Phase problem in a damped RLC circuit

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The discussion revolves around determining the phase angle phi in a damped RLC circuit given specific initial conditions. The derived equations indicate that the current and charge are influenced by both the exponential decay and oscillatory behavior. The initial conditions lead to a relationship where tg(phi) is derived from the damping ratio, suggesting that phi is not zero, contrary to initial assumptions. The presence of damping alters the expected behavior of the circuit, as the charge decreases due to both the exponential decay and oscillation. Understanding this phase shift is crucial for accurately modeling the dynamics of the damped RLC circuit.
anizet
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Homework Statement


The expression for electric charge on the capacitor in the series RLC circuit is as follows: q(t)=A*exp(-Rt/2L)*cos(omega*t+phi)
where omega=square_root(1/LC-R^2/4L^2)
What is the phase phi, if the initial conditions are:
q(t=0)=Q
I(t=0)=0

Homework Equations


The damped oscillator equation is:
second_derivative(q)+R/L*first_derivative(q)+q/LC=0

The Attempt at a Solution


[/B]
I calculated the current (as a derivative of charge):
I(t)=A/square_root(L*C)*exp(-Rt/2L)*cos(omega*t+phi-psi)
where tg(psi)=2*L*omega/R
When I set the initial condtitions, I get:
A*cos(phi)=Q
A/square_root(L*C)*cos(phi-psi)=0
and then I get:
tg(phi)=-1/tg(psi)=-R/(2*L*omega)
which seems really strange to me: shouldn't phi=0? If I start with some charge on capacitor and no current, the charge can only go down, at no moment the charge will be larger, so I think that q(t)=A*exp(-Rt/2L)*cos(omega*t)
Or am I not right?
 
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anizet said:

Homework Statement


The expression for electric charge on the capacitor in the series RLC circuit is as follows: q(t)=A*exp(-Rt/2L)*cos(omega*t+phi)
where omega=square_root(1/LC-R^2/4L^2)
What is the phase phi, if the initial conditions are:
q(t=0)=Q
I(t=0)=0

Homework Equations


The damped oscillator equation is:
second_derivative(q)+R/L*first_derivative(q)+q/LC=0

The Attempt at a Solution


[/B]
I calculated the current (as a derivative of charge):
I(t)=A/square_root(L*C)*exp(-Rt/2L)*cos(omega*t+phi-psi)
where tg(psi)=2*L*omega/R
When I set the initial condtitions, I get:
A*cos(phi)=Q
A/square_root(L*C)*cos(phi-psi)=0
and then I get:
tg(phi)=-1/tg(psi)=-R/(2*L*omega)
which seems really strange to me: shouldn't phi=0? If I start with some charge on capacitor and no current, the charge can only go down, at no moment the charge will be larger, so I think that q(t)=A*exp(-Rt/2L)*cos(omega*t)
Or am I not right?
You would be right if there was no damping, and the charge was q=Acos(wt). But the exponential factor contributes to the current, so there is a phase term in the cosine.
 
OK, so my result was correct. But could somebody explain to me the physical meaning of it? It is not intuitive...
 
Without damping, the time derivative of the cos(wt) function is zero at t=0. But the time derivative of the exponential factor is not.
The charge decreases with two ways on the capacitor. One is the discharge through the inductor and resistor, shown by the exponential factor. The other way is due to the oscillation. The charge decreases faster because of the exponential. If you suppose q=e-kt cos(wt), the current would be -ke-kt cos(wt)-w e-ktsin(wt), different from zero at t=0.
 
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