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In [2] section 2.2 is a comparison of field energy expressions with their circuit equivalents. It’s clearly been too long since I’ve worked with circuits, because I’d forgotten all the circuit energy expressions:
\begin{equation}\label{eqn:averagePowerCircuitElements:20} \begin{aligned} W_{\textrm{R}} &= \frac{R}{2} \Abs{I}^2 \\ W_{\textrm{C}} &= \frac{C}{4} \Abs{V}^2 \\ W_{\textrm{L}} &= \frac{L}{4} \Abs{I}^2 \\ W_{\textrm{G}} &= \frac{G}{2} \Abs{V}^2 \\ \end{aligned} \end{equation}
Here’s a recap of where these come from
Energy lost to resistance
Given
\begin{equation}\label{eqn:averagePowerCircuitElements:40}
v(t) = R i(t)
\end{equation}
the average power lost to a resistor is
\begin{equation}\label{eqn:averagePowerCircuitElements:60} \begin{aligned} p_{\textrm{R}} &= \inv{T} \int_0^T v(t) i(t) dt \\ &= \inv{T} \int_0^T \textrm{Re}( V e^{j \omega t} ) \Real( I e^{j \omega t} ) dt \\ &= \inv{4 T} \int_0^T \lr{V e^{j \omega t} + V^\conj e^{-j \omega t} } \lr{I e^{j \omega t} + I^\conj e^{-j \omega t} } dt \\ &= \inv{4 T} \int_0^T \lr{ V I e^{2 j \omega t} + V^\conj I^\conj e^{-2 j \omega t} + V I^\conj + V^\conj I } dt \\ &= \inv{2} \textrm{Re}( V I^\conj ) \\ &= \inv{2} \textrm{Re}( I R I^\conj ) \\ &= \frac{R}{2} \Abs{I}^2. \end{aligned} \end{equation}
Here it is assumed that the averaging is done over some integer multiple of the period, which kills off all the exponentials.
Energy stored in a capacitor
I tried the same sort of analysis for a capacitor in phasor form, but everything cancelled out. Referring to [1], the approach used to figure this out is to operate first strictly in the time domain. Specifically, for the capacitor where i = C dv/dt the power supplied up to a time t is
\begin{equation}\label{eqn:averagePowerCircuitElements:80} \begin{aligned} p_{\textrm{C}}(t) &= \int_{-\infty}^t C \frac{dv}{dt} v(t) dt \\ &= \inv{2} C v^2(t). \end{aligned} \end{equation}
The v^2(t) term can now be expanded in terms of phasors and averaged for
\begin{equation}\label{eqn:averagePowerCircuitElements:100} \begin{aligned} \overline{{p}}_{\textrm{C}} &= \frac{C}{2T} \int_0^T \inv{4} \lr{ V e^{j \omega t} + V^\conj e^{-j \omega t} } \lr{ V e^{j \omega t} + V^\conj e^{-j \omega t} } dt \\ &= \frac{C}{2T} \int_0^T \inv{4} 2 \Abs{V}^2 dt \\ &= \frac{C}{4} \Abs{V}^2. \end{aligned} \end{equation}
Energy stored in an inductor
The inductor energy is found the same way, with
\begin{equation}\label{eqn:averagePowerCircuitElements:120} \begin{aligned} p_{\textrm{L}}(t) &= \int_{-\infty}^t L \frac{di}{dt} i(t) dt \\ &= \inv{2} L i^2(t), \end{aligned} \end{equation}
which leads to
\begin{equation}\label{eqn:averagePowerCircuitElements:140} \overline{{p}}_{\textrm{L}} = \frac{L}{4} \Abs{I}^2. \end{equation}
Energy lost due to conductance
Finally, we have conductance. In phasor space that is defined by
\begin{equation}\label{eqn:averagePowerCircuitElements:160} G = \frac{I}{V} = \inv{R}, \end{equation}
so power lost due to conductance follows from power lost due to resistance. In the average we have
\begin{equation}\label{eqn:averagePowerCircuitElements:180} \begin{aligned} p_{\textrm{G}} &= \inv{2 G} \Abs{I}^2 \\ &= \inv{2 G} \Abs{V G}^2 \\ &= \frac{G}{2} \Abs{V}^2 \end{aligned} \end{equation}
References
[1] J.D. Irwin. Basic Engineering Circuit Analysis. MacMillian, 1993.
[2] David M Pozar. Microwave engineering. John Wiley & Sons, 2009.