A 1.15 k-ohms resistor and a 560mH inductor are connected in series to a 1450Hz generator with an rms voltage of 12.6 V. A)What is the rms current in the circuit? 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?i would realli appreciate a walk through of this problem and what answer i should expect. thanks
A) the reactance of the inductor is XL 2π*f*L 2π*1450*0.560 5102Ω So the impedance Z sqrt(1150^2 + 5102^2) 5230Ω Therefore the rms current V/Z 12.6V/5230Ω 2.41x10^-3A 2.41 mA B) If the current is reduced to 1/2 then then impedance must double So Z sqrt(R^2 + (XlL- XC)^2) 10460 So 1150^2 + (5102 - XC)^2 10460^2 or 5102 - XC +-sqrt(10460^2 - 1150^2) +-10397 or XC 5102 + 10397 15499 Now XC 1/2πf*C or C 1/(2π*1450*15499) 7.08x10^-9F 7.08nF
