Question

A circuit contains a self-inductance L in series with a capacitor C and a resistor R. This circuit is driven by an alternating voltage V=V0sin(ωt). We have L=0.015 H, R= 80 Ω, C= 5e-06 F, and V0=40 volts.

(a) What is the value (in radians/seconds) of the resonance frequency, ω0?

(b) Consider three separate cases for which ω= 0.25 ω0, ω=ω0, and ω= 4 ω0 respectively. For each case calculate the the peak current I0 in Amperes.

ω= 0.25 ω0: ?
ω= 1 ω0: ?
ω= 4 ω0: ?

(c) Find the energy UC(t) and the energy UL(t) stored in the capacitor and in the inductor, respectively, at time t1=0.0003 seconds for ω=ω0. Express your answers in Joules.

UC(t1): ??
UL(t1):

Answer 1

a. Wo^2=1/LC=1/(0.015*5*10^-6)=1.33*10^7
Wo = 3651 Rad/s.

b. Xl=0.25Wo*L=0.25*3651*0.015=13.7 Ohm
Xc=1/(0.25*3651*5*10^-6) = 219 Ohms.
Z^2 = R^2 + (Xl-Xc)^2
Z^2 = (80)^2 + (13.7-219)^2 = 48,548.1
Z = 220.3 Ohms.
Im = V/Z = 40/220.3 = 0.182A.

Xl = W*L = 3651*0.015 = 54.77 Ohms
Xc = 1/W*C=1/(3651*5*10^-6)=54.77 Ohms.
Z^2 = R^2 + (Xl-Xc)^2
Z^2 = 80^2 + (54.77-54.77)^2=6400
Z = 80 Ohms.
Im = V/R = 40/80 = 0.5A.

Xl = 4Wo*L = 4*3651*0.015 = 219.1 Ohms.
Xc=1/4Wo*C=1/(4*3651*5*10^-6)=13.7 Ohms
Z^2 = 80^2 + (219.1-13.7)^2 = 48,589.2
Z = 220.3.
Im = V/Z = 40/220.3 = 0.182A.

Answer 2

thnx henry !

plz last part is remaining nw....
(c) Find the energy UC(t) and the energy UL(t) stored in the capacitor and in the inductor, respectively, at time t1=0.0003 seconds for ω=ω0. Express your answers in Joules.

UC(t1): ??
UL(t1):