Facing issues in understanding a Purely Inductive Circuit

[Kaushik]

Consider a purely inductive circuit with an inductor of self-inductance ##L## and sinusoidally varying AC source of peak voltage ##V_{m}##.

First of all, Why is ##V_{ac} = ε_{ind}## (where ##ε_{ind}## is the back emf)?

Now, at ## t=0 ##, voltage is increasing at a fast rate. Hence, due to the sudden increase in current, the inductor induces maximum back emf (in the opposite direction) and current is maximum in the opposite direction.

Now, at ## t = T/4## (where ## T## is the time period), Voltage is maximum and not changing much. Now, this is where I think I'm going wrong. I am interpreting this as '##I## is not changing much because ##V##, when maximum, is not changing much (##\frac{dV}{dt} = 0##). Why is it wrong?

Interpreting it in the same way, when ##t = T/2##,## V = 0##. and ##\frac{dV}{dt} = -1##. So, the current is also changing at a fast rate. Due to this, there is a back emf induced that opposes the change in current. But here, Current is reducing rapidly. So, back emf is induced to increase the current in a positive direction.

I feel like I am misinterpreting it here: I am thinking of ##\frac{dV}{dt}## as the rate at which current is changing. And from the graph, it is quite evidently not the case. But, I can't find the reason as to why it is not the case intuitively?

I hope you got my question. If not, I apologise. Feel free to ask questions regarding 'my question'.

I am confused

 

[DaveE]

Sorry, I don't have time right now for a detailed reply. But...

I think you are overthinking this. First, give up on the whole idea of "back emf"; which, BTW, I have only heard of in the context of motors, transformers, coupled inductors, etc. These devices are much too complicated for you to worry about now. First you need to really appreciate the inductor equation V=L⋅(di/dt). This is equivalent to I=(1/L)⋅∫V⋅dt (with integration constants ignored, for now).

Have you studied calculus yet?

 

[Kaushik]

Have you studied calculus yet?

Yes!

 

[DaveE]

The Kahn Academy videos may help:


Plus others...

 

[anorlunda]

It might be easier for you to start with DC.

• Consider first a step change in DC voltage.
• Consider next a step change, then a step in the other direction. Now you have something that looks like a square wave.
• A sinusoid is like an approximation of a square wave. So AC imitates DC.

 

[Kaushik]

I'll share my understanding of it in DC voltage and a AC (but square wave).

• Consider first a step change in DC voltage

When we apply a voltage, the battery tries to send a current. But, when this current begins to flow through the inductor, there is a sudden increase in magnetic flux through the coil (self-induction). Due, to this sudden increase in magnetic flux, an emf is induced in the coil to oppose this change in current. This is the reason why the current increases slowly and then tapers at ##I_{max}##.

At t = 0, the emf induced is at its maximum because of the sudden increase in current through the coils (##\frac{dΦ}{dt}##). Then as time passes, the current stabilises and emf across the inductor reduces. Now,

Consider next a step change, then a step in the other direction. Now you have something that looks like a square wave.

In this, I can understand the behaviour of the current curve. But, I don't get why current is negative initially? Isn't is same as the DC source for the first T/2 time?

 

[DaveE]

I'll share my understanding of it in DC voltage and a AC (but square wave).
View attachment 270221

When we apply a voltage, the battery tries to send a current. But, when this current begins to flow through the inductor, there is a sudden increase in magnetic flux through the coil (self-induction). Due, to this sudden increase in magnetic flux, an emf is induced in the coil to oppose this change in current. This is the reason why the current increases slowly and then tapers at ##I_{max}##.

At t = 0, the emf induced is at its maximum because of the sudden increase in current through the coils (##\frac{dΦ}{dt}##). Then as time passes, the current stabilises and emf across the inductor reduces.Now,

View attachment 270222

In this, I can understand the behaviour of the current curve. But, I don't get why current is negative initially? Isn't is same as the DC source for the first T/2 time?

Everything you described here is an inductor in series with a resistor. If it was just an inductor the current waveform would be a triangle wave. The applied voltage would define a constant current slope di/dt=V/L.

 

[DaveE]

When we apply a voltage, the battery tries to send a current. But, when this current begins to flow through the inductor, there is a sudden increase in magnetic flux through the coil (self-induction). Due, to this sudden increase in magnetic flux, an emf is induced in the coil to oppose this change in current. This is the reason why the current increases slowly and then tapers at ##I_{max}##.

No. This is just wrong. As I said before your idea about self-induction, back emf, etc. doesn't really apply to a simple inductor. You are making this much too complicated. V=L(di/dt), or i=(1/L)∫Vdt. That is all you need to know about simple inductors. Honestly, I don't think there is an easier way to describe it that is sufficient, assuming you know a bit of calculus.

 

[vanhees71]

Summary:: I can understand it mathematically i.e. when I look at the formula ##V## of AC source = ##L*(\frac{dI}{dt})##, it makes sense. But I can't understand it intuitively. I think I know where I am going wrong but I don't know why.

Consider a purely inductive circuit with an inductor of self-inductance ##L## and sinusoidally varying AC source of peak voltage ##V_{m}##.

First of all, Why is ##V_{ac} = ε_{ind}## (where ##ε_{ind}## is the back emf)?View attachment 270193
Now, at ## t=0 ##, voltage is increasing at a fast rate. Hence, due to the sudden increase in current, the inductor induces maximum back emf (in the opposite direction) and current is maximum in the opposite direction.

Now, at ## t = T/4## (where ## T## is the time period), Voltage is maximum and not changing much. Now, this is where I think I'm going wrong. I am interpreting this as '##I## is not changing much because ##V##, when maximum, is not changing much (##\frac{dV}{dt} = 0##). Why is it wrong?

