Why in an RL circuit does the voltage anticipate the current?

  • #1

Main Question or Discussion Point

In my idea the tension change after connecting the circuit to a battery is immediate, while the current takes some time before moving. That should be sufficient to prove my question but, so why in a RC circuit the opposite happens?
 

Answers and Replies

  • #2
7,969
4,651
It's not really true. Ohm's law V=IR is a circular relation. Neither V nor I comes first.
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
In AC circuits, a C gives a 90 degree phase shift, and L gives a 90 degree phase shift in the opposite direction.
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?
V and I shifted by 90deg, what does it mean?
 
  • #4
FactChecker
Science Advisor
Gold Member
5,245
1,864
If there is no charge in a capacitor and a battery is connected across it, there is initially no opposition to the current flowing in to charge the plates. It is only after some charge has accumulated in the plates that resistance to the current builds up. That is when the voltage across the capacitor rises. So initially, the current starts high and the voltage starts low across the capacitor. If there are other components in the circuit, they complicate things.
 
  • #5
tech99
Gold Member
1,775
610
The 90 degree phase shift description only applies if we have a continuous AC supply. It just means that the voltage across L is a quarter of a cycle behind the current. L does not anticipate anything because the current was provided by the previous cycle.
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).
 
  • #6
7,969
4,651
If there is no charge in a capacitor and a battery is connected across it, there is initially no opposition to the current flowing in to charge the plates. I
Careful how you word that, it sounds like it would result in infinite current. The dV/dt term opposes current instantaneously.

But in a circuit what does it mean? V and I shifted by 90deg, what does it mean?
V or I, not V and I. It means we get imaginary power. We call that VAR for volt-amp-reactive.
VARs are very important in the power grid. See this Insights articles.

https://www.physicsforums.com/insights/ac-power-analysis-part-1-basics/

https://www.physicsforums.com/insights/ac-power-analysis-part-2-network-analysis/
 
  • #7
FactChecker
Science Advisor
Gold Member
5,245
1,864
Careful how you word that, it sounds like it would result in infinite current. The dV/dt term opposes current instantaneously.
Good point. It is only good for an intuitive feel for how the current could start high and the voltage low.
 
  • #8
Vanadium 50
Staff Emeritus
Science Advisor
Education Advisor
2019 Award
23,728
6,141
Remember ELI the ICE man.

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.
 
  • #9
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
13,782
5,604
Let's do the math. Using Faraday's Law, integrating along a circuit, including a resistance, an ideal coil and an AC voltage source (described by ##U(t)=U_0 \exp(\mathrm{i} \omega t)##, understanding that the physical real quantities are the real part, because exponentials are much simpler to handle than cos and sin), you get
$$u+L/R \dot{u}=U_0 \exp(\mathrm{i} \omega t),$$
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.
 
  • #10
7,969
4,651
There's an important point that I should have mentioned before.

Provided that we satisfy the assumptions of circuit analysis, then Kirchoff's Laws KVL and KCL are satisfied at every instant of the transient and in the stead state.

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.
 
  • #11
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
13,782
5,604
You don't need electrons to understand circuits. Continuum (fluid) dynamics is more than enough ;-)). Of course, AC circuit theory is based on the quasistationary approximations (of both kinds, electric and magnetic).
 
  • #12
L does not anticipate anything because the current was provided by the previous cycle.
But what about cycle 0?
 
  • #13
tech99
Gold Member
1,775
610
But what about cycle 0?
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.
 
  • #14
FactChecker
Science Advisor
Gold Member
5,245
1,864
At switch-on, we have to calculate the impulse response of the circuit.
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.

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
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

The impulse is composed of all frequencies.
Not only one? for example the 50Hz at home
 
  • #16
FactChecker
Science Advisor
Gold Member
5,245
1,864
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
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.
Not only one? for example the 50Hz at home
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:

a-Unit-step-function-in-the-time-domain-b-Frequency-spectrum-of-the-unit-step.png
 
  • #17
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
13,782
5,604
Of course, if you want to describe the time-dependence of such processes you need to solve the complete differential equation, including the transient states implementing the initial conditions. Of course you can work with Fourier transformations or, since you have an initial-value problem, somewhat more conveniently with the Laplace transformation (the latter being the preferred method by electrical engineers, while physicists usually prefer the Fourier transformation).
 

Related Threads for: Why in an RL circuit does the voltage anticipate the current?

Replies
4
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
3
Replies
50
Views
6K
  • Last Post
Replies
4
Views
916
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
2K
Replies
2
Views
7K
  • Last Post
Replies
6
Views
1K
Top