STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM

[broli]

LarryC your arguments are getting really sad. The calculations are all correct the only question that is left is does current pass resistance or not and for that you either replicate it with the same numbers and do your own heat measuring, or you shut up and let the people that are willing to do it progress.

[LarryC]

You're indeed confused. Notice carefully that that the ABS function does not apply to the current or the voltage. It applies to the time so that the time interval is always positive in case there is offset on the x-axis scale (in most cases there's no offset on the x-axis scale).

You're right, I mismatched a semicolon.

Checked and the 1HZ 61 ohm data does not have the V and C different sign issue, no inductance.

Could you send your excel data for the 700Hz scope shot.

I would still like to understand how the power is increasing, when half the time according to the scope shot, V and C have different signs.

Regards, Larry

[Omnibus]

@LarryC,

I just took these data with the primary coil, bare bones--no core, no secondary coil. The trace you see is a result of averaging of 16 traces taken at 1s interval.

[LarryC]

@LarryC,

I just took these data with the primary coil, bare bones--no core, no secondary coil. The trace you see is a result of averaging of 16 traces taken at 1s interval.

Thanks for the data. I do admit that you are being very open with your output.

But I found an issue with your use of I^2. The value is always positive due to ^2. While V * I is negative if either one of the values are negative.

It should be ((I^2) * sign (I)) to be compatible with V * I.

Edit: I don't think it will correct the whole issue because of the following

I * V             I^2 * sign
-5*5 =-25       -25
5*-5 =-25        25

Regards, Larry

[Omnibus]

Thanks for the data. I do admit that you are being very open with your output.

But I found an issue with your use of I^2. The value is always positive due to ^2. While V * I is negative if either one of the values are negative.

It should be ((I^2) * sign (I)) to be compatible with V * I.

Regards, Larry

In calculating these powers one should follow the physics of the situation. That's what determines the outcome. The voltage and current should be as they are, without messing up with them and the current in the Joule's law is to be squared. So far, as @broli says, the calculations are OK and what has to be determined is if the whole current goes through that resistance R. I've shown data that it really does and I'm working on further definitive proof which, as I said earlier, will post here as soon as I get ready.