8. Damping and the Natural Response in RLC Circuits

Series RLC circuit diagram

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)

Graph of over-damped case solution of RLC Circuit differential equation. t i
`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)

Graph of damping and the natural response in RLC Circuit. t i A
`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)

Graph of under-damped case in RLC Circuit differential equation. t i
`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.

top

Search IntMath, blog and Forum

Online Algebra Solver

This algebra solver can solve a wide range of math problems.

Calculus Lessons on DVD

 

Easy to understand calculus lessons on DVD. See samples before you commit.

More info: Calculus videos

The IntMath Newsletter

Sign up for the free IntMath Newsletter. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!

Given name: * required

Family name:

email: * required

See the Interactive Mathematics spam guarantee.