# Ac circuit with an inductance

1. Jul 20, 2011

### ananthu

In an a.c.circuit containing only inductance, for example, it is said that the current is lagging behind emf by 90 degree. We can understand this that when the emf is zero,the current is at its negative maximum and when the emf reaches its positive maximum, the current becomes zero and so on.. my doubts are:

1. How did they find that one is lagging behind the other when both are rapidly changing with time?

2. How is it possible when the emf is zero the current becomes maximum or vice-versa? To produce current itself in a circuit is not the potential difference required? When pd becomes zero, how the current reaches its maximum?
What all these things actually mean in practical terms?
If any body gives some simple explanation, it will be very useful.

2. Jul 20, 2011

### Staff: Mentor

Here is a long recent thread on the subject:

There is also a thread here in EE "What is Phase" that may be of help to you. It may still be on the top page in EE, or you can find it with a search.

3. Jul 20, 2011

### dlgoff

Last edited by a moderator: Apr 26, 2017
4. Jul 20, 2011

### Evil Bunny

This is something that I struggled with too... This is never explained very well in texts or in the classroom. The whole idea of a voltage leading or lagging a current is pretty confusing when you're trying to wrap your head around it for the first time.

Here's the deal... The sine wave on your scope is measuring voltage. Period. It's not measuring current. If you put one of your scope probes across your source, you see the original, or reference, sine wave. If you put the other scope probe across your inductor, you will see another sine wave, that is 90 degrees ahead of the source sine wave.

Now here is the big secret... at this point, we don't call that source sine wave a voltage anymore... Now we're calling it "the current". It magically became "the current" when we started talking about how inductors and capacitors cause leading and lagging currents and voltages.

I hope that helped. And more importantly, I hope it was right... I'm sure it will be promptly corrected if not.

5. Jul 20, 2011

### 256bits

1. How did they find that one is lagging behind the other when both are rapidly changing with time?
They used the "magical " powers of mathematics.
Care to note that :
for an inductance : v = L di/dt, and
for a capacitance : i = C dv / dt.
It does not matter how rapid the change in time, this holds always.

2. How is it possible when the emf is zero the current becomes maximum or vice-versa? To produce current itself in a circuit is not the potential difference required? When pd becomes zero, how the current reaches its maximum?
You have to jump out of the mindset that V=IR, when working with impedance circuits comprising L and C and R. Comples circuitry implies comples mathematics.

To see the lag for these circuits, maybe draw wave diagrams. Use of a Magic marker is optional.
!. diagram 1 - draw the source voltage e=E sin(t) on the vertical versus time on the horizontal. We will us radian time measure rather than seconds.
You should have a sine wave with 0 at the origin. At time pi/2 e is maximum at value E.
At t=pi, e= 0 crossing the time axis. At t==3/2 (pi) , e=-E below the time axis. And lastly, at t= pi, e again crosses the time axis at 0. Draw several cycles.

2. diagram 2- Draw a cosine wave below the diagram 1. Vertical axis is current and the minimum is -I and maximun is I. The formula is i = I cos (t) or i = I d( sin(t) )/dt.
Note the following :
At t=0, e=0 and i=I
At t=pi/2, e=E and i =0
etc.
One can conclude that the current graph leads the voltage graph by pi/2 or 90 degrees.
The current graph reached its maximum, 1/4 cycle before the sine graph. Draw a 1/4 cycle of the current graph at t=-pi/2 for the upward zero crossing point if you are not convinced about the lead.

You will have to figure out yourself which e and i graphs are for inductance ( and capacitance ).
Hint : for inductance in a circuit Vsource + Vinductance = 0
And Vinductance is actually -Ldi/dt.

6. Jul 20, 2011

### Evil Bunny

Now we're calling the source the "voltage" wave. Is this correct?

Please correct me if I'm straying down the wrong path here...

Because http://www.animations.physics.unsw.edu.au/jw/AC.html#impedance" (wikipedia) are considering the source to be the "current" wave... as far as I can tell.

Am I wrong in saying that the "current" wave is the waveform that we would see if the o-scope probe was placed across the source?

And the "voltage" wave is simply the waveform we would see if we put the o-scope probe across the inductor (or capacitor) which would be the "load" of the circuit?

And whether the voltage was leading or lagging the current, we could determine if the circuit was capacitive or inductive... correct?

In other words... If we had the component (either inductor or capacitor) hidden under a black box and we wanted to determine what it was, we would put the [EDIT: the "A" probe] scope across the source and [EDIT: the "B" probe] across the black box and call the source waveform the "current" waveform and the black box waveform would be the "voltage" waveform...

Yes? No?

Last edited by a moderator: Apr 26, 2017
7. Jul 20, 2011

### I_am_learning

Instead of saying 270 Degree Leading, They say 90 Degree lagging. Now, thats the convention.

8. Jul 20, 2011

### 256bits

Believe me -it takes a while to get the hang of it.
It is quite non-intuitive and after practice of sample circuits it becomes easier.

9. Jul 20, 2011

### Evil Bunny

256bits,

I agree that it's not intuitive, but you indicated that the "source" waveform would be the voltage and the "load" waveform would be the current.

I say this is incorrect. It should be the other way around. Do you agree or disagree?

I mean no disrespect, but the OP was confused about this, and I feel (hope) that my explanation clears this up for him... and your post contradicted mine.

