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Power Dissipation

From the point of view of the power supply,21.5 the circuit is a "black box" that "resists" the applied voltage with a rather weird "back ${\cal{EMF}}$" ${\cal E}_{\rm back}$ given by $R_{\rm eff}$ times the current I; ${\cal E}_{\rm back}$ is given by the sum of all three terms in Eq. (13) or the sum of the three vectors in Fig. 21.2. The power dissipated in the circuit is the product of the real part of the applied voltage21.6 and the real part of the resultant current21.7
 
P(t) = $\displaystyle \Re {\cal E} \times \Re I
= \Re \left( {\cal E}_0 e^{i \omega t} \right)
\Re \left( I_0 e^{i \omega t} \right)
\cr$ (21.14)

which oscillates at a frequency $2 \omega$ between zero and its maximum value

 \begin{displaymath}P_{\rm max} = {\cal E}_0^2 \Re \left( {1 \over R_{\rm eff}} \right)
\end{displaymath} (21.15)

so that the average power drain is21.8

 \begin{displaymath}\langle P \rangle = {1\over2} {\cal E}_0^2
\left[ R \over R^2 + (X_L - X_C)^2 \right] \; .
\end{displaymath} (21.16)

A variation on this algebra yields the practical formula

 \begin{displaymath}\langle P \rangle = {\cal E}_{\rm rms} \; I_{\rm rms} \; {R \over Z}
\end{displaymath} (21.17)

where ${\cal E}_{\rm rms} = {\cal E}_0/\sqrt{2}$,

 \begin{displaymath}I_{\rm rms} = {{\cal E}_{\rm rms} \over Z}
\end{displaymath} (21.18)

is the root-mean-square current in the circuit, R/Z is the "power factor" of the RC circuit and

 \begin{displaymath}Z \equiv \vert R_{\rm eff}\vert = \sqrt{R^2 + (X_L - X_C)^2}
\end{displaymath} (21.19)

is the impedance of the circuit.21.9
next up previous
Up: AC Circuits Previous: The Differential Equation
Jess H. Brewer - Last modified: Mon Nov 16 18:21:16 PST 2015