Another Artificial Capacitor

Looking at the circuit, its equation can be written using Kirchoff's voltage law as follows:

$V + IR = 0 \implies V + RC\ddot{V} = 0$

Given that $R = 2$ and $C = 3$ , the equation can be rewritten as:

$\ddot{V} + \frac{1}{6}V = 0$

The general solution is:

$V = A \cos\left(\frac{t}{\sqrt{6}}\right) + B \sin\left(\frac{t}{\sqrt{6}}\right)$

Plugging in the initial conditions and solving for $A$ and $B$ gives:

$\boxed{V = 10\left(\cos\left(\frac{t}{\sqrt{6}}\right) +\sqrt{6}\sin\left(\frac{t}{\sqrt{6}}\right) \right)}$

Also,

$I = C\ddot{V} \implies I = -\frac{V}{2}$

The period of the circuit is:

$T = 2\pi\sqrt{6}$

Consider the circuit equation:

$V + IR = 0$

Multiplying the equation with $I dt$ gives:

$VI dt + I^2Rdt = 0 \implies \int_{0}^{T}I^2R dt = \int_{0}^{T}-VI dt$

This equation can be interpreted as: 'The energy dissipated by the resistor is equal to the energy supplied to the circuit'. From here, the energy supplied by the active element is concluded to be:

$E = -\int_{0}^{T} VI dt = -\int_{0}^{T} V\left(\frac{-V}{2}\right)dt = \frac{1}{2}\int_{0}^{T}V^2dt$

$E = 50\int_{0}^{T}\left(\cos\left(\frac{t}{\sqrt{6}}\right) +\sqrt{6}\sin\left(\frac{t}{\sqrt{6}}\right) \right)^2dt$ Solving this integral gives:

$\boxed{E = 350\pi\sqrt{6} = 2693.35}$

This is indeed quite interesting. If one looks at a plot of voltage vs. time, it can be seen that the capacitor periodically charges (The voltage across it increases). This is contrary to conventional understanding that a charged capacitor when attached to a resistor only discharges.