Either way... I'm still saying that both waveforms ("load" and "source") are actually voltage waveforms, but for the sake of analyzing circuits, we're just calling the "source" waveform the current.

Is this line of thinking correct or not?

10. Jul 21, 2011

### 256bits

Note the following in a circuit containing an AC source and a pure inductance.
The source voltage and the inductaace voltage are the same.
ie Vsource + Vinductance = 0
So it does not matter which is referred to in this case.

The current is common - it flows from the source thru the inductance.
So the load current = source current = circuit current = i.
I made no mention of source current or load current but current i.

I think it may be better explained by a circuit containing an AC source e= Esin(t), a resiistance R, an inductance L, and a capacitance C, all connected in series.
Vsource+Vresistance+Vinductance +Vcapacitance = 0

The current i will be also a sin wave with a lead or lag from the source voltage. The lead or lag will be somwhere from -90 degrees to + 90 degrees, but this will depend upon the values of L and C. i = Isin(t+phi). phi is the lead or lag.
i = i resistance = i inductance= i capacitance.

For the resistance R: A resistance has no lead,lag with the current
Vr= i R = I sin(t+phi) R

Capacitance C: The current through the capacitance leads the voltage across the capacitance by 90 degrees or pi/2 radians. ie sin( t + phi - pi/2 )

Inductance L : The current through the inductance lags the voltage across the inductance by 90 degrees or pi/2 radians. ie sin( t + phi + pi/2 )

For a parallel circuit ( R, C, L in parallel), the voltage across each compnent is common with the source, but the curent through each will be different.
The lead and lag of current through each component with respect to the voltage across each component still applies.

Google "inductive lag" a some good sites come up
has a bunch of waveform diagrams

Last edited: Jul 21, 2011
11. Jul 21, 2011

### Evil Bunny

You are correct... The voltage of source and load would be the same and you would not see a difference on the scope because they are the same point electrically. In a purely inductive or purely capacitive circuit.

I was way off base on this one. My apologies.

I was thinking of a series RL or RC circuit. In this way you could do what I was saying by putting one probe across the resistor and one probe across the inductor (or capacitor)...

12. Jul 21, 2011

### Evil Bunny

There is no way to get the current wave form on a scope in a parallel RL or RC circuit... is there?

13. Jul 21, 2011

### ananthu

Ohm's law states that current is directly proportional to the p.d.Though this is strictly applicable to d.c. only, it makes logical sense to understand the concept.Only if there is a level difference between two points, water can flow from higher level to the lower one. Other wise water flow will be zero. This should strand true for a.c. also, as there can not be effect without cause.
Applying to the inductor, if we say, there will be a current through it without P.d. across it, does it not contradict the basic common sense on which the Ohm's law was based, thought the law talks only about direct current? Where from the current appear in the first instant with out an emf? This point seems very uncomfortable and difficult to explain while you are teaching the "lagging or ahead" concept to the students. No text books speak about these subtle aspects.

Secondly, during the nineteenth century when Faraday, Lenz, Maxwell and others who contributed to these ideas lived, there was no CRO and other sophisticated devices to display the wave forms. So how did they come to know that at any given instant of time that the voltage is leading the current across an inductor? Was it simply by mathematical method and logical intuition?

While deriving an expression for the current across an inductor, from the step i=-(E_(0 ) Cos ωt)/ωL, why should we arrive at the conclusion as i=-E_(0 )/ωL Sin(ωt-π/2) instead of i=-E_(0 )/ωL Sin(ωt+π/2), which is also mathematically correct?

14. Jul 22, 2011

### 256bits

Good question.
Only way I can think of is by adding a small resistor is series with the L or C, but then that changes the circuit.

15. Jul 22, 2011

### 256bits

Ohms law applies to a resitive circuit for DC and AC. For AC we use the rms ( root mean square) values for voltage and current. Your 120v household voltage is an rms value.

With a circuit comprising R, L, C we use the term reactance. Then V = I Z, similar to Ohms law.

In an inductor, e = L di/dt. The voltage is proportional to the change in current, so if there is no change there is no voltage across the inductor. There can be current, but if it is constant, the voltage = 0.

Water analogies go only so far.
Try a pendulum analogy instead. A pendulum waill swing sinusoidally if the swing is not too great. I swings up to a height h at each end where its velocity becomes zero. Down at the bottom, the height is 0 and the velocity is at the maximum.
The height can thought of as voltage and the velocity can be current. And do not the height and velocity appear to be out of pahse just like in an inductor? Just like an inductor with current and no p.d. as you say, down at the bottom in the pendulum swing there is no height but lots of velocity.
Or a mass hanging from spring anaolgy. Here the force can be the current as it "flows" through the mass and the velocity would be the voltage as velocity needs 2 points - a reference or zero point and the velocity of the component.

How did they do it back then. They knew a lot about mechanical systems, and had a lot of formulas and equations to describe motion and whatnot. A mass on spring - yeah lets see if we use those equations for voltage and current and see what happens. I bet that is what they did - mathematical induction.

16. Jul 26, 2011

### ananthu

Thank you. Your simple pendulum example is quite innovative.

So, shall we say that the emf that is said to lead the current in the text actually means the emf induced across the inductor and the current across it at a given time but not the applied emf across the source in the circuit?