# Phase problem in a damped RLC circuit

• anizet
In summary, the expression for electric charge on a capacitor in a series RLC circuit is q(t)=A*exp(-Rt/2L)*cos(omega*t+phi), where omega=square_root(1/LC-R^2/4L^2). The phase phi can be determined by setting the initial conditions q(t=0)=Q and I(t=0)=0, and solving for tg(phi)=-1/tg(psi)=-R/(2*L*omega). This phase term accounts for the contribution of the exponential factor to the current, and represents the decrease in charge due to both discharge through the circuit and oscillation.
anizet

## 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?

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:

## What is the phase problem in a damped RLC circuit?

The phase problem in a damped RLC circuit refers to the phenomenon where the current and voltage in the circuit are out of phase with each other. This means that the peak of the current does not occur at the same time as the peak of the voltage, resulting in a phase difference between the two.

## What causes the phase problem in a damped RLC circuit?

The phase problem in a damped RLC circuit is caused by the presence of resistance (R) in the circuit. In an ideal RLC circuit, there is no resistance and the current and voltage are in phase. However, in a real circuit, resistance causes energy to be dissipated, leading to a phase difference between the current and voltage.

## How does the phase problem affect the behavior of a damped RLC circuit?

The phase problem affects the behavior of a damped RLC circuit by causing a decrease in the amplitude of the current. This is because the current and voltage are not in phase, resulting in some of the energy being dissipated as heat instead of being used to sustain the oscillations in the circuit.

## What is the significance of solving the phase problem in a damped RLC circuit?

Solving the phase problem in a damped RLC circuit is important for understanding the behavior of the circuit and predicting its performance. It allows us to calculate the phase difference between the current and voltage, which can help in optimizing the circuit for specific applications.

## How can the phase problem in a damped RLC circuit be solved?

The phase problem in a damped RLC circuit can be solved by using complex numbers and the concept of impedance. By representing the resistance, inductance, and capacitance in the circuit as complex numbers, we can calculate the overall impedance and determine the phase difference between the current and voltage.

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