Research website of Vyacheslav Gorchilin
2021-09-01
Parametric RL and RC circuits of the first kind
In this work, we will try to draw the attention of free energy seekers to parametric circuits. This little-known part of radio electronics is still an unpopular and even closed topic. Meanwhile, under certain conditions, such circuits can open up new opportunities for the search for new sources of energy. by increasing Class II efficiency.
Here we will consider chains of the first order, i.e. including RC or RL elements [1]. The difference from the classical approach will be that the capacitance or inductance will behave nonlinearly, or rather, parametrically depend on the current or voltage. The purpose of this note: with the help of classical physics and mathematics, find the conditions under which such circuits can give an energy gain, and also derive formulas for calculating the potentially possible free energy in such chains.
Recall that parametric chains of the first kind are characterized by the following dependencies: $$L = L (I_L)$$, $$C = C (U_C)$$. This type is due to the natural characteristics of the materials from which the investigated element is made. For example, the inductance of a coil depends on the permeability of the core, which depends on the strength of the magnetic field, and therefore on the current passing through it: $$L = L (I_L)$$. Varicap and varicond change their capacity depending on the voltage applied to them: $$C = C (U_C)$$. Further, we will denote such dependencies in a more general form: $$Z = Z (Y)$$.
Let's compose a series circuit consisting of a generator U, which produces periodic oscillations, in principle, of any shape, active resistance R and reactive element Z (Fig. 1a). The latter can be either inductance or capacitance, the magnitude of which depends on the current or voltage across it. In the case of an RL circuit, a parametric inductance acts as a reactive element, the value of which depends on the flowing current, and in the case of an RC circuit, a parametric capacitance, the value of which depends on the voltage applied to it. This approach will allow in this work to simplify the number of formulas and make the proof the most general.
 Fig.1. Parametric RL or RC-circuit (a), and graph of the current in the circuit or voltage on the reactive element (b)
For even greater simplification of formulas and reasoning, we take the value of resistance $$R$$ ??equal to one. Since this is a constant, we can make it different from this value at any time. Then we can write the general differential equation for a first-order chain as follows [1]: $\dot \Psi (t) + Y (t) = U (t) \qquad (1.1)$ $\quad \dot \Psi (t) = \partial \Psi (t) / \partial t, \quad \Psi (t) = Z (Y) \, Y ( t)$ where: $$Y (t)$$ - represents the voltage across the reactive element, if it is an RC circuit, or the current in the circuit, if it is an RL circuit; $$Z (Y)$$ - reactance, the value of which depends on $$Y$$. This element can be capacitance or inductance, depending on the type of circuit: RC or RL, respectively. For example, for an RL-circuit, the dependence $$L (I)$$ is used as $$Z (Y)$$, where $$L$$ is the inductance, and $$I$$ - current. This relationship is described more fully in of this work. By $$\dot \Psi (t)$$ we mean the voltage across the reactive element if it is an RL circuit or current if it is an RC circuit. The expression is obtained by differentiating it simultaneously both by $$Y$$, which depends on time, and by $$Z$$, which depends on $$Y$$. For example, in an RL-circuit, the element $$\Psi$$ is equal to the flux linkage [2], the time derivative of which is equal to the EMF voltage across the inductance.
Due to such a complex dependence, it is not possible to obtain an analytically general formula for the dependence of $$Y$$ on time. Therefore, we will go the other way and get the most general energy ratios for such a scheme (Fig. 1a). But first, let's simplify the further formulas a little: $\Psi = \Psi (t), \quad Y = Y (t), \quad U = U (t), \quad Z = Z (Y)$
RL Chain Energy
Further, we will assume that the generator in the circuit (Fig. 1a) generates periodic oscillations, the shape of which can be of any shape. This means that the current in this circuit will also be periodic (Fig. 1b). To obtain the energy balance of the circuit, we multiply all the terms of formula (1.1) by $$Y$$ (current in this case), and integrate them over one oscillation period: $\int_0 ^ T \dot \Psi \, Y \, \partial t + \int_0 ^ TY ^ 2 \, dt = \int_0 ^ TU \, Y \, dt \qquad (1.2)$
Note. The proof will be the same if we take other intervals of integration, over any number of periods $$T$$.
