Inductive Kickback

[citfta]

Two people believing the wrong thing does not make it true.  I have had this discussion several times before in the past.  Some have learned and some have not.

[MileHigh]

The flyback from an inductor can deliver power to a load,and the value/resistance of that load will determine as to what the voltage and current values will be with a set inductor. E.G-if the resistance of the load is low,then the flyback voltage will be low,and the current high. If the resistance of the load is high,then the flyback voltage will be high ,and the current low.

It should also be noted that the lower the resistive value of the load being placed on the flyback,the longer the magnetic field takes to fully collapse around the inductor. This results in a higher current due to the slower collapsing field,but a lower voltage for the same reason-the slower changing magnetic field with time. The higher the value of the resistive load on the flyback,the quicker the magnetic field around the inductor will collapse. This results in a lower current but a higher voltage due to the faster changing magnetic field with time.

You are right about the current not changing direction but you are not right about the magnitude of the current.

When you disconnect the power source from the coil, whatever the magnitude of the current flowing through the coil is, that's the initial amount of current that will flow through the load.

If you have one amp flowing through the coil and you disconnect your power source and the load is a 100-ohm resistor, then initially the current flow through the load will be one amp and the voltage across the load will be 100 volts.  If the load is 5000 ohms, then initially the current flow through the load will still be one amp and the voltage across the load will be 5000 volts.

[woopy]

Hi Cifta and Synchro and all

If i posted above it is not to get  what i can get from all text books about physic.

My questions are  very clear :

1- at the end of the pulse, on the scope shot, the input current from the power source has totally vanished  yes or not ?

2- The current has built a magnetic field in or around  the inductor and this magnetic field has been " used "  to propel the rotor magnet yes or not ?

3- so the current is not stored in the magneticfield but has been used the give the rotor magnet a kinetic energy yes or not ?

4-So if the magneticfield  collapses and by shrinking it crosses the wires of the coil and redo a new current does it mean that the magneticfield can be used twice yes or not ?

And just for the dessert

A small video showing that the flybackspike can be really powerfull and destroying as  every body knows very well

https://youtu.be/TAx7Y0UIyHA

good night at all

Laurent

[citfta]

You are right about the current not changing direction but you are not right about the magnitude of the current.

When you disconnect the power source from the coil, whatever the magnitude of the current flowing through the coil is, that's the initial amount of current that will flow through the load.

If you have one amp flowing through the coil and you disconnect your power source and the load is a 100-ohm resistor, then initially the current flow through the load will be one amp and the voltage across the load will be 100 volts.  If the load is 5000 ohms, then initially the current flow through the load will still be one amp and the voltage across the load will be 5000 volts.

Don't you want to rethink that last paragraph?  You have just given (if true) the formula for OU.  In your first example you are putting 100 watts into the load.  In your second example you are putting 5000 watts into the load.  Where did that extra 4900 watts come from?

[verpies]

If you have one amp flowing through the coil and you disconnect your power source and the load is a 100-ohm resistor, then initially the current flow through the load will be one amp and the voltage across the load will be 100 volts.  If the load is 5000 ohms, then initially the current flow through the load will still be one amp and the voltage across the load will be 5000 volts.

That's why they call it a current source - constant current regardless of the load resistance.

P.S.
That Igor guy and Synchro are wrong about the current through the inductor reversing when the inductor is opened.
How do I know? -  I did the lab work and I can distinguish when my scope is measuring current vs. when it is measuring voltage.