Phase problem in a damped RLC circuit

  • #1
anizet
13
0

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?
 

Answers and Replies

  • #2
ehild
Homework Helper
15,543
1,915

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.
 
  • #3
anizet
13
0
OK, so my result was correct. But could somebody explain to me the physical meaning of it? It is not intuitive...
 
  • #4
ehild
Homework Helper
15,543
1,915
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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