Question

A 1.15-kΩ resistor and a 520-mH inductor are connected in series to a 1450-Hz generator with an rms voltage of 13.9 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? answer in nF

Answer 1

here,

the resistance , R = 1.15 Kohm = 1150 ohm

the inductance , L = 520 mH = 0.520 H

the frequency of generator , f = 1450 Hz

rms voltage , V = 13.9 V

a)

the impedance of circuit , Z = sqrt(R^2 + (2*pi*f*L)^2)

Z = sqrt(1150^2 + (2*pi * 0.52 * 1450)^2)

Z = 4872.76 ohm

the rms current , Irms = V / Z

Irms = 13.9 /4872.76 A

Irms = 2.85 * 10^-3 A

b)

the new rms current , Irms' = Irms /2 = 1.426 * 10^-3 A

let the capacitance of the capacitor be C

the new rms current , Irms' = V /sqrt(R^2 + (2*pi*f*L - 1/(2*pi*f*C))^2 )

1.426 * 10^-3 = 13.9 /sqrt(1150^2 + (2*pi*1450*0.52 - 1/(2*pi*1450*C))^2 )

solving for C

C = 7.61 * 10^-9 F

C = 7.61 nF

the capacitance is 7.61 nF