Interesting Relationship

There is a little typo in this, which does not change the result. In the integral, the current term needs to be to the power of two. In the next step this is corrected. This is a valid solution otherwise.

First, let's determine the current within this RC circuit. By Kirchhoff's Voltage Law:

$V = Ri(t) + [i_{C}(0) + \frac{1}{C} \int_0^t i(\tau) \, d\tau]$

which differentiating both sides with respect to time $t$ yields $0 = R\frac{di}{dt} + \frac{1}{C} \cdot i \Rightarrow \frac{di}{dt} = -\frac{i}{RC} \Rightarrow i(t) = I_{0}e^{-\frac{t}{RC}} = (\frac{V}{R})e^{-\frac{t}{RC}}.$

Now the voltage across the capacitor is measured according to $V_{C}(t) = V(1 - e^{-\frac{t}{RC}})$ , and $V_{C} \rightarrow V$ as $t \rightarrow \infty.$ The limiting energy stored in the capacitor is just $E_{C} = \boxed{\frac{CV^2}{2}}.$

The power dissipated across the resistor for $t \in [0, \infty)$ equals $P(t) = Ri^{2}(t) = \frac{dE}{dt}$ , and solving for the energy yields:

$E_{R} = \int_0^\infty P(t) \, dt = \int_0^\infty Ri^{2}(t) \, dt = \int_0^\infty R(\frac{V^2}{R^2})e^{-\frac{2t}{RC}} \, dt = -\frac{CV^{2}}{2} \cdot e^{-\frac{2t}{RC}} = -\frac{CV^2}{2}(0 - 1) = \boxed{\frac{CV^2}{2}}.$

It turns out $\boxed{E_{R}(t) = E_{C}(t)}$ as $t \rightarrow \infty$ , which adheres to the Law of Conservation of Energy. Choice C is correct.