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): ???
4 answers
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.
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