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.

Q) 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): ???

plz help !

V=V₀sinωt,

ω= ω₀=1/sqrt(LC)=1/sqrt(0.015•5•10⁻⁶)= =3652 rad/s.

V(t)=40sin3652t.
q=CU,
q(t) =CV₀sinωt,
q(t) =5•10⁻⁶•40sin3652t=2•10⁻⁴sin3652t,
I(t) =dq/dt =d(CV₀sinωt)/dt= CV₀ωcosωt,
I(t)= 5•10⁻⁶•40•3652cos3652t =
=0.73cos3652t.

UC=CV²/2=CV₀²sin²ωt/2=
=(5•10⁻⁶•40²/2) sin²3652t=
=0.004 sin²3652t,

UC(t₁) = 0.004 sin²(3652•0.0003)=
=0.0032 J.

UL =LI²/2 = L(CV₀ω)²cos²ωt/2 =
=(0.015•(0.73)²/2)cos²3652t = =4•10⁻³cos²3652t,

UL(t₁) = 4•10⁻³cos²(3652•0.0003)=
=8.37•10⁻⁴J.

am not getting these answers marked correct ! plz if u can check it again ! thnx :)

anyone get this

To find the energy stored in the capacitor (UC) and the inductor (UL) at a specific time (t1), we need to use the formula for energy stored in each component.

Energy stored in the capacitor (UC) can be calculated using the formula:

UC = 0.5 * C * V^2

Where C is the capacitance and V is the voltage across the capacitor.

Energy stored in the inductor (UL) can be calculated using the formula:

UL = 0.5 * L * I^2

Where L is the inductance and I is the current flowing through the inductor.

To find the energy at time t1, we need to find the voltage across the capacitor and the current flowing through the inductor at that time.

Given:
L = 0.015 H
R = 80 Ω
C = 5e-06 F
V0 = 40 volts
t1 = 0.0003 seconds
ω = ω0 (unknown value)

To find the voltage across the capacitor (Vc) at time t1, we can use the equation:

Vc(t1) = V0 * sin(ωt1)

To find the current flowing through the inductor (I) at time t1, we can use the equation:

I(t1) = V0 / √(R^2 + (ωL - 1 / ωC)^2)

Now, let's calculate the values using the given information.

First, we need to determine the value of ω0. Since it is not given, we can solve for it using the equation:

ω0 = 1 / √(LC)

Substituting the given values:

ω0 = 1 / √(0.015 * 5e-06) = 13416.41 rad/s

Now, substitute the known values into the formulas to find the voltage across the capacitor (Vc) and the current flowing through the inductor (I) at time t1:

Vc(t1) = 40 * sin(ωt1) = 40 * sin(13416.41 * 0.0003) = 40 * sin(4.024928) = 39.999648 V (approximately)

I(t1) = 40 / √(80^2 + (ωL - 1 / ωC)^2) = 40 / √(80^2 + (13416.41 * 0.015 - 1 / (13416.41 * 5e-06))^2) = 40 / √(6400 + (201.2469 - 14806.77)^2) = 40 / √(6400 + (-14505.5231)^2) = 40 / √(6400 + 210663339.9977) = 40 / √(210663979.9977) = 40 / 14505.53 ≈ 0.002756 A

Now, let's calculate the energy stored in the capacitor (UC) and the inductor (UL) at time t1:

UC(t1) = 0.5 * C * Vc^2 = 0.5 * 5e-06 * (39.999648)^2 ≈ 0.003999 kJ ≈ 3.999 J

UL(t1) = 0.5 * L * I^2 = 0.5 * 0.015 * (0.002756)^2 ≈ 2.697e-08 kJ ≈ 2.697e-05 J

Therefore,

UC(t1) ≈ 3.999 J (approximately)
UL(t1) ≈ 2.697e-05 J (approximately)