Let us introduce the notation for each term of this equation: $W_r + W_a = W_g \qquad (1.3)$ Here $$W_r$$ is the energy in the reactive element, $$W_a$$ is the energy in the active element, and $$W_g$$ is the generator energy. Consider the integral of reactive energy from (1.2): $W_r = \int_0^T \dot \Psi\,Y\, \partial t = \int_0^T Y\, \partial (Z\, Y) = Z\, Y^2 \bigg |_{Y(0)}^{Y(T)} - \int_{Y(0)}^{Y(T)} Z\, Y\, \partial Y \qquad (1.4)$ After transformations, we got an expression that does not depend on the time coordinate, which is a very important result for further calculations. Pay attention to this formula - it contains potentially achievable energy in parametric reactances. Now, if you look at the limits of integration of expression (1.4), its meaning immediately becomes clear: at the beginning and at the end of the period $$Y$$ is equal to zero (Fig.1b), which means that this whole integral is also equal to zero: $\int_0 ^ T \dot \Psi \, Y \, \partial t = 0, \quad W_r = 0 \qquad (1.5)$ The result obtained is quite obvious: one part of the period the reactive element accumulates energy, and in the other part it gives it away. The sum is zero. But here one more important point was proved, which is that this principle also applies to coils in which the inductance changes depending on the flowing current - $$Z (Y)$$, and the type of this dependence is completely unimportant.
Now we can return to formula (1.2) and write it like this: $\int_0 ^ T Y ^ 2 \, \partial t = \int_0 ^ T U \, Y \, \partial t \qquad (1.6)$ If we rewrite it for an RL-chain, replace $$Y$$ by $$I$$ and remember the unit $$R$$, we get: $R \int_0 ^ T I ^ 2 \, \partial t = \int_0 ^ T U \, I \, \partial t \qquad (1.7)$ But this expression represents the balance of energies in our circuit, meaning that for any full number of periods, all the energy of the generator is completely dissipated on the active resistance $$R$$: $W_a = W_g$ From here we can draw the following conclusion:
In a parametric RL circuit of the first kind, for any total number of periods, it is impossible to obtain an increase in energy from the parametric inductance for any character of its parametric dependence.
Let's check that the integral $$W_r$$, over the full period, is equal to zero even if we split it into several sections. To do this, we divide the period T into two such segments: $$T_1$$ and $$T_2$$ (see Fig.1b), and for them we obtain the sum of integrals: $\int_ {Y (0)} ^ {Y (T)} Z \, Y \, \partial Y = \int_ {Y (0)} ^ {Y (T_1)} Z \, Y \, \partial Y + \int_ {Y (T_1)} ^ {Y (T_2)} Z \, Y \, \partial Y \qquad (1.8)$ Since the current is periodic, $$Y (T_2) = Y (0)$$, therefore: $\int_ {Y (0)} ^ {Y (T_1)} Z \, Y \, \partial Y + \int_ {Y (T_1)} ^ {Y (T_2)} Z \, Y \, \partial Y = \int_ {Y (0)} ^ {Y (T_1)} Z \, Y \, \partial Y - \int_ {Y (0)} ^ {Y (T_1)} Z \, Y \, \partial Y = 0 \qquad (1.9)$ The zero sum is proved in the same way for the first term on the right-hand side of expression (1.4).
Here we do the same, only in this case we multiply (1.1) by $$\dot \Psi$$, which represents the current in the circuit here: $\int_0 ^ T \dot \Psi ^ 2 \, \partial t + \int_0 ^ TY \, \dot \Psi \, dt = \int_0 ^ TU \ , \dot \Psi \, dt \qquad (1.10)$ Also, by analogy, we express each term of this expression in terms of the corresponding energies: $W_a + W_r = W_g \qquad (1.11)$ Let's turn our attention to reactive energy: $W_r = \int_0 ^ TY \, \dot \Psi \, \partial t = \int_0 ^ TY \, \partial (Z \, Y) = Z \, Y ^ 2 \bigg | _ {Y (0)} ^ {Y (T)} - \int_ {Y (0)} ^ {Y (T)} Z \, Y \, \partial Y \qquad (1.12)$ Thus, we have obtained an expression that fully corresponds to formula (1.4), and hence the property that it possesses: the reactive energy in the circuit, for any full number of periods, is zero, which means that all the energy of the generator is completely dissipated on the active resistance $$R$$: $W_a = W_g$ Therefore, we immediately draw the following conclusion: