For the correct calculation of electric circuits, we need a model of the electrostatic capacitor (ESC), details of which we discussed in the previous part of this work. Unfortunately, to get a model is not possible because the ESC, and the associated reference system, previously simply was not considered. Let's calculate it yourself.

Model dvuhkletevogo conventional capacitor is shown in figure (3a), and its reactance in the steady circuits is given by: \[X_{Cs} = {1 \over \Bbb{i} \omega C_S} \qquad (2.1)\] where: \(\Bbb{i}\) is the imaginary unit, indicating a phase shift of 90 degrees, \(\omega\) angular frequency, which is associated with the usual way: \(\omega = 2\pi f\), and \(C_S\) is the capacitance of this capacitor.

From previous part we know that the HSC is composed of three tanks, one conventional and two — solitary. For the calculation of electronic circuits it is necessary that all the capacity was doobsledovanie, so our model will consist of a capacitor of \(C_S\), whose capacity equal to the capacity between two of its plates, and capacitor \(C_1 C_2\), which in the scheme as well — duroblade, but has a secluded capacitance value of the first and second electrode, respectively (Fig. 3b). Also, there will be a source of charges that the scheme is designated as source voltage \(U_g\), connected in series with \(C_2\). In this model, we combine two independent reference systems and assume that for lone tanks the second surface is Land, so the schema specifies the mandatory grounding. It will be assumed in the idealized calculations, though not explicitly shown.

Fig.3. Model conventional and electrostatic capacitor for the calculation of diagrams (a-c). Double-circuit scheme for ESCOs (d).

Remembering the formula (1.2), and the conclusion from the previous section on the phase shift, we can enter a specific value for a new source of tension: \[U_g = \Bbb{i} U_1 \frac{C_1}{C_2} \qquad (2.2)\] Recall that \(\Bbb{i}\) is the imaginary unit, indicating a positive phase shift of 90 degrees. This is different calculation schemes of the alternating current with fixed settings by using complex numbers: in order to show a positive phase shift, a sufficient parameter (current, voltage or resistance) multiplied by \(\Bbb{i}\), and to show negative on \(\Bbb{i}\) with a minus sign. Shift 180 degrees to reflect the even easier: it is necessary before setting a minus sign [1,2].

Interestingly, this additional source of charge is never taken into account in the calculation of the diagrams in theoretical electrical engineering, although it is known that electrostatic induction can often be the cause of failure of some elements of the radio equipment. Next, we will try to turn this disadvantage into an advantage and use this effect on purpose.

Let's check the power ratio of such a source. Imagine independent from the further presentation of the picture and consider a capacitor of \(C_2\) and the source of \(U\) separately from the rest of the circuit (Fig. 3c). Let the capacitor is discharged, and the source changed my voltage jump from zero to \(U_g\). Then the energy of full charge the capacitor will be known formula: \[W = \frac{C_2 |U_g|^2}{2}\] the Energy released at the active load \(r\) is exactly the same. In relation to the algorithm shown in figure (2) from the previous section, we know that \(|U_g| = |U_1| \frac{C_1}{C_2}\) and charge \(C_2\) is the same as for \(C_1\): \(q_1 = C_1 |U_1|\), from which we obtain finally the formula for the energy of full charge of the capacitor: \[W = \frac{C_1^2 |U_1|^2}{2 C_2} = \frac{q_1^2}{2 C_2}\] If now the source of \(U\) will change its voltage in the opposite direction: from \(U_g\) to zero, then the resistor again clears the same energy. Energetically, it repeats proposed in the previous section algorithm. In the real scheme proposed below, the role of the load will be to carry multiple items, including inductance, thanks to kotorym we will be able to get a positive feedback.

Note that the model ESCs (Fig. 3b) can be used for any other circuit variants, but we focus on the most simple of dual.

