MH's ideal coil and voltage question

[picowatt]

The voltage being measured is the instantaneous voltage-Vmax.
If the CEMF was equal to the EMF at T=0,then there should be no current flow,so why would there be a voltage drop across the resistor at T=0
Remember,we are reading the maximum voltage values on each channel,and so this is the voltage at T=0.

Your input square wave looks rather soft, but all in all what I am seeing in your scope shot is very little, if any, current flow at T=0 as evidenced by there being little drop (difference between channels) at that time.  However, it is hard to see any detail at your scope settings and as I said, your input waveform looks soft.  But, all in all, I see very little current flow at T=0 with current rising at T>0.

I disagree.
If the CEMF and EMF are equal at T=0,then there should be no current flow at that time,and there for there should be no voltage drop across the resistor.

From what I see in your scope shot, there is no Vdrop across the resistor at T=0, so no current is flowing at T=0.  However, you should consider using a stiffer Vsource for your input waveform for this type of testing.  The CEMF=EMF when the current flow reaches the appropriate rate of change. 

The difference between the EMF and CEMF i am seeing,can only be due to the coils winding resistance.

How are you "seeing" a difference between the EMF and CEMF when nothing in your test setup will allow you to see that? 

The drop across R1 can be used to calculate the rate the current is changing, which can act as a proxy to allow you to see the results of the CEMF.  The CEMF and EMF however are both contained in the scope channel that is directly across the inductor.

However, if you believe your two scope channels are measuring EMF and CEMF and if you believe that difference is due to winding resistance, that should be a piece of cake to prove or disprove by using R1's value and the DCR of the coil.  But if the observed drop it is due to the DCR of the coil as you say, why is the current waveform shaped like it is?

As we have no winding resistance in an ideal coil,then i would agree that the CEMF is equal to the EMF--and there we have our problem.
This feed back system you talk about,must have a loss in the negative feed back in order for it to no be the same that induced it in the first place.

No dissipation is required.  But if that is where you are getting hung up, then as .99 attempted to do, perhaps steering the discussion towards an inductor with some amount of resistance would be helpful.

At T=0,every change that the EMF tried to make to the current flow,the CEMF would counteract this change with 100% efficiency-and so no change takes place. As you you have said that the drop in voltage in my scope shot is due to current flow,and that is the two instantaneous voltage value's,then current could only flow instantly if the instantaneous CEMF value was less than that of the applied EMF. As i stated,i believe this to be true when the coil has winding resistance.

I am unable to follow this.  Your scope shot appears to show no current current flow at T=0 and then the expected increase in current at T>0

I am yet to see(as partzman said),anything that confirms that the CEMF is equal to the applied EMF at T=0. So far,i have seen the opposite.

I have repeatedly stated that immediately following T=0, and as soon as the rate of change reaches .8amps per second, the CEMF becomes equal the EMF.  This can appear as being instantaneous as .99 stated, but I am more comfortable with the wording "as soon as the rate of change reaches .8 amps per second" the CEMF=EMF.  This need only happen in a very very short period of time and involve only a minuscule amount of current flow, but again, as I see it, the CEMF will not equal the EMF until the appropriate rate of change is achieved.  With 4 volts applied to 5H, that is .8 amps per second  That is what the time tested formulae tell us.

PW

[tinman]

 author=picowatt link=topic=16589.msg487598#msg487598 date=1467430986]

How are you "seeing" a difference between the EMF and CEMF when nothing in your test setup will allow you to see that?

The test setup allows exactly that,and now allows us to see the effects of the electric field and capacitance of the windings at work,before current starts to flow.

Your input square wave looks rather soft, but all in all what I am seeing in your scope shot is very little, if any, current flow at T=0 as evidenced by there being little drop (difference between channels) at that time.  However, it is hard to see any detail at your scope settings and as I said, your input waveform looks soft.  But, all in all, I see very little current flow at T=0 with current rising at T>0.
From what I see in your scope shot, there is no Vdrop across the resistor at T=0, so no current is flowing at T=0.  However, you should consider using a stiffer Vsource for your input waveform for this type of testing.  The CEMF=EMF when the current flow reaches the appropriate rate of change.

