Natural response of series RLC system
R = Ω
C = Farrad
L = Henry
 
Initial conditions: iL(t=0+) =
vC(t=0+) =
Natural Response:
Characteristic equation:
Characteristic equation solutions:
Natural response examples RCL parameters change effect
Natural response of second order system ▲
RLC circuit:
In RLC circuit, energy may be stored in the inductance or in the capacitor. Voltage E accumulates on the capacitance C when switch 1 is closed (2 is open).
With the switch 1 open and 2 is closed the circuit equation become:
(Second order linear system)
Assuming an exponential solution of the form   i = Aest   then equation (1) is:
which is satisfied when:
(Called characteristic equation)
The roots of the characteristic equation are:
Letting
The roots are: s1 = α + β and s2 = α − β
s1 and s2 can be either real and unequal roots, real and equal roots or complex roots, depending on the value of β.
Case 1: Two real unequal roots:       (Overdamped case)   →   β is real
The most general solution of equation (1) is:
(4)
Case 2: Real equal roots:       (Critically damped case) →
From equation (3)
The most general solution of equation (1) is: (5)
Case 3: Two complex conjugates roots:      (Oscillatory case)    →    β complex
The most general solution of equation (1) in this case is:
NOTE: The value of eαt = e(-R/2L) t tends to 1 as t increases, A and B are constants to be avaluated from the initial conditions.
Examples of natural response of second order systems ▲
The graph of the resulting current.
Case 1:     real s1 and s2         Overdamped case
Example: Find the current response as a function of time after the switch 2 is closed from position 1 (see sketch above) if   R = 4Ω,   L = 1H,   C = 1/3 F
and V = 10V at t = 0.
Characteristic equation (from eq. 1) s2 + 4s + 3 = 0
The roots of the Characteristic equation. s1 = − 1           s2 = − 3
The current is from (4): i(t) = A1 e− t + A2 e− 3t
From initial condition:
        i(t = 0+) = i0 = 0		 0 = A1 e0 + A2 e0 = A1 + A2 → A2 = −A1
the voltage iR across the resistor is
zero and vL = vC and   L di/dt = V0 = 10
from initial conditions:
A1 = 5     A2 = − 5
And the final current i(t) is: i(t) = 5e− t − 5e− 3t
The graph of the resulting current.
Case 2:     real s1 = s2         Critically damped case
Example: Find the current i(t) in the circuit of a resistor of 4Ω, a capacitor of 0.25F and a coil of 1H at time t = 0 the capacitor has a voltage of 5V.
Characteristic equation (from eq. 1)
The roots of the Characteristic equation. s1 = s2 = − 2
The current is from (5): i(t) = e− 2t (A + Bt)
From initial condition:
Before the switch is closed, the current is zero: i(0+) = (0−) = 0		 i(t = 0) = A   →   A = 0
Also, from initial condition the voltage on capacitance is 5V:
at (t = 0)     5 = s1A + B     →     B = 5
And the final current i(t) is: i(t) = 5te− 2t
The graph of the resulting current.
Case 3:     complex s1 and s2         Oscillatory case
Example: The capacitance of 0.2 farad is charged to 10 volts for t < 0, when
t = 0 the switch is moved to position 2, find the current i(t) in the circuit, if
R = 2Ω and L = 1 Henry.
Characteristic equation (from eq. 1) s2 + 2s + 5 = 0
The roots of the Characteristic equation. s1 , s2 = − 1 ± j2
α = − 1         β = ±j2
The current is from (6): i(t) = e− t(Ae2tj + Be− 2tj)
i(t) = e− t(Acos2t + B sin2t)
From initial condition:
Before the switch is closed, the current is zero: i(0+) = (0−) = 0		 i(t = 0+) = 0 = (A∙1)   →   A = 0
And the current is:      
i(t) = e-tB sin2t
Also, from initial condition the voltage on capacitance is 10V:
10 = − 1∙0 + 1∙2B     →     B = 5
And the final current i(t) is: i(t) = 5e− t sin2t
RCL parameters change effect ▲
L,C and V = Constant
R -> Changes
R,L and V = Constant
C -> Changes
R,C and V = Constant
L -> Changes
L,C and V = Constant
R -> Changes
R,L and V = Constant
C -> Changes
R,C and V = Constant
L -> Changes