Potencia en AC Potencia en AC. Corrección del Factor de Factor de ...
Potencia en AC Potencia en AC. Corrección del Factor de Factor de ...
Potencia en AC Potencia en AC. Corrección del Factor de Factor de ...
Create successful ePaper yourself
Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.
θ<br />
S<br />
P R<br />
Q C<br />
Q L<br />
ZRL RL= = 426.5 426.5Ω∟62.05 62.05º=199.9+j376.75<br />
º=199.9+j376.75<br />
ZC= = 800.8 800.8∟-90 90º º = -j800.8 j800.8<br />
ZT=ZRL// RL// ZC=(341541.2 =(341541.2∟-27.95 27.95º)/ )/(469.03 (469.03∟-64.77 64.77º)<br />
ZT= = (728.18∟36.82<br />
(728.18 36.82º)<br />
I IT= = (110 (110∟0º)/ (110 (110∟0º)/ º)/ (728 (728.18 (728.18∟36.82 18 18∟36 36.82º) 36 82 82º)<br />
IT= = 0.15 0.15∟-36.82 36.82º<br />
IL= = 0.258 0.258∟-62.05 62.05º<br />
I T<br />
Resonancia <strong>en</strong> <strong>AC</strong>.<br />
Respuesta <strong>en</strong> Frecu<strong>en</strong>cia.<br />
S Q L<br />
θ<br />
36.87º Q´<br />
P R<br />
Q C<br />
θ =62.05º-> Cos Cosθ= 0.468<br />
QL=I =I 2 XL=25.09 25.09 VAR<br />
Determine el Capacitor para que<br />
el factor <strong>de</strong> pot<strong>en</strong>cia sea 0.8<br />
Cos Cosθ´=0.8 -> θ´=36.87º<br />
tg tgg θ´= Q´/P Q / R<br />
Q´= = PRtg tg θ´=13.31 13.31tg tg 36.87º= 9.98 VAR<br />
QC= QL-Q´= = 25.09 25.09-9.98= 9.98=15.11 15.11 VAR<br />
I IT QC= = V V2 /X /XC C -> > XC= = V2 /Q C<br />
XC= = 800.8 800.8Ω -> > C=3.31x10 -6F IT= = 0.15 0.15∟-36.82 36.82º<br />
IL= = 0.258 0.258∟-62.05 62.05º<br />
IC= = 0.14 0.14∟90 90º<br />
I L<br />
Circuito Resonante Serie <strong>en</strong> <strong>AC</strong>.<br />
Resonancia!!!<br />
E<br />
2