Question

A 1.15-kΩ resistor and a 515-mH inductor are connected in series to a 1150-Hz generator with an rms voltage of 14.1 V .

A.)What is the rms current in the circuit? in mA

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? in nF

Answer 1

A)

the reactance of the inductor is

XL = 2π*f*L

XL = 2π x 1150 x 0.515

XL = 3721 Ω

So the impedance Z = sqrt(1150^2 + 3721^2)

Z = 3895 Ω

Therefore the rms current = V/Z

Irms = 14.1V/ 3895Ω

Irms = 3.62 x 10^-3 A

Irms = 3.62 mA

B)

If the current is reduced to 1/2 then then impedance must double

Z = sqrt(R^2 + (XL- XC)^2) = 7790

1150^2 + (3721 - XC)^2 = 7790^2

3721 - XC = sqrt(7790^2 - 1150^2) = 7874

XC = 3721 + 7874 = 11595

Now XC = 1/2πf*C

C = 1/(2π*1150*11495)

C = 12 nF