Transmission of energy in a dc line (Передача энергии в линиях постоянного тока)

All the energy entering the volume V through the surface s is released as heat in the receiver. The energy of the electromagnetic field released in the conductor in the form of heat penetrates them through the surface of these conductor. In this case we shall not neglect the resistivity of the wires themselves like we have done in the previous case. In such a case when we shall consider only some part of the transmission line, only some sample. And we can say that the energy which is dissipated inside the wire should be exactly equal to the energy which crosses the area of the cylinder which is a part of the wire. The right part its energy that dissipated inside. It might be expressed as squared current and resistance of this part of the wire. Here we assume that the current is distributed uniformly of the cross section of this wire and this power losses should be identical to this:

Magnetic field, induced by the wire, circulate around this wire, the field intensity, if there is some voltage applied to this wire, may be split into two parts. One part is the field which is normal to the surface of this wire, and the component of electric field which is parallel to the wire. This component must exist because the Ohm’s Law here is E_τ is equal to J/γ and so we shall consider both vectors: H, just circulate around the surface, that’s why in this point it is directed like it shown here, and E_τ, and we can see that these two vectors initiate a flow of the energy just inside the wire. So, in such a case energy crosses the cylindrical border of this wire and enters and then disperse inside the wire, it is transformed to the several (наверное имеется в виду рассеивание на несколько энергий) energy. In principle, it is easy to find exact expression of these vectors if you transform these reflections into the formulas. S - the Poynting’s vector, is E cross H (S=E x H) they are normal to each other that’s why we can simply multiply H and E_τ. H is the field which is induced by very long wire its current I/(2πr) (r₀ is a radius of this wire). I·R/l – is the resistance of the wire. I²·R is Power, 2π·r₀·l is an area of this surface. So, finally we get relations which is – energy which crosses the surface of the wire is the energy which is dissipated inside the same wire.

24. Transmission of energy in 2-wire line loaded on a R-L-C circuit (передача энергии по 2-проводной линии, нагруженной RLC)

Transmission of energy in a DC line (Передача энергии в линиях постоянного тока)

Similar consideration may be made for the load which consists of resistor, capacitor, and inductance. We should simply split the total energy which crosses this surface into several parts.

First of them, ∂W_m/∂t, the energy of the magnetic field which crosses this area corresponds to the magnetic field which is induced by inductance and is equal to

In any time moment, this power should not be equal to zero, but if the inductivity coil is ideal, of course, this power integrated, for example, by period will be equal to zero in any case.

A similar consideration can be done for the capacitor. Finally, we shall come to the same conclusion – it’s equal to the power which correspond to the capacitor.

And from the previous consideration we can also conclude that power in the resistor which is dissipated here is equal to such a product.

Now, this energy comes to the load from the external electromagnetic field and may be expressed as integral over the surface from E cross H, ds.

It is ∂W/∂t and then plus the energy which is dissipated in resistors. The energy ∂W/∂t is dependence of the energy stored in the inductance and capacitor and it is equal i·u_L + i·u_C and also i·u_R. We can transform this to i times sum of these voltages. It is power. Again, we can find out that energy that crosses this area is equal to the power, which now is not dissipation of the energy but the power of the energy transformation (into the heat, electric field, magnetic field). The power of the flow of electromagnetic energy entering the surface S is equal to the total power consumed in the circuit between ab terminals.