Question

A 1.15 series to a resistor and a 520 - mH inductor are connected in 200 - Hz generator with an rms voltage of 13.2 V

A1.15-k resistor da 520 m inductor are connected in series to 1200-H generator with an is voltage of 13.2 V Part A What is the ma current in the circuit? 180 AEG Submit Request new Part в What capacitore must be red in series with the restorand inductor to reduce theme current to the value found in part ? ΑΣ 2 Submit

Answer 1

f= 200 Hz L=0.52 Hz R=1150r X = wharafl = 27 (200) (052) I 653.12 2 = J R² + x ² = Jso? + 653,12² 1322.5n isus Kons 13.2 v て = 0.01 Amp Ans. 1322.5 if we add capacitor, z' = J 1150 + (653.12-Xc) irms 2 = 22 Z (213² 42² 1180² + (653-12-*c)² - 4 ( 1158 + 653 12) 4 Xco-1728.8n ur X c = 3035.12 ( - 3035) >> zafc 0.26x10 of 2(200) (3035:1) = 0.26 u F Am