Question

pretty sure part A is correct really need help on part B

A 1.15-k resistor and a 566-mH inductor are connected in series to a 1500-Ha generator with an mms voltage of 12.5 V Part A What is the current in the circuit? I... 2.3 m Previous Answers Completed Part B What capacitance must be inserted in series with the resistor and inductor to reduce the rms current to half the value found in part ? 80 AEC C- F Submit Request Answer

Answer 1

Given resistance (R) (R) = 1.15k_2= 1-15 X 103_2 Inductance (L) = 555m H = 555 x 10-3 H Frequency (f) = 1500 Hz sms voltage Ving 12-5 v = Since we know that inductive reactance- X = 20 fL so x = 2x 3.142.1500X555_X10-3_4 XL = 5.228.1 Now from Ohms Law Verns Irms = 12.5 Jams А- 5228.1 Irms = 2.39 XLO A = 2.39 MA Now value of rms cussent when Capacitos. is insented in circuit - Irins = 9.39 x 10-3 A = 1.195 X10-3 A 2

12.5 14.45x1032 + 52284 94206 1.195 X 10-3 taking square of both sides - 12 1.3225 X 106+5228.1- 99200 = (10.4 10.46 X 103 2 5 228.1- 34200 109.42 x 106 1.3225x106 = taking squas root both sides- of 5228.1 108.09x106 992oc 5 298.1 10396.6 = 9420C 9420X 5168.5 9 c = 20.54 x 10 C = 70.54 nF so capacitance of insented capacitor is 220.540f.