Research website of Vyacheslav Gorchilin
2017-08-27
Free energy in a parametric RLC-circuits of the first kind of the second order
In this work we consider the electric circuit containing the nonlinear reactive elements: the capacitor and inductance. Their nonlinearity is determined by the parametric dependence: the capacitance — voltage on it, and inductance from the current flowing through it. The resistance is constant, but even if its value changed, for example, from the time the proof is not affected by. According to the classification here will be considered generators of the first kind and second order. In real devices the parametric dependence can be only one element capacitance or inductance, but here we show the General case and prove that these elements are independent from each other to influence change in the efficiency of the second kind.
We present here the proof is more General in comparison with the similar circuits of the first order. And to consider we will for serial RLC-circuit; parallel — the proof is derived the same way.
According to Kirchhoff's law, the sum of the stresses for each element a series circuit will equal the supply voltage $$U(t)$$: $U_C + U_L + U_R = U(t) \qquad (5.1)$ For the proof we will also need the following well-known formula that can adjust the current and voltage with each other via the active and reactive resistance: $U_R = R\,I, \quad U_L = L(I)\,\dot I_t \qquad (5.2)$ Substituting (5.2) in (5.1), we obtain the characteristic equation $U_C + L(I)\,\dot I_t + R\,I = U(t) \qquad (5.3)$ damnosa all members of which the current $$I$$ we get: $U_C\, I + L(I)\, I\,\dot I_t + R\,I^2 = U(t)\, I \qquad (5.4)$ Remembering what is the current in the capacitor can be found through the derivative of the voltage $$I = C(U_C)\,\dot U_{Ct}$$, we get the equation of instantaneous power in the circuit: $C(U_C)\,U_C\,\dot U_{Ct} + L(I)\, I\,\dot I_t + R\,I^2 = U(t)\,I \qquad (5.5)$ integrating, we get which is necessary to prove the equation of energy balance: $\int_{U_C(0)}^{U_C(T)} C(U_C)\,U_C\, dU_C + \int_{I(0)}^{I(T)} L(I)\,I\ dI + R \int_{0}^{T} I^2\, dt = \int_{0}^{T} U(t)\,I\, dt \qquad (5.6)$ Note the first two terms — under certain conditions, this is the expression for the free energy in a parametric circuit. An interesting property of these two integrals is their complete independence from the time coordinate $$t$$. Despite the apparent complexity of equation (5.6) one can notice that it consists of a simple sum of kinetic and potential energy: $W_{FC} + W_{FL} + W_R = W_E \qquad (5.7)$ thus, $$W_E$$ is the energy of the power source is a known value $$W_R$$ — energy dispersion in the active resistance, which is unknown to us, and $$W_{FC}$$, $$W_{FL}$$ — potential energy in capacitance and inductance, which can be found using the integrals in (5.6).
The proof for the full cycle (FCC)
Recall that in this case, in the reactive elements of the energy missing at the beginning and at the end of the cycle. Therefore, the boundaries of the integrals $$W_{FC}$$, $$W_{FL}$$ will be the same, therefore: $W_{FC} = 0, \quad W_{FL} = 0 \quad \Rightarrow \quad W_R = W_E \qquad (5.8)$

In chains of the second order, in its full cycle, it is impossible to gain energy even if the reactive elements is parametric. It does not depend on the nature of parametric dependencies, nor from the selected interval time.

The increment of energy in PCCIE
Partial cycle differs from the FCC the availability of energy in the reactive elements either at the beginning (PCCIE), or late (PCCFE). Another feature PCCIE is the lack of an external power source, and the initial energy is taken from capacitance or inductance, which accumulated her to the beginning of the studied cycle.

Recall that in the case PCCIE we do not was seen how the capacity or inductance have accumulated this energy, our task is how efficiently this energy will be dissipated in the active resistance.

Thus, equation (5.7) in this case looks like this: $W_R = - W_{FC} - W_{FL} \qquad (5.9)$ and accumulated to the beginning of this cycle, the energy in the reactive element will be expressed, obviously, so: $W_0 = \frac{C_0\, U_{C0}^2}{2} + \frac{L_0\, I_0^2}{2} \qquad (5.10)$ where $$C_0, L_0, U_{C0}, I_0$$ — circuit parameters at the beginning of the cycle. Consequently, the ratio of the increment of energy will be, as the ratio of the energy scattering on the resistance to the initial energy $$W_0$$: $K_{\eta 2} = \frac{W_R}{W_0} = -\frac{W_{FC} + W_{FL}}{W_0} \qquad (5.11)$ the Upper limit of integration for the first two integrals in (5.6), obviously, will be zero, and the lower one denoted as $$U_C(0)=U_{C0}$$ and $$I(0)=I_0$$. Then, swapping the upper and lower bounds, we find the increment of energy in PCCIE: $K_{\eta 2} = 2 { \int_\limits{0}^{U_{C0}} C(U_C)\,U_C\, dU_C + \int_\limits{0}^{I_0} L(I)\,I\ dI \over C_0\, U_{C0}^2 + L_0\, I_0^2} \qquad (5.12)$ From the equation immediately shows that if the capacity and inductance nonparametric, increment of power and $$K_{\eta 2} = 1$$.
Conditions for PCCFE
PCCFE from the definition it follows that equation (5.7) remains unchanged, but the lower border of the first two integrals of (5.6) is equal to zero. We rewrite this equation as $W_R = W_E - W_{FC} - W_{FL} \qquad (5.13)$ it is Obvious that to obtain the energy surplus energy is allocated to resistance should be more than consumed by the power source, and this means that the sum of terms $$W_{FC}, W_{FL}$$ must be less than zero: $W_{FC} + W_{FL} \lt 0 \quad \Rightarrow \quad \int_{0}^{U_C(T)} C(U_C)\,U_C\, dU_C + \int_{0}^{I(T)} L(I)\,I\ dI \lt 0 \qquad (5.14)$ Possible physical and mathematical sense of the inequalities described here, and how to find the increment of efficiency, we will show next. To do this, compare the energy dissipation of active resistance to the cost of the power: $K_{\eta 2} = {W_R \over W_E} = {W_E - W_{FC} - W_{FL} \over W_E} = 1 - {W_{FC} + W_{FL} \over W_E} \qquad (5.15)$ from Here we immediately see that if we keep the condition (5.14), then $$K_{\eta 2}$$ is greater than one. A more complete formula for efficiency at PCCFE will be like this: $K_{\eta 2} = 1 - {1 \over W_E} \left( \int_{0}^{U_C(T)} C(U_C)\,U_C\, dU_C + \int_{0}^{I(T)} L(I)\,I\ dI \right) \qquad (5.16)$
Insights
Insights on circuits of the second order similar to those done on the circuits of the first order. However, repeat them.
In this work we proved that it is impossible to gain energy in parametric circuits of the second order in the full cycle (FCC) because the energy dissipated in the resistance is always equal to the energy expended by the power formula (5.8). But if the cycle is incomplete, the receiving gain becomes achievable task. If reactive elements contain potential energy in the beginning of the cycle (PCCIE), the gain can be found using the formula (5.12). If the energy in the reactive elements is at the end of the cycle, then the conditions for receiving allowances, we can find from formula (5.14), and the increment of efficiency by (5.16).
You need to understand that there is a mathematically strictly proved potentially achievable values of the increment of energy, part of which, in the real reactance, can be spent inefficiently, for example, on heating. However, on the basis of evidence about the energy increment in the fractional cycles, one can obtain special cases for engineering calculations, which, in turn, will allow you to build a real device with high efficiency.
Additional materials and some special cases with examples from real wireless components you can see here.
Go to the next section which describes the parametric circuits of the second kind can here.

The materials used