Looking ahead, we can say that without the presence of inductances to obtain the necessary us a positive feedback schemes with ESCOs impossible, so they are included in the following scheme (Fig. 3d). It depicts two linked circuits, each of which represents an active resistance, inductance and capacitance. The connection between the circuits is performed through the capacitor is \(C_S\). The currents from the two voltage sources are indicated as follows: from \(E\) -> \(I_1\) and \(I_2\), and \(U_g\) -> \(Ig_1\) and \(Ig_2\). Thus, it is possible to consider these two chains as independent and then calculate their total currents through \(r_1\) and \(r_2\), and after them — the balance of power. We will not torture our chitateli long derivation of the formula, and immediately present the results. Anyone can always double-check their conclusion, moreover, we also recommend the excellent work of circuits [3].

The reduction calculations are applied to the following: \[k = \frac{C_2}{C_1}, \quad k_s = \frac{C_1}{C_S}, \quad k_r = \frac{r_2}{r_1} \quad \Delta = \omega C_1 r_1 \] \[X_{C1} = {1 \over \Bbb{i} \omega C_1} = {r_1 \over \Bbb{i} \Delta}, \quad X_{C2} = {1 \over \Bbb{i} \omega C_2} = {r_1 \over \Bbb{i} \Delta k}\] \[X_{L1} = {\Bbb{i} \omega L_1}, \quad X_{L2} = {\Bbb{i} \omega L_2}, \quad X_S = {1 \over \Bbb{i} \omega C_S} = {r_1 k_s \over \Bbb{i} \Delta}\] \[T_1 = r_1 + X_{L1}, \quad T_2 = r_1 k_r + X_{L2}\] Then, according to [1] find the total resistance for the first frame of reference related to the source voltage \(E\): \[Z_1 = T_1 + {r_1 \over \Bbb{i} \Delta} {1 + k_s k \over 1 + k + k k_s}\] \[Z_2 = T_2 + {r_1 \over \Bbb{i} \Delta} {1 + k_s \over 1 + k + k k_s}\] \[Z_C = {r_1 \over \Bbb{i} \Delta} {1 \over 1 + k + k k_s}\] \[A = Z_1 Z_2 - Z_C^2\] Where will immediately find the currents: \[I_1 = E {Z_2 \over A}, \quad I_2 = E {Z_C \over A}\] the same will do for the second reference system, associated with a source \(U_g\): \[Z_{g1} = X_{C1} + {T_1 (X_S + T_2) \over T_1 + T_2 + X_S}\] \[Z_{g2} = X_{C2} + {T_2 (X_S + T_1) \over T_1 + T_2 + X_S}\] \[Z_{gC} = {T_1 T_2 \over T_1 + T_2 + X_S}\] \[A_g = Z_{g1} Z_{g2} - Z_{gC}^2\] Where we find the currents: \[I_{g1} = U_g {Z_{gC} \over A_g}, \quad I_{g2} = U_g {Z_{g1} \over A_g}\] Connecting these two systems of reference voltage \(U_1\) (see Fig. 3d) is thus: \[U_1 = I_1 X_{C1} {k A_g (X_S + X_2) \over (X_S + X_2 + X_{C1}) (k A_g - \Bbb{i} Z_{gC} X_{C1}) } \qquad (2.3)\] Here and then apply two more cuts: \[X_1 = {T_1 X_{C1} \over T_1 + X_{C1}}, \quad X_2 = {T_2 X_{C2} \over T_2 + X_{C2}} \] Then the total current flowing through the resistance \(r_2\), which is our load can be found as: \[I_{22} = I_2 + I_{g2} {X_S + X_1 \over X_S + X_1 + T_2} \qquad (2.4)\] Yet we ponadobitsa the total current flowing simultaneously through \(r_1\) and power \(E\): \[I_{11} = {E - U_1 \over T_1} \qquad (2.5)\] Recall also that due to (2.2): \[U_g = \frac{\Bbb{i}}{k} U_1 \qquad (2.6)\] Now we come to the main — balance of power. It is the ratio of output power to input \[K2 = {P_2 \over P_1} \qquad (2.7)\] and the power, respectively, so: \[P_1 = E |I_{11}|, \quad P_2 = |I_{22}|^2 r_1 k_r \qquad (2.8)\] note that the currents here are taken modulo.

In the next part of this work, on the basis of these formulas, we give a detailed calculation of generator ESK with a spherical capacitor.