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?
DH
Favorite Answer
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