* Hi .Thanks for detailed returning. Yes , right .We already use momentary I and V I(t), V(t) , not avg or RMSHi,
some considerations.
You said "I".
* it needs to be I(t). The momentary I.
* you can´t use I(avg)
You said "R = 1 Ohms".
* be sure the shunt is resistive in the rage of frequency you expect.
* better avoid wire wound resistors (they have high series inductance)
You said "DC current for a very short time"
* this contradicts itself.
* if it is DC then it needs to be continous. No ON and no OFF
* if it is "for a short time", then it includes a lot of frequencies (somehow the opposite fo DC)
You said: "2-How much energy spent for COIL HEAT LOSSES?"
* this only works for true DC (see above)
* but as soon as you work with pulses the current is also used to increase/decrease the magnetic field, which also is energy. 0.5 x L x I x I is stored as magnetic energy.
Klaus
Because when we give current to coil , coil attracts the magnet .Then Magnet produces back EMF and this changes the shape of waves like this . I made the same measurement for control experiment without magnet and waves were the same which you explained. Wawes are different here like you see in the pictureHi,
yellow is current: Then the scope picture makes no sese to me.
I don´t see why the current goes up 3 divisons within a couple of milliseconds.
I don´t see why the current should drop after some tens of milliseconds.
I expect the current to startt at zero and then start to rise. The starting rise rate should be about V/L.
I expect the current to rise ... then approximate a horizontal line, never drop. (as long as the switch is closed)
Klaus
I dont want to make the questions more complex so i didn't tell all the details. Yes magnet produce back EMF ... In fact L mini is only for able to measure the last velocity and kinetic energy as a part of this experiment .L mini is becaming a part of circuit for the last 1 cm and we can measure avg last velocity and calculate the kinetic energy.Here when you look nearer to picture you can see the peak of time for mini coil start.Second point is when magnet touches to coil . The distance off two point is 1 cmWhat did you measure for L.main and L.mini and DCR?
I expect dI/dt=L/R R=DCR+1ohm (sense)
There is also kinetic energy being measured expect some back EMF rising at the end to reduce current.
Yes , you are right about names but in here our circuit is not equal to a basic magnet and coil circuit. We already give current to coil first like electrical motor. Yes ,we normally give them different names .If we move a magnet for example with hand near to a immobile coil , it induce Voltage and we name it as induced voltage.Here there is not and opposite current or voltage so we dont name it as opposing induced voltage or Back EMF .But if give electrical current to this coil first , coil will already attract magnet without moving with hand , and magnet will aproach to coil again , and the same amount of voltage will be induced again. But this time the direction of induced current is opposite with the direction of given current and we name it as Back EMF (Back or opposing to given current ).Of course we see and measure the sum off given and back EMF. In summary, If the distance and velocity are same , in theory induced voltage and back EMF will have same amount and same direction . Only their names are different according to circuit .Hi,
you say "magnet produce back EMF".
I don´t know whether to agree or disagree. Because the magnet itself causes no back EMF. Induced voltage is caused in a coil.
Now there are two different things to consider:
* back EMF voltage of an inductance (without moving core) is negative
* back EMF voltage of a motor (with moving core) is positive
Which back EMF are you talking about?
Indeed - due to stray inductance - I expect both to happen.
Klaus
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?