# Phase problem in a damped RLC circuit

Tags:
1. May 25, 2016

### anizet

1. The problem statement, all variables and given/known data
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

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

3. The attempt at a solution

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?

2. May 25, 2016

### ehild

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. May 26, 2016

### anizet

OK, so my result was correct. But could somebody explain to me the physical meaning of it? It is not intuitive...

4. May 26, 2016

### ehild

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: May 26, 2016