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):

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.

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):

To find the value of the resonance frequency (ω0), we will use the given values of the inductance (L) and capacitance (C).

(a) Resonance frequency (ω0) is given by the formula:

ω0 = 1 / √(LC)

where L is the inductance and C is the capacitance.

Plugging in the values, we have:

ω0 = 1 / √(0.015 H * 5e-06 F)

Calculating this expression gives us the value of ω0 in radians/seconds.

(b) To find the peak current (I0) for different values of ω, we can use the formula:

I0 = V0 / √(R^2 + (ωL - 1/(ωC))^2)

For ω = 0.25 ω0, ω = ω0, and ω = 4 ω0, we can substitute the respective values into the formula.

For ω = 0.25 ω0:

I0 = 40 V / √((80 Ω)^2 + (0.25 ω0 * 0.015 H - 1/(0.25 ω0 * 5e-06 F))^2)

For ω = ω0:

I0 = 40 V / √((80 Ω)^2 + (ω0 * 0.015 H - 1/(ω0 * 5e-06 F))^2)

For ω = 4 ω0:

I0 = 40 V / √((80 Ω)^2 + (4 ω0 * 0.015 H - 1/(4 ω0 * 5e-06 F))^2)

Calculate each of these expressions to find the respective peak currents (I0) in Amperes.

(c) To find the energy stored in the capacitor (UC) and the inductor (UL) at time t1 = 0.0003 seconds for ω = ω0, we can use the formulas:

UC(t) = 0.5 * C * V0^2 * sin^2(ωt)
UL(t) = 0.5 * L * I0^2 * sin^2(ωt)

Plug in the values and calculate the energy UC(t1) and UL(t1) to express the answers in Joules.