Interpreting it in the same way, when ##t = T/2##,## V = 0##. and ##\frac{dV}{dt} = -1##. So, the current is also changing at a fast rate. Due to this, there is a back emf induced that opposes the change in current. But here, Current is reducing rapidly. So, back emf is induced to increase the current in a positive direction.

I feel like I am misinterpreting it here: I am thinking of ##\frac{dV}{dt}## as the rate at which current is changing. And from the graph, it is quite evidently not the case. But, I can't find the reason as to why it is not the case intuitively?

I hope you got my question. If not, I apologise. Feel free to ask questions regarding 'my question'.

I am confused

Maybe you are confused, because of the figure, which is wrong. If you switch on your EMF at ##t=0## as indicated you have
$$\mathcal{E}=\mathcal{E}_0 \sin(\omega t) \; \quad t \geq 0.$$
The equation for your current reads
$$L \dot{i}=\mathcal{E}_0 \sin(\omega t),$$
and you have to solve this differential equation under the initial condition ##i(0)=0##. Here, you only need to integrate
$$i(t)=\frac{\mathcal{E}_0}{L} \int_0^t \mathrm{d} t' \sin(\omega t') = \frac{\mathcal{E}_0}{\omega L}[1-\cos(\omega t)].$$
What the textbook writer, however, had in mind is something else, because usually in AC circuit theory you are not so much interested in the transient state when switching on the EMF but only in the state the circuit is in after a long time.

Now this example is a bit artificial, because you have assumed a coil with 0 resistance, but in the real world there is always a finite resistance, which implies that the initial transients decay due to Ohmic loss in the resistor. This more realistic case is described by (using Kirchhoff's rule for a circuit consisting of the series of a resistor and an ideal coil)
$$L \dot{i}+R i=\mathcal{E}_0 \sin(\omega t).$$
Now you have a real differential equation. It's a linear one, and thus the general solution is given by the solution of the homogeneous equation, i.e., the equation with the right-hand side set to 0 + a special solution of the equation.

Now the homogeneous equation reads
$$L \dot{i}+R i=0 \; \Rightarrow \dot{i}=-\frac{L}{R} i.$$
The solution is a decaying exponential, i.e.,
$$i(t)=A \exp \left (-\frac{L}{R} t \right)$$
with some constant ##A##.

So what you are really after is only a special solution to the inhomogeneous equation which survives after a long time, and intuition tells us that the current should be some harmonic motion with the same frequency as the driving EMF. Thus the right ansatz is
$$i(t)=A \sin(\omega t +\varphi)$$
with constants ##A>0## and ##\varphi \in [-\pi,\pi]##. Plugging this into the homogeneous equation shows that you can always find ##A## and ##\varphi##, but it's a bit tedious. Since you are after the case ##R=0## anyway, we just neglect ##R## to find this stationary solution and plug in our ansatz:
$$L \dot{i}=L A \omega \cos(\omega t+\varphi)= \mathcal{E}_0 \sin(\varphi).$$
This gives ##A=\mathcal{E}_0/(\omega L)## and ##\varphi=-\pi/2##. So the solution for the stationary state is
$$i(t)=\frac{\mathcal{E}_0}{\omega L} \sin(\omega t-\pi/2).$$
This means the phase of the current is always behind by ##\pi/2## compared to the phase of the EMF.

Intuitively now the stationary solution is clear: The current has the same frequency as the EMF but the current's phase is always behind by ##\pi/2##. That can be understood due to Lenz's law: When the current is changing with time, the magnetic field inside the coil changes in a way such as to counter this change of the current in the coil and thus the change of the current in a coil is always behind the change of the EMF. For ##R=0## this leads to the phase shift ##\varphi=-\pi/2##.

 

[anorlunda]

In this, I can understand the behaviour of the current curve.

OK. But that curve you posted introduce a resistance R, so it is no longer a purely inductive. That's better because nonzero R is the usual case.

You can see in those curves, that current reaches its peak positive and negative values at the time when voltage crosses zero. That I think is what you were looking for in your OP. Why the phase difference?

So if we rounded the corners of the square wave, it eventually looks like the sin wave. Does that help to resolve your confusion in your original question?

But, I don't get why current is negative initially? Isn't is same as the DC source for the first T/2 time?

If you calculated those curves instead of pasting, you would see the difference. That step function is defined such that V=0 for time<0. But the square wave was defined for ##\infty< time < \infty## but the plot only shows time>=0. That is why current starts as negative at t=0. If you defined that square wave to begin at t=0 so that V=0 for t<0, and I=0 at t=0 initial condition, then the curves would look different.

 

[vanhees71]

It's the same problem as in the OP. In the figure with the square waves obviously the stationary state is shown not the transient state of the situation when switching on the circuit at ##t=0##.

 

[artis]

@Kaushik the current is not negative as such in an inductor when you first apply voltage to it, the current just lags behind voltage, in your own posted images there is shown a diagram of square wave, now because current lags voltage in an inductor the moment the voltage goes positive the current takes some time to "adjust" so that is why you see a positive square voltage while current is still "coming back up" from the negative side.
In your diagram just imagine the -V is zero volts and +V is some arbitrary voltage above zero volts, then you will see that there is no negative current just a current that lags behind voltage and increases and decreases in amplitude.

 

[rude man]

The best description I've seen of how an inductor reacts to an applied voltage is given in the ARRL handbook.

(@berkeman, how you like that? )

 

[vanhees71]

Do you have a link?