Question 1.5: A capacitor of 1.0 µF is charged to a voltage of 20 V and th...
A capacitor of 1.0 µF is charged to a voltage of 20 V and then connected to an inductor of 0.20 H and a resistor of 100 Ω in series. a) Determine the damping constant, the quasi-frequency, the relaxation time and the quality factor. b) Write the expressions of the charge and the current intensity. How long it takes for the amplitude of the intensity to be reduced to 1% of its initial value. How long it takes for the average energy to be dissipated up to 99%.
a) The natural frequency ω_o of this circuit is ω_o = 1/\sqrt{LC} = 2200 rad/s. The damping coefficient, the pseudo-frequency, the relaxation time and the quality factor are respectively:
\beta =R/2L=250s^{-1}, \quad \widetilde{ω}=\sqrt{ω_0^2-\beta^2}=2 \ 222 \ rad/s \quad \Rightarrow \widetilde{v}= 354 \ Hz\tau = 1/\beta =4.00 \times 10^{-3}s \quad \text{and } f_q= \pi \tau /T_0=½ \tau ω_0 =4.45
b) The charge and the current intensity are given by the expressions:
Q=Ae^{-\beta t}\cos(\widetilde{ω}t+ \phi ), \quad \quad I= Ae^{-\beta t}[-\beta\cos(\widetilde{ω}t +\phi )- \widetilde{ω}\sin (\widetilde{ω} t +\phi) ].The initial conditions Q = Q_o = CV = 20 µC and I = 0 are verified if A \cos \phi = Q_o and A(- \beta \cos \phi -\widetilde{ω} \sin \phi)=0. By taking the amplitude A as positive, the first equation gives cos Φ > 0, thus -π/2<Φ< π/2 and the second equation gives tan Φ=- \beta/ \widetilde{ω} =-0.1125 , thus Φ=-0.112 rad and A = 20.1 µC. The intensity may be written as
I= Ae^{-\beta t}\sqrt {\widetilde{ω}^2+\beta^2 }[\sin \phi \cos(\widetilde{ω} t +\phi )- \cos \phi \sin (\widetilde{ω}t + \phi )]= Aω_0e^{-\beta t }\sin (\widetilde{ω}t )The amplitude of the intensity Aω_o e^{\beta t } is reduced to 1% of its initial value at time t such that Aω_o e^{\beta t } = 10^{-2} Aω_o, thus e^{\beta t} = 10^{-2} and t = 18.4 ms. The average energy decreases according to the equation < U_{EM }> = U_{EM}(0)e^{-2\beta t }. It is dissipated up to 99% in the resistor if < U_{EM} > = 10^{-2}U_{EM}(0), i.e.
e^{-2\beta t} = 10^{-2} \Rightarrow \beta t = \ln 10 \Rightarrow t=9.21 \ ms.