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Lewin's NCF Experiment and Lecture

Started by poynt99, April 24, 2016, 10:20:07 AM

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0 Members and 5 Guests are viewing this topic.

minnie




   So even though the volt meter is in effect an open circuit there will be a small
   displacement current and thus an emf.
   Can't wait to see the "real answer"
   I've had a couple of real gruelling days at the office and I enjoy this bit of
   light relief.
   I'm depending on poynt to put me right,here.

poynt99

Quote from: minnie on April 24, 2016, 05:29:17 PM


   So even though the volt meter is in effect an open circuit there will be a small
   displacement current and thus an emf.
   Can't wait to see the "real answer"
   I've had a couple of real gruelling days at the office and I enjoy this bit of
   light relief.
   I'm depending on poynt to put me right,here.
Not sure if I'm getting to where you're going, but I'll give it a try:

- The 1V induced E field is present regardless if a piece of wire is there or not.
- The 1V induced emf is present in the circuit regardless if a meter is placed across the resistors or not.
- The combined loop voltage across the resistors is 1V.
- The meter leads if wrapped around the solenoid and shorted together will measure an induced emf close to 1V.
- If the meter had an infinite impedance, the leads would still have an induced voltage of 1V across them.
- I don't see any reason displacement current (current across the plates of a capacitor) would be present anywhere. Where do you see it?
question everything, double check the facts, THEN decide your path...

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tinman

Ok,well my test results turned out to be the same as i would have expected,where there is very little difference at all between the wave forms shown on the scope.

I had to settle for a 680R instead of the 900R,as i dont have one. But the results should have shown the difference we were looking for-but did not.
A slight difference in peak voltage was all that was seen,but the polarity remains the same,and also the wave forms them self are near identical.

Below is a picture of the setup-along with the circuit as it is,and the scope capture to go along with the test.
As Poynt said,the scope tip and ground were rapped around the coil,and joined together. The voltage to the cap was then increased until i had my 1v potential seen on the scope trace.
The end resulting voltage will of course be lower,as i used a 680R resistor instead of the 900R.

So i do not get the results shown by Lewin,or as predicted by Poynt.
I will parallel a couple of resistors,and get the required 900R,and see if that makes a difference. But i see only the result being the same,but where the peak voltage may rise.


Brad

tinman

Quote from: poynt99 on April 24, 2016, 10:57:18 AM
What voltage will the left meter indicate?

Ok,have an answer to this one.
The left meter in this situation reads very close to 0v.
I would expect the slight gap due to the resistor as being the reason for a slight voltage reading across the meter/scope.

Brad

TinselKoala

Brad, look at your schematic, and consider the direction of conventional current as you have indicated with the arrows. Look at the locations of the probe Tips and Grounds with respect to the current flow, hence the voltage drop across the resistors.   Also consider that the probe ground references are connected together at the scope chassis.

I think it might be interesting to make these two measurements separately. That is, do a shot with only CH1 connected, save the waveform, then do a shot with only CH2 connected. Then compare the waveforms. Also you might try the CH2 measurement (separately) with probe tip connected to the "current in" side of the resistor and ground connected to the "current out" side.The CH1 probe is already connected this way, with tip on "current in" and ground on "current out".