The input is fine,and the square wave is also very defined-as you will see in the scope shot below.
We can reduce the time per division to 50 microseconds,and clearly see when current starts to flow--remember,this is at a low frequency of 10 Hz.
At the very point current starts to flow,both the V/in and V across the inductor should be exactly the same. There should be no voltage drop across the resistor until current starts to flow,so the test setup is valid to see the voltage across the inductor before current starts to flow.

Below is a scope shot with the time base now at 50uS per division.
We can clearly see that the voltage is a vertical rise,and so the square wave being delivered is quite clean. We can also see clearly when current starts to flow,as both trace values begin to fall. We can also see that the transition between no current flowing,and current starting to flow is very clean and definitive. If the CEMF was the same as the applied EMF,then we should see the blue trace start to fall from the same maximum value that the yellow trace reaches--that being our EMF.
But as we can clearly see,the voltage across the inductor is not that of the supply voltage before current starts to flow,and we know we do not get a voltage drop across a resistor until such time as current starts to flow through that resistor. So why is the voltage across the inductor less than the applied voltage at T=0--the very moment current starts to flow?.


Brad

[picowatt]

author=picowatt link=topic=16589.msg487598#msg487598 date=1467430986]




The test setup allows exactly that,and now allows us to see the effects of the electric field and capacitance of the windings at work,before current starts to flow.

Your "test setup" is measuring the current using R1 as a CSR.

The input is fine,and the square wave is also very defined-as you will see in the scope shot below.

Your previous scope capture clearly showed that your input "square" wave had a large amount of tilt.

We can reduce the time per division to 50 microseconds,and clearly see when current starts to flow--remember,this is at a low frequency of 10 Hz.

As we previously discussed, the frequency content of your squarish waveform with a fast rising edge is well beyond the 10Hz repetition rate of the waveform.  Take a look using your scope's FFT function...

At the very point current starts to flow,both the V/in and V across the inductor should be exactly the same. There should be no voltage drop across the resistor until current starts to flow,so the test setup is valid to see the voltage across the inductor before current starts to flow.

Yes, the CH1 and CH2 levels should be similar at T=0 (there should be little current flow hence less drop across your CSR), unless, of course, your series R and your inductor's capacitance are producing an RC related error.  What is the value of the resistor you are using?  What is the capacitance of the inductor? 

Below is a scope shot with the time base now at 50uS per division.
We can clearly see that the voltage is a vertical rise,and so the square wave being delivered is quite clean. We can also see clearly when current starts to flow,as both trace values begin to fall. We can also see that the transition between no current flowing,and current starting to flow is very clean and definitive. If the CEMF was the same as the applied EMF,then we should see the blue trace start to fall from the same maximum value that the yellow trace reaches--that being our EMF.
But as we can clearly see,the voltage across the inductor is not that of the supply voltage before current starts to flow,and we know we do not get a voltage drop across a resistor until such time as current starts to flow through that resistor. So why is the voltage across the inductor less than the applied voltage at T=0--the very moment current starts to flow?.

The difference between peak levels is much greater in this capture than in the previous.  What else besides the time base has been changed?

Without knowing more details about your test setup, I am going to guess that you are using an inductor with a rather large parasitic capacitance.  That would account for current flow at T=0.  You can also see signs of ringing on the voltage across the inductor.

If you want to pursue these kinds of tests, we could discuss improvements to your test setup.

PW

[3Kelvin]

11.3
Energy Stored in Magnetic Fields
Since an inductor in a circuit serves to oppose any change in
the current through it, work must be done by an external source such as a
battery in order to establish a current in the inductor.  From the  work
-energy  theorem,  we conclude  that  energy  can  be  stored  in  an 
inductor.  The  role  played  by  an  inductor  in  the magnetic  case  is  analogous  to  that  of  a
capacitor in the electric case.

The power, or rate at which an external emf extε
works to overcome the self-induced emf Lε and pass current
I in the inductor is .....  => read more at page 10 of the pdf

http://web.mit.edu/viz/EM/visualizations/coursenotes/modules/guide11.pdf

Love + Peace
3K

[minnie]

  An ideal inductor would not behave like a capacitor?
        John.