Holcomb Energy Systems:Breakthrough technology to the world

[bistander]

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The fact that Voltage (vortex electric field) turns into current strength (vortex magnetic field) can be emphasized from this material of mine:

https://rakatskiy.blogspot.com/p/ampere-force.html

Hi rakarskiy,
I do not believe in your vortex theory for B and E.

Also, you offer no satisfying response for your contradiction of Ohm's Law.

P = I²R = 1.3A²*3ohms = 5.07 watts.
not 13.5W as you show from V*I.
bi

[rakarskiy]

Believe it or not, it's entirely up to you. The second why try to calculate the power in all three ways and you will be surprised: the voltage at the generator terminals is 10.4V; Current in the circuit 1.3A; Loop resistance 1+3=4ohm.

By the way, how much power will be dissipated in the resistor? - with these indicators.
Well, I started to worry about you.

[bistander]

Believe it or not, it's entirely up to you. The second why try to calculate the power in all three ways and you will be surprised: the voltage at the generator terminals is 10.4V; Current in the circuit 1.3A; Loop resistance 1+3=4ohm.

By the way, how much power will be dissipated in the resistor? - with these indicators.
Well, I started to worry about you.

Hi rakarskiy,

You say "Current in the circuit 1.3A; Loop resistance 1+3=4ohm." Implying that the current sense resistor (meter shunt) is 1.0 ohm. So then I²R = 1.3A²*4ohms = 6.76W. Still does not satisfy circuit equations. Besides who uses current instrument shunt of that high resistance? Typical would be on order of 5 milliohms. The common analog ammeter typically has full scale deflection of 50, 75, or 100 millivolts. When the power in the circuit branch yields different values using using I²R, I*V, or V²/R, there is mistake.
bi

[rakarskiy]

You have not made a full calculation! Loop resistance R=3Ω (load); r0=1Ω (source resistance, generator winding).

You did not make calculations from the table, for all options, I will do it for you:

1) P=I*U= 10.4*1.3= 13.52W

2) P=I^2*R = 1.3^2*3= 5.07W (This is what will be released on the resistor in the form of heat.)

3) P=U^2/R =10.4^2/3=36W

Now we will bring to the truth 3 formula:

P=U^2/(R+r0) =10.4^2/(3+1)=27W
P=1/2(U^2/(R+r0)) =0,5*(10.4^2/(3+1))=13.5W

The voltage in the circuit, taking into account the drop with the load (You need to know this to calculate the current in the circuit, more precisely, what part of the electric field will go into current strength):

U =E-(Ui=I*(R+r0) 

I = (E - U) / (R+r0)

But why is that? You do not know?
Then, what is the difference between the generator sources with an inductive winding and the battery, and what is the external EMF for calculating the current in the circuit with the battery source.


You surprised me a lot, I'm already very worried about you.


Although there are no such trifles in general educational materials:

https://drb-m.org/eb1/4-Ohm's%20Law,%20Power,%20and%20Energy.pdf

[bistander]

You have not made a full calculation! Loop resistance R=3Ω (load); r0=1Ω (source resistance, generator winding).

You did not make calculations from the table, for all options, I will do it for you:

1) P=I*U= 10.4*1.3= 13.52W

2) P=I^2*R = 1.3^2*3= 5.07W (This is what will be released on the resistor in the form of heat.)

3) P=U^2/R =10.4^2/3=36W

Now we will bring to the truth 3 formula:

P=U^2/(R+r0) =10.4^2/(3+1)=27W
P=1/2(U^2/(R+r0)) =0,5*(10.4^2/(3+1))=13.5W



The voltage in the circuit, taking into account the drop with the load (You need to know this to calculate the current in the circuit, more precisely, what part of the electric field will go into current strength):

U =E-(Ui=I*(R+r0) 

I = (E - U) / (R+r0)

But why is that? You do not know?
Then, what is the difference between the generator sources with an inductive winding and the battery, and what is the external EMF for calculating the current in the circuit with the battery source.


You surprised me a lot, I'm already very worried about you.


Although there are no such trifles in general educational materials:

https://drb-m.org/eb1/4-Ohm's%20Law,%20Power,%20and%20Energy.pdf

Hi rakarskiy,

You must have your own unique interpretation of Kirchhoff's laws.

Also, you say:

1) P=I*U= 10.4*1.3= 13.52W

2) P=I^2*R = 1.3^2*3= 5.07W (This is what will be released on the resistor in the form of heat.)

3) P=U^2/R =10.4^2/3=36W

1), 2) & 3) must all be numerically equal. There is only one power at the generator terminals.

And the these two equations, seemingly both for the same power:

P=U^2/(R+r0) =10.4^2/(3+1)=27W
P=1/2(U^2/(R+r0)) =0,5*(10.4^2/(3+1))=13.5W

Where did the 1/2 factor come from?

Obviously we speak different languages. Good luck.
bi