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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:

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 \\

Here’s a recap of where these come from

Energy lost to resistance

v(t) = R i(t)

the average power lost to a resistor is

&= \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
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.

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

&= \int_{-\infty}^t C \frac{dv}{dt} v(t) dt \\
&= \inv{2} C v^2(t).

The \( v^2(t) \) term can now be expanded in terms of phasors and averaged for

&= \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.

Energy stored in an inductor

The inductor energy is found the same way, with

&= \int_{-\infty}^t L \frac{di}{dt} i(t) dt \\
&= \inv{2} L i^2(t),

which leads to

= \frac{L}{4} \Abs{I}^2.

Energy lost due to conductance

Finally, we have conductance. In phasor space that is defined by

G = \frac{I}{V} = \inv{R},

so power lost due to conductance follows from power lost due to resistance. In the average we have

&= \inv{2 G} \Abs{I}^2 \\
&= \inv{2 G} \Abs{V G}^2 \\
&= \frac{G}{2} \Abs{V}^2


[1] J.D. Irwin. Basic Engineering Circuit Analysis. MacMillian, 1993.

[2] David M Pozar. Microwave engineering. John Wiley & Sons, 2009.