MH's ideal coil and voltage question

[poynt99]

Very simple Poynt.

As we have agreed on,while the ideal voltage source has a value of 0 volts,the current will continue to flow through this ideal inductor loop.
At T=5 seconds,the voltage source wants to instantly reverse that current flow direction.
So we have an ideal 5 Henry coil,with current flowing in say a CW direction around the loop.
At T= 5 seconds,the coil is instantly supplied negative 3 volts from an ideal voltage source.
At that instant,you have to infinite current values trying to flow in opposite directions.

Why do you believe the current will be infinite? In your posts following this one, you seem to be on board with the idea that the current after the first 3s is 2.4A, correct?

You also mentioned that at T=5s, the circuit would explode. This is not the case. Does the circuit explode when the circuit connection is first made? I can tell you that the two scenarios are no different, and the inductor simply adjusts for a change in current. In other words, going from 0V to 4V, and the resulting final value of 2.4A, is no different than going from 0V to -3V with a starting current of 2.4A. The circuit current simply begins adjusting/changing as a result of the new voltage.

Have you drawn out the circuit current as I suggested? I suspect not. Here is a hint: at t=0s, the current ramps up linearly to the final value of 2.4A at t=3s. Between t=3 and t=5, the current remains at 2.4A. Now, what do you suppose happens to the current between t=5 and t=7? Between t=7 and t=13? From t=13 onward?

[poynt99]

Brad,

Here is an inverted color version of the graph in case you want to print it out and finish the current plot with a pen a ruler.

[tinman]

Why do you believe the current will be infinite? In your posts following this one, you seem to be on board with the idea that the current after the first 3s is 2.4A, correct?

You also mentioned that at T=5s, the circuit would explode. This is not the case. Does the circuit explode when the circuit connection is first made? I can tell you that the two scenarios are no different, and the inductor simply adjusts for a change in current. In other words, going from 0V to 4V, and the resulting final value of 2.4A, is no different than going from 0V to -3V with a starting current of 2.4A. The circuit current simply begins adjusting/changing as a result of the new voltage.

Have you drawn out the circuit current as I suggested? I suspect not. Here is a hint: at t=0s, the current ramps up linearly to the final value of 2.4A at t=3s. Between t=3 and t=5, the current remains at 2.4A. Now, what do you suppose happens to the current between t=5 and t=7? Between t=7 and t=13? From t=13 onward?

I do not agree with this. The energy stored in the magnetic field cannot just disappear. That energy has to be dissipated before the opposite magnetic field can be induced--which is what the-3 volts is trying to do. The startup current is dampened by the induction process due to the CEMF. But there is now a steady current flow through the looped inductor,and a magnetic field that is storing energy. The -3 volts is applied,and the current being produced is trying to collapse that already built magnetic field,which cannot be collapsed due to the steady state current that is flowing that keeps it built. In a real world situation,that energy would be dissipated as heat,but as we have an ideal inductor loop,then the energy cannot be dissipated.


Brad.

[poynt99]

The energy in the inductor does not "disappear", some is returned to the source during that period.

[MileHigh]

I honestly disagree with the hand-holding at this point.  There are others that were interested in this and I suggested they brainstorm.  Magluvin may also give it a shot when he comes back.

The big "breakthrough" is apparently Brad now accepts that the current ramps up to 2.4 amps after three seconds.  What about the other people that disagreed vehemently about this, what do you have to say?