Question

An RLC series circuit with 150-ohm resistor, 25-mH inductor, and 2 microfarad-capacitor is

powered by an AC voltage source with a peak voltage of 340V and a frequency of 660 Hz. (a)

Find the peak current that flows in the circuit. (b) Determine the phase angle of the source

voltage relative to the current. (c) Determine the peak voltage across the resistor and its phase

angle relative to the voltage source. (d) Find the peak voltage across the inductor and its phase

angle relative to the voltage source. (e) Find the peak voltage across the capacitor and its phase

angle relative to the voltage source.

Answer 1

R = 150 Ω; L = 0.025 H; C = 2 x 10^-6 F

Peak voltage, V = 340 V; frequency of AC, f = 660 Hz ;

angular frequency of AC, ω = 2πf = 2 x 3.14 x 660 = 4144.8 rad/s

inductive reactance, X_L = ω L = 4144.8 x 0.025 = 103.6 Ω

capacitive reactance, X_C = 1/ω C = 1/4144.8 x 2 x 10^-6   = 120.6 Ω

impedance of the circuit, Z = {R² + (X_C – X_L)²}^1/2 = {150² + (120.6 – 103.6)²}^1/2 = 153 Ω

(a) the peak current in the circuit, I = V/Z = 340/153 = 2.22 A

(b) if ɸ is the phase angle, Cos ɸ = R/Z = 150/153 = 0.9803

the phase angle, ɸ = 11.4⁰

( c ) peak voltage across the resistor, V = I R = 2.22 x 150 = 333 V. it is in phase with current

( d) peak voltage across the inductor, V = I X_L = 2.22 x 103.6 = 230 V . it is leading the current by π/2

(e ) peak voltage across the capacitor, V = I X_C = 2.22 x 153 = 339.7 V. it is lagging the current by π/2