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Andrea Vironda

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- Thread starter Andrea Vironda
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Andrea Vironda

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anorlunda

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There is also the complex form of Ohm's law that we use for AC circuits, V=IZ where V, I and Z are all complex and Z can include R and L and C components.

When we have an inductor, there is a term ##L\frac{dI}{dt}##.

When we have a capacitor, there is a term ##C\frac{dV}{dt}##.

Neither of those create a first/next relationship.

On the other hand, for basic DC circuit analysis, I see no harm in visualizing V or I to come first. You can think that we need a V to create an I in RL circuits, and you need an I to create a V in RC circuits. Whatever turns you on, but the true answer is that V=IR is circular and simultaneous.

In AC circuits, a C gives a 90 degree phase shift, and L gives a 90 degree phase shift in the opposite direction.

- #3

Andrea Vironda

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At this point i understand the phase practically speaking, for example when some AC motor is not well connected, it will turn in the wrong direction. But in a circuit what does it mean?In AC circuits, a C gives a 90 degree phase shift, and L gives a 90 degree phase shift in the opposite direction.

V and I shifted by 90deg, what does it mean？

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tech99

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If we connect L to a battery, the current grows linearly towards infinity!

An inductor behaves like a heavy mass - its hard to get it moving and hard to stop it. To simulate AC, imagine pushing and pulling the mass back and forth and notice when it has velocity (current) and when it requires a push (voltage).

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anorlunda

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Careful how you word that, it sounds like it would result in infinite current. The dV/dt term opposes current instantaneously.If there is no charge in a capacitor and a battery is connected across it, there is initially no opposition to the current flowing into charge the plates. I

V or I, not V and I. It means we get imaginary power. We call that VAR for volt-amp-reactive.But in a circuit what does it mean? V and I shifted by 90deg, what does it mean？

VARs are very important in the power grid. See this Insights articles.

https://www.physicsforums.com/insights/ac-power-analysis-part-2-network-analysis/

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Good point. It is only good for an intuitive feel for how the current could start high and the voltage low.Careful how you word that, it sounds like it would result in infinite current. The dV/dt term opposes current instantaneously.

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Vanadium 50

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I work with big magnets, and they are essentially big inductors. I think in terms of energetics: when you first turn it on, you don't get a steady-state until all the energy that belongs in the field is there. Then the device looks resistive. Likewise, when you shut it off, current doesn't stop flowing until the energy is exhausted.

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$$u+L/R \dot{u}=U_0 \exp(\mathrm{i} \omega t),$$

where ##u## is the voltage drop along the resistor.

Usually in circuit theory one considers only the stationary state and ignores the "transients" which decay exponentially on time scales ##\tau=L/R##. The stationary state is then the one, where ##u## is oscillating with the external angular frequency ##\omega##. Thus we make the ansatz

$$u(t)=u_0 \exp(\mathrm{i} \omega t),$$

where ##u_0## is a complex constant. Plugging this Ansatz into the differential equation, we get

$$(1 + \mathrm{i} \omega L/R)u_0 \exp(\mathrm{i} \omega t)=U_0 \exp(\mathrm{i} \omega t),$$

leading to

$$u_0=\frac{U_0}{1+\mathrm{i} \omega L/R} = \frac{1-\mathrm{i} \omega L/R}{1+\omega^2 L^2/R^2} U_0.$$

Now you can write

$$\frac{1-\mathrm{i} \omega L/R}{1+\omega^2 L^2/R^2}=r \exp(\mathrm{i} \varphi)$$

with

$$r=\frac{1}{\sqrt{1+\omega^2 L^2/R^2}}, \quad \varphi=-\arccos \left (\frac{1}{r} \right)<0,$$

which implies that

$$u(t)=r \exp(\mathrm{i} \omega t+\varphi)$$

or

$$\mathrm{Re} u(t)=r \cos(\omega t+\varphi).$$

Since ##\varphi<0## the phase of ##u## stays behind by this phaseshift. The current is of course

$$i(t)=\frac{1}{R} u(t)=\frac{r}{R} \cos(\omega t+\varphi).$$

Thus the phase of the current also lacks behind the source's phase by this same phaseshift.

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anorlunda

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Provided that we satisfy the assumptions of circuit analysis, then Kirchoff's Laws KVL and KCL are satisfied

Particularly important in this case is the assumption that the propagation of EM waves from one side of the circuit to the other at near light speed happens in negligible time. If the R and the L in the circuit were separated by 10000 km of wire, ordinary circuit analysis can't be used.

To have a first/next cause/effect view of voltage or current would violate KVL or KCL. I suspect that those who insist that voltage comes first, secretly harbor the mental model that electrons are like billiard balls. Thinking of electrons is very unhelpful understanding circuits.

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- #12

Andrea Vironda

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But what about cycle 0?L does not anticipate anything because the current was provided by the previous cycle.

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tech99

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We can only use the 90 degree phase shift concept when the AC supply is continuous. At switch-on, we have to calculate the impulse response of the circuit.But what about cycle 0?

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The impulse is composed of all frequencies, with very high frequencies dominating the initial time t=0. At that time, the capacitor acts approximately like a simple wire for those very high frequencies. That should give some intuitive feel for the initial transient response.At switch-on, we have to calculate the impulse response of the circuit.

PS. I think @Andrea Vironda should clarify whether the subject is an RL circuit, as the title specifies, or an RC circuit, as the initial post specifies.

- #15

Andrea Vironda

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Not only one? for example the 50Hz at homeThe impulse is composed of all frequencies.

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If you short-circuit a battery, there is a very small resistance in the shorting wire and the voltage drop across the shorting wire is small. (There is still an internal resistance in the battery, but we can ignore that for now.) The current is very high.In my opinion the problem is similar for both circuits. When i switch on I and V should have the same phase. And then, after some time, they would be dephase by inductor or condensator

No. An impulse or a step input both have components of frequencies from low to very high. It take very high frequencies to give the sharp increase. Here is the unit step function in the time domain and its associated frequency spectrum:Not only one? for example the 50Hz at home

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