1/u = du/dt -- capacitor constant power discharge

Yesterday, I have discovered the magnitudes which change at their value speed: v=v′. Today, playing with capacitor discharging equations, a solution of 1/u=u′ has surprisingly came.

There are three types of discharge I have analyzed (plots to be added):

1. Via a constant resistor: U_r(t) = U₀*e^-t/RC or dU/dt = - kU (minus is because of discharge)


2. Constant current: U_i(t) = U₀ - I*t/C or dU/dt = - k


3. Constant power: ? or dU/dt = - k/U

The latest is not known to the general public (like me). It is the fastest discharge because the constant resistor in (1) drops both capacitor voltage and discharge current current (slope becomes less vertical), while the constant current discharge (2) reduces the voltage and thus the power, the energy extracted from the capacitor per time unit, P = U*I, drops too.

How do we compute the U_p(t)? The energy comes down at const power rate P:
E(t) = E₀ - Pt
The capacitor's Energy is defined by its U: E_cap = 1/2 CU². So,
1/2 CU² = 1/2 CU₀² - Pt
Moving 1/2 C to the left and taking the root we derive the U(t) sought:
U_p(t) = √‾(U₀² - 2Pt/C)‾

Initially, I have made a bad U(t) plot here drawing y=y₀ -√‾x rather than y=√‾(y₀ -x). This turned me to think about the current. The first nuance I have mentioned is that voltage must start discharging slowly and speed up to I= ∞ when U drops to zero. I knew that the current is the speed of voltage change I = -CU′. At the same time, the power formula is P = UI, where P is constant. Therefore, the differential equation that naturally comes:
I = P/U = -CU′ or U′ = -P/C * U

Suddenly, we have got u(t)=sqrt(U0² + 2kt) as a u′ = k/u diff equation solution. This extends my considerations of capacitor (dis-) charging at electronix.ru.

Philosophic conclusions: we have discovered how i connects sin/cos (mutual growth) with exp (self-growth) and sqrt (inverse growth).