Question

A 180 Ω resistor is in series with a 0.105 H inductor and a 0.400 μF capacitor.

(A) Compute the impedance of the circuit at a frequency of f1 = 500 Hz and at a frequency of f2 = 1000 Hz .

(B) In each case, compute the phase angle of the source voltage with respect to the current.

Answer 1

here,

resistance , R = 180 ohm

inductance , L = 0.105 H

capacitance , C = 0.4 uF = 4 * 10^-7 F

a)

for f1 = 500 Hz

the impedance of the circuit , Z1 = sqrt(R^2 + (2*pi*f1*L - 1/(2*pi*f1*C))^2)

Z1 = sqrt(180^2 + (2*pi*500*0.105 - 1/(2*pi*500*4 * 10^-7))^2)

Z1 = 499.47 ohm

for f2 = 1000 Hz

the impedance of the circuit , Z2 = sqrt(R^2 + (2*pi*f2*L - 1/(2*pi*f2*C))^2)

Z2 = sqrt(180^2 + (2*pi*1000*0.105 - 1/(2*pi*1000*4 * 10^-7))^2)

Z1 = 317.75 ohm

b)

for f1 = 500 Hz

the phase angle , theta1 = arccos(R/Z1)

theta1 = arccos(180/499.47) = 19.81 degree'

for f2 = 1000 Hz

the phase angle , theta2 = arccos(R/Z2)

theta2 = arccos(180/317.75) = 55.5 degree