Question

1. Compute the impedance of a series R-L-C circuit at angular frequencies of ω1= 1000 rad/s , ω2= 710 rad/s and ω3= 455 rad/s . Take R = 170 Ω , L = 0.935 H and C = 2.40 μF . What is the phase angle of the source voltage with respect to the current when ω = 1000 rad/s?

2. A series R–L–C circuit of R = 150 Ω , L = 0.915 H and C = 2.05 μF carries an rms current of 0.245 A with a frequency of 100 Hz . a. What is the average rate at which electrical energy is converted to heat in the resistor? b. What average power is delivered by the source? c. What is the average rate at which electrical energy is dissipated (converted to other forms) in the capacitor and in the inductor?

3. An ac series R-L-C circuit contains a 130 Ω resistor, a 2.0 μFcapacitor, and a 5.2 mH inductor. a. Find the resonance angular frequency. b. Find the length of time that each cycle lasts at the resonance angular frequency.

4. A series circuit consists of an ac source of variable frequency, a 130 Ω resistor, a 1.10 μF capacitor, and a 4.48 mH inductor. a. Find the impedance of this circuit when the angular frequency of the ac source is adjusted to the resonance angular frequency. b. Find the impedance of this circuit when the angular frequency of the ac source is adjusted to twice the resonance angular frequency. c. Find the impedance of this circuit when the angular frequency of the ac source is adjusted to half the resonance angular frequency.

Answer 1

1)Given,

R = 170 Ohm ; L = 0.935 H and C = 2.40 μF

for w1 = 1000 rad/s

We know that

Z = sqrt (R^2 + (Xl - Xc)^2)

Z1 = sqrt (170^2 + (1000 x 0.935 - 1/(1000 x 2.4 x 10^-6))^2) = 545.5 Ohm

Hence, Z1 = 545.5 Ohm

w2 = 710 rad/s

Z1 = sqrt (170^2 + (710 x 0.935 - 1/(710 x 2.4 x 10^-6))^2) = 186.62 Ohm

Hence, Z2 = 186.62 Ohm

w3 = 455 rad/s

Z3 = sqrt (170^2 + (455 x 0.935 - 1/(455 x 2.4 x 10^-6))^2) = 518.96 Ohm

Hence, Z3 = 518.96 Ohm

we know that the phase angle is given by:

phi = tan^-1 (Xl - Xc)/R

Xl - Xc = (710 x 0.935 - 1/(710 x 2.4 x 10^-6) = 518.33

phi = tan^-1 (518.33/170) = 71.84 deg

Hence, phi = 71.84 deg