Question

A 1.15-kΩ resistor and a 580-mH inductor are connected in series to a 1500-HzHz generator with an rms voltage of 12.7 V
Part A: 2.274 mA

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?

Answer 1

Inductance of the inductor X = 2142 = (27x 1600x"58'0x200 x 203) 5463.6in Here Vesistance R 1.15 kV = 1150 u Total Vesistoner of the cireuit. z = Řt xq (1150.) 7 (5463-67% 1322500 +29850924.96. ~ 2009 Z = 31173424.196.14 5583.31 Series circuit : current in the I Vimi Z 12.7 5583.3 )4 = 3.274 x1034 J 2:274 mA C 7 :) then capacitative Capacitanee Let the industanee X at fe now 25 ✓ R+ (XL-*=)? -3 I : Corrent 12.7 Z I'= { x 2.274x10²4 : 11BEX26 11.2x10? 1.137*10*3 125.4 X 106 Ý R4+1(*2-vejs = 125.4 x 105 Il ZY

|(x-x.) 125.4x10^ - 1.82x106 5 124.1 x 10 x106 3 → X j 3 9 1X2 - xxl. 11.18703 X 11.1x103 5.46*103 - 11.1*10 5.64*103 → Zale = 5.64*10 3 tie = 3 -> -7 e = 2+x.1500x 5.64*103 1: 88. X. 10-8 F e