content: Quote from: ElectronicsGeek on July 12, 2017, 09:25:28 pmQuoteWatch your manners. With a handful of posts here, an obvious superiority complex based on apparently very limited knowledge, I think Facebook et al is more like your field. Civility and respect would bring you further... folk.