# I Inductor behaviour in an AC power supply

1. Jul 28, 2017

### pranav p v

what is actually happening when an Inductor is connected with an ac source.(i mean at the very instant of connection is made).i know emf will be induced due to change in current.(curl of E is =-time derivative of B)here my confusion is,inorder to induce emf,current should flow first,then who limit the magnitude current(is it infinity at the very first time?).....and how will be the current wave form at the very instant of connection. i want to know what is the physics behind it not mathematical equations..plz helpme

2. Jul 28, 2017

### cnh1995

3. Jul 28, 2017

### anorlunda

The voltage across the inductor is not proportional to current, it is proportional to the rate of change of current.

So what happens when you connect it to an AC supply? The answer depends on what time in the cycle you close the switch. Let's say you close the switch when V=0. Then rate of change of current is zero. As V builds to maximum, rate of change of I increases. When V decreases to zero, rate of change of I will be zero, and thus I will be at maximum.

You don't want equations, how about time plot curves?

4. Aug 3, 2017

### pranav p v

Here two things are known,one is applied voltage,that is sinusoidally varrying..other thing is emf induced across the pure inductor is L times time derivative of current,..from this current equation will be the integration of voltage by L..if v(t) is Vm sin wt then I(t)=- Imcos(wt)/wL....so mathematically current lags behind voltage by 90 degrees..I am ok with this things..if I close switch at t=0,how current can flow in the reverse direction of voltage polarity?(at the very beginning)..inductor behaves similar to mass right?voltage is force and current is velocity of mass..then if force is sinusoidally varying then at the very beginning of the force is applied velocity will be in the same direction of force right?( if force is reversed in the second half cycle means then velocity will not be the direction of force because due to inertia)....I want to know what is exactly happening in the very first half cycle...I have asked this question to my teachers but they answered like this,first they draw a sinusoidal current wave form and at the maximum value change in current is zero so emf is zero like that...but here initially we don't know anything about current behavior right?,how even can draw sinusoidal wave form of current? (Consider during first half cycle as the voltage(force) increases ,current ( velocity) increase with time derivative of current( acceleration) increases when voltage reaches maximum value ,for that instant it is constant so time derivative of current is also constant(acceleration is constant) when voltage starts to decreases then current increases ( force is there in the same direction but decreasing)but change in current iis less than previous value( acceleration decrease)..so at the end.of first half cycle voltage is zero but current is maximum( from here we can say current is lagging )..........is this happening at the very first half cycle????? Please give me a correct answer for this ...I can't sleep because of this..help.me

5. Aug 3, 2017

### cnh1995

Did you see the waveforms I linked in #2?
What did you not understand about it?

6. Aug 3, 2017

7. Aug 7, 2017

### pranav p v

i think my question is not cleared for you.. Consider a pure inductor is connected across a voltage source v(t)= vm sinwt.inductor behaves like mass and current is like velocity, voltage is similar to force..inductor opposes change in current like mass opposes change in velocity..if I apply a sinusoidally varying force or voltage velocity or current will be proportional to the integration of force or voltage...then I(t)= constant* ∫vm sinwt...here the problem,integration of sin x is - cos x right? Then from the wave form we can see at the very instant of time of closing the switch,v(t)=0, current is -ve value how it is possible..?if I apply limit to the integration from 0 to some t( because here my voltage wave form starts from 0) then I(t) is proportional to (1-cost)...if I plot (1-cos t) wave form I am happy because,here when t=0 ,I(0)=k*(1- cos 0)=0, and for the first positive half cycle I (t) increase from 0 to a maximum value...when t= T/2 then the value of 1- cosx will be 2*k...it makes sense..here actual problem is why integration of sin x is - cosx ...if I add all values of sin x from 0 to some x ( if x< 180) then it will be + ve only right? How integration of sin x starts from - ve value.... Confused with mathematics

8. Aug 7, 2017

### anorlunda

You do not have sin(wt) from $-\infty$ to $+\infty$. You have the forcing function zero from $-\infty$ to $0$ and $sin(wt)$ from $0$ to $+\infty$. What is the integral of that at t=0?

9. Aug 7, 2017

### pranav p v

"What is the integral of that at t=0?" .....

