Research website of Vyacheslav Gorchilin
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Parametric variation of the resistance, capacitance and inductance in RLC-circuits
1. Parametric variation of the resistance
Consider a simple RC-chain, but with one addition — the resistance \(R\) here parametrically varies with time: \(R = R(t)\). Capacity \(C\) are constant in time, but is pre-charged to a voltage \(U_{0}\). RC-цепочка с параметрическим R зависящим от времени I.e. at the initial moment of time, before the circuit key SW: \[U(0)=U_{0} \qquad (1.1)\]
Make the equation for this circuit assuming that the key SW is closed at time \(t=0\): \[ {dU(t) \over dt} + {U(t) \over C \ R(t)} = 0 \qquad (1.2) \] we have Before us a homogeneous differential equation of the first order. After his decision [1] and finding of the initial coefficients, we obtain the final dependence of the voltage on the capacitor from time to time: \[ U(t) = U_{0} \ exp \left( -{1 \over C} \int {dt \over R(t)} \right) \qquad (1.3) \]
But we need to get the value of energy that will get resistance by discharge of the capacitor. To do this, recall the school formula for power is: \[ I(t) = {dQ(t) \over dt} \] where \(Q\) is the charge located in the condenser. Also recall that: \(Q(t) = C\,U(t)\). To find the total energy \(W_{R}\) that passes through \(R\), we apply the well-known formula in integral form: \[ W_{R} = \int_0^\tau I(t)^{2} \, R(t) \, dt \qquad (1.4) \] where \(\tau\) — discharge time. Substituting all the previously obtained formulas, differentiating and making the substitution, we obtain the final result: \[ W_{R} = { C\,U_{0}^2 \over 2 } \left( 1 - exp \left( -{2 \over C} \int_0^\tau {dt \over R(t)} \right) \right) \qquad (1.5) \]
Now we have to find the energy \(W_{c0}\), which we had to pre-charge the capacitor, subtract from it remaining in the case, if we don't expect full charge, and then compare the expended energy \(W_{C}\) and stress \(W_{R}\).
The total potential energy of the capacitor is on the well-known formula: \[ W_{c0} = {C\,U_{0}^2 \over 2}, \] but if you don't wait for the full discharge of the capacitor and open circuit key SW at the time \(\tau\), then the remaining energy will be so: \[ W_{c\tau} = {C\,U(\tau)^2 \over 2} \qquad (1.6) \] the voltage on the capacitor, when opening the key, is given by (1.3): \[ U(\tau) = U_{0} \ exp \bigg( -{1 \over C} \int_0^\tau {dt \over R(t)} \bigg) \qquad (1.7) \] Thus, the spent energy will be, as the difference \(W_{c0}\) and \(W_{c\tau}\): \[ W_{C} = W_{c0} - W_{c\tau} = { C\,U_{0}^2 \over 2 } \left( 1 - exp \left( -{2 \over C} \int_0^\tau {dt \over R(t)} \right) \right) \qquad (1.8) \] Comparing formulas (1.5) and (1.8) we see that \(W_{C}\) and \(W_{R}\) equal, and it does not depend either on the properties of parametric resistance, or when opening a key.

Thus, in an RC circuit with a constant capacitance it is impossible to gain energy, even if the resistance is parametric!

The proof for RL-circuit is similar, only instead of voltage is substituted for current \(I\) instead of \(C\) is the inductance \(L\) and \(R\) is moved from the denominator to the numerator: \[ {dI(t) \over dt} + {I(t) \, R(t) \over L} = 0 \] Is a proof the reader can infer on their own, but it also shows that the case of constant inductance, RL-circuits also cannot be receiving increase energy, even if \(R\) — dimensional.