# 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 coefficientof the circuit`omega_0 = sqrt(1/(LC)`is the

resonant frequencyof the circuit.

m_{1}andm_{2}are called thenatural frequenciesof the circuit.

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

There are three possibilities:

## Case 1: *R*^{2} > 4*L*/*C* (Over-Damped)

Graph of overdamped case.

Here both *m*_{1} and *m*_{2} 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: *R*^{2} = 4*L*/*C* (Critically Damped)

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: *R*^{2} < 4*L*/*C* (Under-Damped)

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*.

Answer

Two ways of solving this problem are shown here. You can choose whichever one makes more sense to you, or seems easiest.

#### Solution: Alternative 1

We will solve this in the same way as the previous section, 2nd Order Linear DEs.

A.E. `m^2+4m+4`

Solutions are a repeated root: `m_1=-2`, and `m_2=-2`

The response is critically damped, since the roots are equal.

`i(t)=(A+Bt)e^(-2t)`

`i(0)=2` implies `A=2`

So we can now write:

`i(t)=(2+Bt)e^(-2t)`

Differentiating gives:

`(di)/(dt)=(2+Bt)(-2e^(-2t))+Be^(-2t)`

`[(di)/(dt)]_(t=0)=4 `

This implies:

`(2)(-2)+B(1)=4 `

So `B=8`

Therefore `i(t)=(2+8t)e^(-2t)`

#### Solution: Alternative 2

Using the variables given in the damping theory above:

`L=1`, `R=4`, `1/C=4`, so `C=1/4`

`R^2=16`

`(4L)/C=4/(1"/"4)=16`

So `R^2=(4L)/C` and therefore we have critical damping.

Now `m_1=m_2=-R/(2L)=-4/2=-2`

The general solution is given by `i(t)=(A+Bt)e^(-Rt"/"2L`

So `i(t)=(A+Bt)e^(-4t"/"(2xx1))` `=(A+Bt)e^(-2t)`

This is the same solution we have using Alternative 1. The rest of the solution (finding *A* and *B*) will be identical.

#### Solution using Scientific Notebook

We need to set up the 2nd order DE with initial conditions as follows.

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

`i(0)=2`

`i^'(0)=4`

[Then go to **Compute** menu, **Solve ODE...**, **Exact**]

The graph of our solution is:

Graph of `i(t)=(2+8t)e^(-2t)`, a critcally damped case.