Research website of Vyacheslav Gorchilin
2016-07-29
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 $$U_{0}$$. 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.