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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.