Phase problem in a damped RLC circuit

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Homework Help Overview

The discussion revolves around determining the phase angle phi in the context of a damped RLC circuit, specifically analyzing the expression for electric charge on the capacitor and the implications of initial conditions on the behavior of the circuit.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between charge and current in a damped RLC circuit, questioning the implications of initial conditions on the phase angle. There is an attempt to derive expressions for current based on charge and to understand the role of damping in the system.

Discussion Status

Some participants have provided insights into the relationship between the exponential decay and oscillatory behavior of the circuit. There is an ongoing exploration of the physical meaning behind the phase angle and its implications in the context of damping.

Contextual Notes

Initial conditions include a specific charge on the capacitor and zero initial current, which raises questions about the expected behavior of the circuit under these constraints.

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.
 
Last edited:

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