# 8. Damping and the Natural Response in RLC Circuits

Consider a series RLC circuit (one that has a resistor, an inductor and a capacitor) with a constant driving electro-motive force (emf) E. The current equation for the circuit is

L(di)/(dt)+Ri+1/Cinti\ dt=E

This is equivalent: L(di)/(dt)+Ri+1/Cq=E

Differentiating, we have

L(d^2i)/(dt^2)+R(di)/(dt)+1/Ci=0

This is a second order linear homogeneous equation.

Its corresponding auxiliary equation is

Lm^2+Rm+1/C=0

with roots:

m_1=(-R)/(2L)+(sqrt((R^2-4L"/"C)))/(2L)

=-alpha+sqrt(alpha^2-omega_0^2

m_2=(-R)/(2L)-(sqrt((R^2-4L"/"C)))/(2L)

=-alpha-sqrt(alpha^2-omega_0^2

Now

alpha=R/(2L) is called the damping coefficient of the circuit

omega_0 = sqrt(1/(LC)is the resonant frequency of the circuit.

m1 and m2 are called the natural frequencies of the circuit.

The nature of the current will depend on the relationship between R, L and C.

There are three possibilities:

## Case 1: R2 > 4L/C (Over-Damped)

A+B

Graph of overdamped case.

Here both m1 and m2 are real, distinct and negative. The general solution is given by

i(t)=Ae^(m_1t)+Be^(m_2t)

The motion (current) is not oscillatory, and the vibration returns to equilibrium.

## Case 2: R2 = 4L/C (Critically Damped)

t=(2L)/R - A/B

Graph of critcally damped case.

Here the roots are negative, real and equal,

i.e. m_1= m_2= -R/(2L).

The general solution is given by

i(t)=(A+Bt)e^(-Rt"/"2L)

The vibration (current) returns to equilibrium in the minimum time and there is just enough damping to prevent oscillation.

## Case 3: R2 < 4L/C (Under-Damped)

sqrt(A^2+B^2)\ e^(-Rt"/"2L)
-sqrt(A^2+B^2)\ e^(-Rt"/"2L)

Graph of RLC under-damped case.

Here the roots are complex where

m_1=alpha+jomega, and m_2=alpha-jomega

The general solution is given by

i(t)=e^(-alpha t)(A\ cos\ omegat+B\ sin\ omegat)

where

\alpha = R/(2L) is called the damping coefficient, and omega is given by:

omega=sqrt(1/(LC)-R^2/(4L^2)

In this case, the motion (current) is oscillatory and the amplitude decreases exponentially, bounded by

i=+-sqrt(A^2+B^2)\ e^(-Rt"/"2L)

as we can see in the diagram above.

When R = 0, the circuit displays its natural or resonant frequency, omega_0=sqrt(1/(LC)).

### Example

In a series RCL circuit driven by a constant emf, the natural response of the circuit is given by

(d^2i)/(dt^2)+4(di)/(dt)+4i=0

for which the initial conditions are i(0) = 2 A and (di)/(dt) at t = 0 is 4.

State the nature of response of the current and hence solve for i.

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