I have to know the answer of this same question.. here indefinite integral gives - cos x and if I apply definite integration with lower limit as zero and upper limit as a variable t then it will be (1- cos x)

10. Aug 7, 2017

### anorlunda

You need definite integrals.

Or just remember the underlying definition of integral as area under the curve. Your sin(wt) curve begins at t=0, what is the area under that curve at t=0+? Not Ve.

11. Aug 7, 2017

### pranav p v

"what is the area under that curve at t=0+?" Is it positive or negative?

12. Aug 7, 2017

### cnh1995

It is clear to me.

Let's talk about a purely inductive circuit, so R=0.
Now, the forcing function is sinusoidal and its equation will depend on the angle at which you close the switch. Let's say you closed the switch when the sinusoidal voltage is at positive zero crossing.
So,
Ldi/dt= Vmsin(ωt)
So,
Ldi=Vmsin(ωt) dt

Integrating, we get
Li=-Vmcos(ωt)/ω+c......1)

Now at t=0, current through the inductor is zero because there is no initial current as the voltage is switched at t=0.

So,
1) becomes,
0=Vm/ω+c... (cos 0=1)
I.e. c=Vm

So, i(t)=Vm/ωL-Vmcos(ωt)/ωL

It will look like this.

Last edited: Aug 7, 2017
13. Aug 7, 2017

### cnh1995

@pranav p v, I edited and fixed some typos in my earlier post.

Now, as they teach you in ac circuits class,
the current lags the voltage by 90 degrees but only when you close the switch at the voltage peak.

Let's analyse that mathematically.

Ldi/dt=Vmsin(ωt).

∴Ldi=Vmsin(ωt)dt

∴Li=-Vmcos(ωt)/ω+c

At t=π/2ω i.e. at voltage peak, current is 0.
So,
0=-Vmcos(π/2)/ω+c
∴c=0.

So,
i(t)=-(Vm/ωL)cos(ωt).

Last edited: Aug 7, 2017
14. Aug 7, 2017

### pranav p v

"Integrating, we get
Li=Vmcos(ωt)+c......1) "

Integration of sin x is - cos x

So i(t)= - (Vm/ XL)cos (wt)+ c XL = inductive reactance

initially no current at t=0 ,so c=Vm/XL

That implies i(t)=Vm/XL- Vm/XL cos wt
i( t)=Vm/ XL( 1-cos wt)

This only I mentioned in #7,#9....

So you are saying that there is an offset dc component is present there ??waveforms u showed here is the actual wave form right?in that how to say current is lagging?

15. Aug 7, 2017

### cnh1995

I had forgotten to divide by ω and made a couple of sign errors. Fixed it now.

Yes. The dc offset is present initially. In R-L circuits, it dies out exponentially because of the resistance and after a few cycles, only sinusoidal current remains. In case of purely inductive circuit, this dc offset does not die out and is present there all the time. The only time you have zero dc offset is when you close the switch at the voltage peak.

The dc offset is maximum when you close the switch at voltage zero crossing and is zero when you close it at the voltage peak.

16. Aug 7, 2017

### cnh1995

You can't say the current is lagging unless it is a symmetrical waveform. The terms lagging and leading are used for sinusoidal signals (sin and cos) only, with no dc offset.

Hence, you can say current lags the voltage only in the second case i.e. when there is zero dc offset.

17. Aug 7, 2017

### pranav p v

Okay.. But I haven't heard before that we have to close the switch at peak value of voltage,then only current lags voltage..can you suggest any book which talk about this things

18. Aug 7, 2017

### cnh1995

Only for purely inductive circuits. In R-L circuits, the dc offset dies out and a steady state is reached where current lags the voltage by the power factor angle.

Google 'ac transients'. I don't know about any physics books but in electrical engineering books, you may find some material in network analysis.

19. Aug 7, 2017

### pranav p v

So if I use a simulator for this purpose ,is it show current waveform with that offset value?( simulator because ideal inductor is no there).... And for this condition what will be the avg power?

20. Aug 7, 2017

### cnh1995

For purely inductive circuits, average active power is always zero no matter when you close the switch. You can see it from the graphs.

Yes, you can use a simulator. The waveforms I've posted are actually the screenshots of the simulations I ran on my phone.

Last edited: Aug 7, 2017