6. Substitute and Differentiation
Differentiate both sides for equation for
current
Substitute dQ/dt for I in original equation and
arrive at second order differential equation
Current is the rate of change of charge (in
Coulombs) with respect to time
Substitute voltage drops into equation
KVL )(tVVVV CLR
)(
1
tVQ
C
RI
dt
dI
L
)(tI
dt
dQ
)(
1
'" tVQ
C
RQLQ
)('
1
'" tVI
C
RILI
7. General forms of equation
Which is the same as:
Divide by L:
0)(
1)()(
2
2
ti
LCdt
tdi
L
R
dt
tid
0)(
)(
2
)( 2
02
2
ti
dt
tdi
dt
tid
8. Calculate Parameters for Substitution
0
1
''' I
LC
I
L
R
I
5.660272
L
R
92
0 100899.1
1
LC
0100899.1'5.66027'' 9
III
0100899.15.66027 92
RR
0)(
1)()(
2
2
ti
LCdt
tdi
L
R
dt
tid
9. Solve Characteristic Equation for R
62.32
76.33013
62.3276.33013
)1(2
)100899.1)(1(45.66027
5.66027
2
4
0)100899.1(5.66027
92
2
92
iR
iR
R
a
acb
bR
RR
10. Select Natural Response of Circuit
Three forms:
Over-damped: Two negative real roots
Under-damped: Two distinct complex roots
Critically-damped: Two real, distinct roots
12. Substitute Parameters into Underdamped
Equation
t
t
d
etCtCti
L
CR
etCtCti
d
d
62.32
21
2
0
22
0
0
0
21
))57.23cos()57.23sin(()(
57.231
30339999997452.0
2
62.32
76.33013
))cos()sin(()(