A 1.15 kilo Ohms resistor and a 590 mH inductor are connected in series to a 1350 Hz generator with an rms voltage of 12.1 V.What is the rms current in the circuit in 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 in nF?
A) the reactance of the inductor is XL 2π*f*L 2π*1350*0.590 5005Ω So the impedance Z sqrt(1150^2 + 5005^2) 5135Ω Therefore the rms current V/Z 12.1V/5135Ω 2.36x10^-3A 2.36 mA B) If the current is reduced to 1/2 then then impedance must double So Z sqrt(R^2 + (XlL- XC)^2) 10270 So 1150^2 + (5005 - XC)^2 10270^2 or 5005 - XC +-sqrt(10270^2 - 1150^2) +-10205 or XC 5005 + 10205 15210 Now XC 1/2πf*C or C 1/(2π*1350*15210) 7.75x10^-9F 7.75nF
