Question

A 1.15-kΩkΩ resistor and a 595-mH inductor are connected in series to a 1350-Hz generator with an rms voltage of 12.3 V .

Part A: What is the rms current in the circuit? (Express answer in mA)

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 A? (Express answer in nF).

Answer 1

Resistance R = 1.15 k.2 = 115x10 3 2 Inductance L = 595 o H = 595x103H frequency f= 135042 Bong loltage Vars = 12:37 Jadotive greactance XL = 21SL 2291x 1350X595x18? = 50442 Impedance 2=2122+ x2 (1-15x103)2 + (504) -5176. un Roms current in the cine ait Vooms Irms 2 12.3 5176.4 2.376 m A Panto) Hore Xc = capacitive reactance √ R2 + (xe-*.)? R? +(*c - x)? 2xXrms 12 Irms 2x 12.3 2.376 x 163 = 107.2 X106 " (Xc-x2) 107.2 X106 X - XE = 10289.7 10289.7 + 5047 CSScanned with CamScanner Vnons Irons 2 (-15x103)? Xc =

Xe - 15336.7 Now, Xe = atite =15336.7 ca 21 x 13 50 X 153 36.7 C = 7.69x109 F = 7.69 n F - The capa citance le c=7.69 cscanned with CamScanner