Question 12.52: In Fig. 12.91(a), C=100 μF, ν(t)=Vcos(ωt) with f=60 Hz an...
In Fig. 12.91(a), C=100 \mu F , \nu(t)=V \cos (\omega t) with f=60 Hz and V=15 V . The dissipation factor for the capacitor is 0.03 . The ESL and parallel resistance R_p can be neglected. Find the power dissipated in the capacitor (in the ESR ).
We may determine the current through the model and the power dissipated in the ESR , as follows: we refer to the model in Fig. 12.91(b), where (see Fig. 12.90)
\begin{aligned} X_s=-\frac{1}{\omega C} &=-\frac{1}{2 \pi(60 Hz )(100 \mu F )} \cong -26.5 \Omega \\\\ R_s &=-X_s D=(26.5 \Omega)(0.03) \cong 0.80 \Omega \end{aligned}
Thus
Z_s=(0.80-j 26.5) \Omega \cong 26.5 \angle-1.54 \Omega
and
\tilde{I}=\frac{\tilde{V}}{Z_s}=565 \angle 1.54 mA
The power dissipated in the ESR is
P=I_{r m s}{}^2 R_s=\frac{|\tilde{I}|^2}{2} R_s=127 mW
D=\frac{1}{Q}=\omega R_s C_s =\text{ dissipation factor.} (12.122)
D=\omega R_s C_s = \frac{R_s}{-X_s}=-\tan (\delta) \cong \cos (\theta_Z) . (12.123)
