Question

A series RLC circuit with a resistance of 122.0 Ω has a resonance angular frequency of 3.4 ✕ 10⁵ rad/s. At resonance, the voltages across the resistor and inductor are 60.0 V and 40.0 V, respectively.

(a) Determine the values of L and C.

L	=	............. H
C	=	..................F

(b) At what frequency does the current in the circuit lag the voltage by 45°?
......................Hz

Answer 1

(a)

Impedance of the series RLC circuit,

Z = (R^2 + (wL - 1 / wC)^2)

At resonance,

Z = R

Voltage across resistor, V = 60 V

Voltage across the inductor = V*wL / Z = 40

(V / R)*wL = 40

(60 / 122) * 3.4*10⁵*L = 40

L = 0.232 mH

At resonance, wL = 1 / wC

C = 1 / w^2*L

C = 1 / [(3.4*10⁵)²*0.232*10^-3]

C = 3.72*10^-8 F

(b)

the circuit lag the voltage by 45°,

tan(phi) = (1 / wC - wL) / R = -1

(1 / w*3.72*10^-8 - w*0.232*10^-3)/ 122 = -1

0.268*10^8 / w -  w*0.232*10^-3 = -122

By solving,

w = 5.26*10^5 rad/s